2014-G12-Physics-E

19.4 if the speed of light were infinite, what would the equations of special theory of relativity reduce to?,19.5 Since ... m a s s t h a n sit a n bv * ruhsm 19.6 As a solid is heated and begins toglow, why does itfirst appear red&nmetoi 19.7 dWohuabtlehda?ppen■s'•t»o\"«ptontMalorvandii*amtiountdforwom; •a b.laycik••bo•di.yroifitiotsyioabesrDoliusitoeetqeamopfleTrature •is19.8 A beam of red light andVlJeamo^ fciue'tfght have exactlyl!^^melenergy. Which beam contains the grealqr ovtn^berofpbotons?, ■ ^fiqioaw elerlT19.9 Which photon, red. grew . or. blue;cacrie!Sflhe...(rK>%t f a h w y m and (b) # momentum? * \"loetr.i 'iirtftoeaeib’6p9i19.10 Which has the lower energy quanta? Radiowayes.qr X-rays19.11 Does the brightness of a beam of light primarily depends on the frequency of photonsoronthenurtlberbfphdtort's'7 1‘ *>[ ‘>ifo86aiYbod*.>GMA •19.12 When ultraviolet light falls on c e r t ^ dvps'ly'Sih^'light is eitttlpcf.jMv'iJod? this not19.13 Whailplpberingwhthleignhitnefrjaercetdmlio9grhet feallelsc•ot«rtno>onthqseqfsxreo5mdy^tajeosme?nebtal surface th...a..n..:dimrpiuna^o^MiB«Q0hl of the samecolour? OKt, mom e moil ?noi»ooi-. to ndT •19.14 Will higher frequency light ejectgreater numberdf electronsthan lowfrequency light?19.15 When lightshines on a surface, is momenturivtfdnsteWed to the metal surface? •19.16 Why can red light be used in a photographic dark room when developing films, but a blue or white light cannot?19.17 PhotonA has twice the energy of photon B. What is the ratio ofthe momentum ofA to thatof B? ’ °' »2<>to jomoo noitiioq >■ 'toriw •19.18 Why don't we observe a Compton effect ^ith visitttefigriW'' ',f1,n \" l0,or1<119.19 Can pair production take place invacuum^Ekpl^fh nuinemom bnsnoihao9 •19 20 Is itpossible to create a single electron19.21 Ifelectrons behaved onlyUke particles, tyha&palteoi wouldyouexpeclon Ihe screen after the electrons passes through the double slit?19.22 Ifan electron and a proton have the S8me de Broglie wavelength, which particle has greater speed?19.23 We do not notice the de Broglie wavelength for a pitchedcricket ball. Explain why?19.24 Ifthe following particles have the same'efteflflyT tfi? ^ r f t ^ ^ ^ b le n g t h ? Electron, alpha particle, neutron, proton. noqu19.2.5 When does light behave as a wave? Wbeo doesrtbqfoiyq asa eadste'feeoa19.26 What advantages an electron microscope has over-anoptlcafmlcrolsedp&?i>19.27 if measurements ^ q w .a precisejx>sitk)n for an u q o r l j l w y m f e j -ex SobnIIel\" i*e l h b g eol kxi aoo

19.1 A partide called the pton lives on the average only about 2.6 x 1(T* s wnen'at rest in the laboratory. It then changes to another form. How long would such a particle live when shooting through the space at 0 95 c? [Ans. 8.3 x 10'* s)19.2 What is the mass A £ V j k ?m a rTl& QJ<QfVWKr»Aiveling at 0.8 c from us as measured fromEarth? [Ans. 116.7 kg]19.3 F_ind theenergy ofphotonin 3 9 v ii3 9 [d 0 D n im B 9. (b) Radiowave ofwavelengttv I00n»d !tiwalnebula9<tjlelqcrOaid! 1obnrt -<rti u (c) GreenlightofwavelengtnSSOntnpoihyrt toeioBtleJnsrrmeqxewonX (d) X-ray withwavelength 0?2hrff’.ooibyrl toaeMkiteoq *’irio8 edinesr did ihrfoatonoilBtoiqietnia’eheoia-ebodnoasO eV ] t>19.4 Yellow light of 577 nm wavelertgtWl® mci4am on H cesium surface. Thai Stopping voltageis foupdjgb£0r25y> to nodiaoq pnibiepo’ ylmcl'oonu odito^eC Pi (a) the M a x iin e q tX ^lo fth e d h o tootfetnona .notouboiq erUbnr.iaiahnUaoteJa e(b)^f•^•logWMlcRmdloffofceWuW^sirne auoenetnoqa amiel eriJ edroeeO QVfJ Jioiainyni noilRlguoqbnR19.5 X-rays of wavelengtli 22 pm are scattered f ^ ^ c b o p t t e f ^ r i ^ s c a tte re d radiation being viewer] ^r bfiV 'ItoSton* ^ mP{9«?<*ttenr>g,opc3saobuionftfxJ m*.wevotengUi. oJ.men c i photoo, and, the wavelength of the scattered(i) bellsaJbOlonfdrscarttflrmqanglecfffl»30t'(b46flf>«rnol6 yd beoubowj stioeqe toybuH» fidoeqaonihoetmowbt It>ni[AMP.**®&T#f1(ft/) <»«*,Prtf'(#Y'f3'pm]ASbBnilrVWteti is thernwxlfnerrf’wavelength of the twb photons.’produced whenla positron mBOitooeqq^af fei bmfieiot dsi ?teewrt*i»1o#tot *a6eficQl?'T hx$r.f «ofnl m laotsosiiidSDWaTiWIJndilNafbeebWm O t f l W ^ u 3,^^ S ^ R6* >r«\"'?m ]19.8 Calculate the wavoiength of A 5 ! T 0 3 9 3 O IM O T A h.OS.belioxetflbRtui^^^flafwWfftoviriQatjfWfflSfiri) &?.-• im tc iuoq6v x) asp oimol6 ns noriwrrttooiuratwp.rr^rJy3ueyqwja&fj,r^twjtd6droilho- MvJrytd§_fiSl_tftgdSrWYfte^4dQ&j^ldfr ^i e.iip*wn«ettw»evoRhwtWoeoleiftloneeqpaninraaiafsiRhqeyosdayrnlliRRntarwu m“ ° ' BOrrnptr- MM08rkn muilo-ir;.-oruUi. o: r erlijRrIjnoefift1 |An8*1.12<j«rV0;fn»]d M M ) Art eiectforvis placedw a box«bout the suJe ofmn ktoro matte abooirWOX>t0 Hiff.2tnemeW»f^19»te9^Sr?bf mekfectrbr^ ertt . tiR rt tv mb edt ni <201

d ie iiiq ) 2 0 Learning Objectives At the end ofthis chapter the studentswill be able to: 1. Know experimental facts of hydrogen spectrum. 2. Describe Bohr's postulates ofhydrogen atom. 3. Explain hydrogen atom interms ofenergy levels. 4. Describe de-Brogile’s interpretation of Bohr'sorbits. 5. Understand excitation and ionization potentials. 6. Describe uncertainty regarding position ofelectron inthe atom. 7. Understand the production, properties and uses ofX-rays. 8. Describe the terms spontaneous emission, stimulated emission, metastable states and population inversion. 9. Understand laser principle. 10 Describe the He-Ne gas laser. 11. Describe the application of laser including holography. H h e branch of physics that deals with the investigation of wavelengths and intensities of electromagnetic radiation emitted or absorbed by atoms is called spectroscopy. It includes the study ofspectra produced by atoms. In general there are three types of spectra called (i) continuous spectra, (ii) band spectra, and (in)discrete or line spectra. Black body radiation spectrum, as descnbed in chapter 19 is an example of continuous spectra: molecular spectra are the examples of band spectra while the atomic spectra, which we shaUinvestigate in detail inthis chapter, are examples ofdiscreteor linespectra.20.1 ATOMIC SPECTRA When an atomic gas or vapour at much less than atmospheric pressure is suitably excited, usually by passing an electric current through it. the emitted radiation has a spectrum, which contains certain specific wavelengths only. An idealized arrangement for observing such atomic spectra is shown in Fig. 20.1. Actual spectrometer uses diffraction grating for better results. The impression on the screen is in the form of lines ifthe slit infront ofthe source S is narrow rectangle. Itis forthis reason that the spectrum is referred to as line spectrum. The fact that the spectrum of any element contains wavelengths that exhibit definite regularities was utilized in the second haIfof the 19*century in identifying different elements. 202

Fig. 20.1 line *pec(rum of hydrooenThese regularities were classified into certain groups called 1 IIIthe spectral series. Th e first such series was identified by J .J 2 * 51Balmer in 1885 in the spectrum of hydrogen. Th is series,called the Balmer series, is shown in Fig.20.2. and is in the Red GBoUwen Blue Veto! Wvisible region of the electromagnetic spectrum. Fig. 20.2Th e results obtained by Balmer were expressed in 1896 byJ .R Rydberg in the following mathematical form For Your Information Different type* o< epectr* - = R. (20. 1) ( » ) Ccnsnuou# ipecmxnw here R „ is the R yd berg's constant. Its value is1.0974 x 1 0 'm ’. Since then many more series have beendiscovered and proved helpful in predicting the arrangementof the electrons in different atoms.Atom ic Spectrum of HydrogenT h e Balmer series contain wavelengths in the visible portionof the hydrogen spectrum. T h e spectral lines of hydrogen inthe ultraviolet and infrared regions fall into several otherseries. In the ultraviolet region, the Lyman series contains thewavelengths given by the formulawhere n = 2 . 3 .4 ........ (b ) Line epecfrum (c) Bend ipeOrumIn the infrared region, three spectral series have been foundwhose lines have the wavelengths specified by the formulae 203

Paschon.seriesF o r Y o u r Inform ation ■f ) ....................... (20.3) where n = 4 ,5 .6 ........ |*43 Brackett series /\ ( 2 0 -4 ) where n = 5 . 6 .7 ........ Pfund series •'i.w*;:m w d o n e '• i 'f T .'•! ^«ciT noQoft“r rt6 ,m ,.;TW9q« o«iJ m 288f n. ieml&8 T h e existence of these regularities m the hydrogen spectrum together with simitar: regulanties In the spectra of more complex elements, proposes a definite test for any theory of atomic structure. '■** 20.2 BOH- R'S - - - '•0S> ATOM In order to explain the empirical results obtained by Rydberg. NeSs Bohr, in 1913. formulated a model of hydrogen atom utilizing classical physics and Planck's quantum theory Thfe sem i classical theory is based 6n the followtag thrde postulates: P o s tu la te I: An electron, bound to the nucleus in an atom. radiating. These o.b;ts 3rc called the discrete stationary Baimef m o m i*j n Dm StatpspT^e.atOfn- • mini bns letomntiu erfts p e c tu m P o s tu la te If: O nly those siationary drbits ard allbwed'for which orbital angular momentum is equal to an integral IM U IV 'iV / l. h. liV .,P1B P B m v r = —2n (20.6) where n - 1 , 2 . 3 . . . . . . . and n is called the principal quantum number, m and U a re the m ass and velocity of the orbiting electron respectively, and h is Planck's constant. 204

Postulate III: Whenever an electron makes a transition,that is. jumps from high energy state E , to a lower energystate E .a photon ofenergy h f\s emitted so that h'= ^ r £ (20-7)where f= c/X is the frequencyoftttfera&ation emitted. Oo You Know?de-Broglie's Interpretation of Bohr's OrbitsAt the time ol ftxmn ation of Bohr**.theory, therewas no H«*um w m ri W Svnjustification for the fist two postulates,,yhile Postulate III had ovtorciogvwweadornaotaarxtny before it w stsome rootsin P a c k 's thesis. Later on wtththe developmentofde Broglie's hypothesis, some justification could be seen inPostulate11esnxosstnedbelow. \"oec. Unw ,r. n o - no £ 00000* _ I- (a) Fig. 20.3 Stafconvyw nvo fo rn * o n a iln n gConsider a string oflength t as shown in Fig. 20.3 (a). Ifthis isput into stationary vibrations, we musfhave I = n>. where n isan integer. Suppose that the string is bent into circle of radiusr. as demonstrated for n = 3 and n = 6 in Fig. 20.3 (b) and (c).sothat': I = 2 r.r= n>.or 2xr (20.8),, X=~ iTFrppxjte Broglie's hypothesis ? h nh ” p mvgrit ni noitoele edl to ^ ^p^#if> letot erij sirIuoIro won au iHlHfbnR 3 >1 vB>fnrnvi >n t„ orlt a ,3 .Jicho 3rto8 n h e ' U ygiene leitnmou Postulate II. 2? yA |+'\.vm ^ = U ♦ .3 .X r 3 20fiC

Quantized Radii Consider a hydrogen atom invv+nchelectron moving with velocity v. isinstationarycircularorbit of radiusr„. From Eq. (20.6), v. x z2-*Pm- —r„■ (20.9) For this electron to stay in the circular orbit, shown in Fig. 20.4, the centripetal force Fc = —m v“l is provided by the Coulomb's force F , *—keyJ, where e is thf°e magnitude ofcharge on electron as well as on the hydrogen nucleus consisting of a single proton. Thus. ( 2 0 .10)For Your Information '• rl where constant k isequal to — 5— . 4xc0 Aftersubstituting for v, from Eq .20.9, we have (20” ) where r, =4s,2h,k2mej2 = 0.053 nm This agrees with the experimentally measured values and is called the first Bohr orbit radius of the hydrogen atom. Thus according to Bohr's theory, the radii of different stationary orbitsof the electrons in the hydrogen atom are given by r,® r,.4r„9r,,16r,........th e W B o w t x l r> the hydrogen Substituting the value of r, from Eq. 20.11 in Eq. 20.9, the speed ofelectron inthe nth orbit isalom has a rad u s r , * 5 3 > lO 'mT h e se c o nd a n d t v d B o h r oTOd* v .= —2*nkh—e2 (20.12)have radu r ,« 4 « , and r, =9r.r«pect«eV Quantized Energies Let us now calculate the total energy E. of the electron in the Bohr orbit; E„ is the sum of the kinetic energy K.E. and the potential energy U. i.e.. j£ , = K.E. ♦ U = ^ mv/ + ............ <20- 13) 206

B y rearranging E q. (20.10). we get (20.14) -21 m v ,‘ ■ k2- e—r„2 Do You Know?then *C B ke2 ke2 ■ - ke2 (20.15) r„ 2 Ti nv ** v, ritai a*,1 - ■- - - - - k.K ll* --—------- ^ - — •mourn 01 «n «rg * * atwre as frae 2r0 alacfroni may have any amount o> •nargyB y substituting the value of r, from E q. (20.11). we have 1 ( 2x*kW ) E. J= (2016)where E„= 2'r 2 k^2m a4 *constant = 13.6 eVwhich is the energy required to completely remove an electronfrom the first Bohr orbit. Th is is called ionization energy. Th eionization energy may be provided to the electron by coftsionwith an oxtemal electron. Th e minimum potential throughwhich this external electron should be accelerated so that itcan supply the requisite ionization energy is known asionization potential. Thus for n = 1. 2. 3......... we get theallowed energy levels of a hydrogen atom to be ...... 4 9 16Th e experimentally measured value of the binding energy ofthe electron in the hydrogen atom is in perfect agreementwith the value predicted by Bohr theory.Normally the electron in the hydrogen atom is in the lowestenergy state corresponding to n = 1 and this state is calledthe ground state or normal state. W hen it is in higher orbit, it issaid to be in the excited state. Th e atom m ay be exited bycollision with externally accelerated electron. T h e potentialthrough which an electron should be accelerated so that, oncollision it can lift the electron in the atom from its groundstate to some higher state, is known as excitation potential.H yd ro ge n E m issio n SpectrumTh e results derived above for the energy levels along withPostulate III can be used to arrive at the expression for thewavelength of the hydrogen spectrum. Suppose that theelectron in the hydrogen atom is in the excited state n with 207

energy £, and makes a W»sWorttfe(^kM»W energyr£(.wtiere £,<£,. then f os)' Do You Know? h / = £ .^ = VmSPhcaon muM h M onc*yy « u c « y h Mi :**e , m§ & 3 neriJ•bqcufoao*onto t rt*ooo»hivf»$igay(ofO.j4wlwatrteiooocoo hence a' 6rt,'“ - J l j («'*DgniiutiradueySo l an atom but «n f f t t w * mcgarenaatxercttohtahtetgla» sraMtowrxr.*Oft tf*T*fa«ai ^ubs^uting for f = c/>. and rearranging eiertw 1 cwpKMel . xlools oe vomo- ^ ttaC(aV) ySberg conslantgiven by theequation ^- 02$ , ,.1.tv o - ^ ^ u f 0974x10'mu ’ . ..... (2 0 .1 # 1 hrrworul «*i ytftd?* n u l l .. iJviuptn •>’ 9■•0O3M$ whlklV’^gfees wotl With the latest medsur&J vStoe for F«g.20.3 E w j i (oral i * / a hydrogen atom. Hj ' \" ’’ n\" •’< '■••‘*Ygv \"*eb the fryijrogon «lo>« Eq. 20.17 reduces tothe emptrical result denved by Rydberg and given by Eq 20.1. provided that we substitute p = 2 and 05 gnprgyleveis c o n e s g p q ^ g ^ HI . . •fit toitoolo erU Example 20.1: Findtftui Speed of the electron'IfHhtfW&r fojr. nop. tfyrt ydl • «rii yllKrmoU Solution:1'1^ •*! t • ' - n o : , *>6le Y0**>f* Thy«et ts»tii^xf:>ctu»j>hydu^mEmbo'l.t(.^-.tr1tT2 )w i t h n = 1. is vmbn\"dwOl0bit- v,,n .?5jfcTs2y!j 1^.Aw,<1crNm>c *) l (1.6' 1 0 ^ 0 * llo tto led1k bolt v t 6.63*10tf\l*M bnuoi( V,*2.19x10‘ rhs-' ! mo it noteilioo .l*Mlrv 20.3 IN N E R S H E L L T R A N S IT IO N S C H A R A C T E R IS TIC X -R A YS small w ol olcctrons m the hydrogen, qr o l ^ r , ^ ienvpsionof spectral linesiptlipjofowi^ri of electromagnetic specbumdue. |o, iqpie transition levels. .. Arji\Q 20«r.

In he a vy atoms, the electrons are assum ed to be arranged inconcentric shells labeled a s K . L. M. N . O etc.. the K shellbeing closest to the nucleus, the L shell next, and so on(F ig . 2 0 .6 ). T h e inner shell electrons are tightly b o und andlarge am o unt of en ergy is required for their displacem entfrom their normal energy levels. After excitation, w he n anatom returns to its normal state, photons of larger e n e rg y areemitted. T h u s transition of inner shell electrons in heavyatoms gives rise to the emission of high energy photons or X -rays. Th e s e X-rays consist of series of specific wavelengthsor frequencies and hence are called characteristic X-rays.Th e study of characteristic X -ra y s spectra has played a veryimportant role in the study of atomic structure and theperiodic table of elements.P ro d u ctio n of X -ra ys IW9h«n*ry,rFig. 2 0 .7 show s an arrangem ent of producing X -ra y s . Itconsists of a high vacuum tube called X -ray tube. W hen Pig. JO «the cathode is heated by the filament F. it emits electronsw hich are accelerated towards the anode T. If V is thepotential difference between C a n d T the kinetic e n e rg y K Ewith w hich the oloctron strike the target is g ive n byK.E. = Ve .............. (2 0 .1 9 )Suppose that these fast m oving electrons of energy Vo stnkea target m ad e of tungsten o r a n y other h eavy elem ent. It ispossible that in collision, the electrons in the innermostshells, such as K or L . will be knocked out. S u p p o s e that oneof the electrons in the K shell is rem oved, thereby producinga va ca n cy o r hole in that shell. T h e electron from the L shell 209

jumps to occupy the hole in the K shell, thereby emitting a photon of energy h called the K. X-ray given by h ............. (20.20) It is also possible that the electron from the M shell might also jump to occupy the hole in the K shell. The photons emitted are K , X-ray with energies hf»= E „-E , (20.21) these photons give rise to K , X-ray and soon. ■ The photons emitted in such transitions i.e.. inner shell transitions are called characteristic X-rays, becauso their onergies depend upon the type of target material.Wavatongtn (nrn) Th e holes created in the L and M shells are occupied by transitions of electrons from higher states creating more no 20* X-rays. Th e characteristic X-rays appear as discrete lines on a continuous spectrum as shown in Fig. 20.8. The Continuous X-ray Spectrum Th e continuous spectrum is due to an effect known as bremsstrahluno or braking radiation When the fast moving electrons bomoard the target, they are suddenly slowed down on impact with tf\p target. W e know that an accelerating charge emits electromagnetic radiation. Hence, these impacting electrons emit radiation as they are strongly decelerated by the target. Since the rate of deceleration is so large, the emitted radiation correspond to short wavelength and so the bremsstrahlung is in the X-ray region. In the case when the electrons lose all their kinetic energy in the first collision, the entire kinetic energy appears as a X-ray photon of energy h f ^ . i.e.. K.E. = h / L The wavelength in Fig. 20.8 corresponds to frequency f ^ . Other electrons do not lose all thoir energy in the first collision. They may suffer a number of collisions before coming to rest. This will give rise to photons of smaller energy or X-rays of longer wavelength. Thus the continuous spectrum is obtained due to deceleration of impacting electrons. Properties and Uses of X-rays X-rays have many practical applications in medicine and industry. Because X-rays can penetrate several centimetres 210

into a solid matter, so they can be used to visualize theinteriors of the m aterials opa que to ordinary light, such asfractured bones or defects in structural steel. T h e object to bevisualized is placed betw een an X -ra y source and a largesheet of photographic film; the darkening of the film isproportional to the radiation exposure. A crack or air bubbleallows greater amount of X -ra ys to pass. This appears as adark area on the photographic film. S h ado w of bonesappears lighter than the surrounding flesh. It is due to the factthat bones contain greate r proportions of elem ents with highatomic num ber and so they absorb greater amount ofincident X -ra y s than flesh. In flesh, light elements like carbon,hydrogen and oxygen predominate. Th e se elements allowgreater am ount of incident X -ra ys to pass through them.C A T -S c a n n e rIn the recent past, several vastly improved X -ra y techniques 'mCAT*c*nning«T*nned'Out’ aiT»yhave been d e ve lope d. O n e w idely used system is oT X-f*y baam* * araetod Woughco m p uterized axial tom ography;* the corresponding tbo coOont from a nurobaf ordtfarantinstrument is called C A T-S c a n n e r. T h e X -ra y sourceproduces a thin fan-shaped be am that is detected on the odontabonsopposite side of the subject by an array of several hundreddetectors in a line. E a c h detector m easures absorption ofX -ra y along a thin line through the subject. T h e entireapparatus is rotated around the subject in the plane of thebeam during a few seconds. T h e changing reactions of thedetector are recorded digitally; a com puter processes thisinformation and reconstructs a picture of different densitiesover an entire cross section of the subject. Densitydifferences of the order of one percent can be detected withC A T -S c a n s . Tu m o rs , a n d other anom alies m uch too small tobe seen with older techniques can be detected.B io lo g ica l Effects o f X -ra ysX -ra y s cause da m age to living tissue. A s X -ra y photons areabsorbed in tissues, they break molecular bonds a n d createhighly reactive free radicals (s u c h as H and O H ), w hich inturn can disturb the m olecular structure of the proteins andespecially the genetic material. Young and rapidly growingc e lls a re p a rtic u la rly s u s c e p tib le ; h e n ceX -rays are useful for selective destruction of cancer cells. O nthe other hand a cell m ay be dam aged by radiation butsurvive, continue dividing and produce generation ofdefective cells. Th u s X -ra ys can cause cancer. E ve n whenthe organism itself s h o w s no apparent dam age, excessive 211

radiation exposure can cause changes in their productivt system that will affect the organism 's offspring. 20.4 U N C E R TA IN TY W ITHIN TH E ATO M O n e of the characteristics of dua l nature of matter is < fundamental limitation in the a ccu ra cy of the simultaneou: m easurem ent of the position and m om entum of a particle. H eisenberg show ed that this is g ive n by the equation VH ApAx> — L* 2k <t» H ow ever, these limitations are significant in the realm o atom. O n e interesting question is w hether electrons art(• ) This tw cw Jim oni-oruf C A T scan present in atom ic nu- iei. A ? the typical nuclei <: > 'oss thano ( a bram rovoals a large rtra cra n t*tum or (colored c u p le ). (b ) Th re o - 10 m e diameter, for an electron to be confined within s u thd rre n s c o a l C A T scans are n o » nucleus, the uncertainty in its position is of the order of 10 '* na ^a A a U o a n d this axam plo revests an T h e corresponding uncertainty in the electron's momentum ■arachncM cyst ( colored yeaow )w B v n a sfcul. I n both p h o to g ra p h s the hco lo rs a re artificial h a v n g b e en Apt* —c o m p u te r g e n e ra te d to a id ind s tn g u n h m g tn M o m c a l features. AX 6 .63x1 0-* Js 10'T « 'm * 6 .6 3 x 1 0 * kg ms ' As Ap = m Av Jr7.3 x1 0 wm s ' Hence 6.63x10 * kg ms Av = 9.11 x 1 0 \"JI kg H e n ce , for the electron to be confined to a nucleus, its speec w ould have to be greater than 10' m s i.e.. greater than the speed of light. B ecause this is impossible, w c m ust conclude that an electron c a n never be found inside of a nucleus. Bui can a n electron reside inside the atom ? T o find this, w e again calculate the speed of an electron and if it turns to be less than the speed of light, w e have reasonable expectation of finding the electron within the atom but outside the nucleus. T h e radius of the hydrogen atom is about 5 x 1 0 '' m . A pplying the uncertainty principle to the m om entum of electron in the atom w e have h Ap* — ^ AX As A p = m w Therefore, h Av= m \x 212

:o r an atom a x is g iven a s 5 x 10 ” m_ 6 .6 3 x 1 0 ^ Js A V s £ .1 1 x 1 0 J1k g x 5 x 1 0 ’ ’ mT lU S = 1 .4 6 x 1 0 'm s ’\"his speed of the electron is less than the speed of light,hereforo. itcan exist in the atom but outside the nucleus.>0.5 L A S E Raser is the acronym for Light Amplification by Stimulated•mission of Radiation. A s the nam e indicates, lasers aretsed for producing an intense, monochromatic, andmidirectional coherent be am of visible light. To understandie working of a laser, terms such as stimulated emission and■opulation inversion m ust be understood. pontaneous and S tim ulated E m iss io n sConsider a sam ple o f free atom s s o m e of w hich are in the ■if iJround state with energy E , and som e in the excited state E t Induced ib to ip M ns shown m Fig. 20.9. Th e photons of energy hf * £ , - E . are E,icident on this sam ple. T h e s e incident photons can interact Sponl*ocoo» *mi»»ionnth atom s in two different w ays. In Fig. 20.9 (a ) the incidenth oto n is absorbed by a n atom in the ground state E . . thereby•aving the atom in the excited state £ ,. T h is process is called <•>timulated o r induced absorption. O n c e in the excited state,no things can happen to the atom , (i) It m ay de cay bypontanoous em ission as sh ow n in Fig. 20 .9 (b ). in which thetom emits a photon of energy h f = E s - E , in any arbitrary (b)irection.h e other alternative for the atom in the excited state E , is to I t - vOu v -£- J = S k=e c a y by stimulated o r induced em ission as show n in E.ig.20.9 (c ). In this case the incident photon of energy <«) Induced emission f = E , - £ . induces the atom to decay b y emitting a photon Fig 20.9lat travols in the direction of the incident photon. Fo r eachicident photon w e will have two photons going in the sam eirection thus w e have accom plished two things: a n amplifieds well as a unidirectional coherent beam . From a practicaloint this is possible on ly if there is m ore stimulated orKluced emission than spontaneous emission. Th is can bechieved as described in the next section. 213

P o pula tion In ve rsio n and Laser A c tio n Let us consider a simple case ot a material w hose atoms can reside in three different states a s s h ow n in Fig. 20.10 . state -to*# N, > « , . IV* 69*3 nm —* O pO ctlpum pM g -V \ / V — e. Op0c«l(*fnp*»w d) ------------------------ N. Fl*. 20.10 ILarger anotgy) £ , w hich is g round state; the excited state £ ,. in which the • C K '^O O O O O - atom s can reside only for 10* s and the metastable state £(Sm*l*< e n v y ji Nom ul populMlon in w hich the atom s can reside for - 1 0 ’ s. m uch longer thar 10* s. A metastable state is an excited state in which ar A normal p o p U te o n Of M o m c gxcited electron is unusually stable and from w hich the •norgy itata. «w> m o ra a tsm c in Bw electron spontaneously falls to low er state only aftei tow«r «fw yy Mat* E. man in m* relatively longer time. T h e transition from o r to this state are difficult a s com p are d to other excited states. H e n ce , insleac E, (J O O O O V O O of direct excitation to this state, the electrons are excited tc higher level for spontaneous fall to metastable state. A lso le PopwlMton lnv**»ioo us a ssum e that the incident photons of e n e rg y h f = £ , - £ raise the atom from the ground state £ . to the excited state £ , A popu to ten n v o rtlo n . in w fn tfith a haghar e n erg y u a to h a t a greener but the excited atom s do not de cay back to £ ,. T h u s the onh population than lh a kxro r on o rgy alternative for the atom s in the excited state £ , is to decai spontaneously to state £ .. the atom s reach state E , mucf faster than they leave state £ ,. T h is eventually leads to th< situation that the state £ , contains m ore atom s than state £ , T h is situation is k now n as population inversion O n c e the population inversion has been reached, the lasinj action of a laser is simple to achieve. T h e atom s in th« metastable state E , are bom barded by photons of energy h f = £ , - £ ,. resulting in an induced om ission, giving ar intense, coherent beam in the direction of the inciden photon. 214

T h e emitted photons must be confined in the assem bly longenough to stimulate further em ission from other excitedatoms. Th is is achioved by using mirrors at the two ends ofthe assem bly. O n e end is m ade totally reflecting, and theother end is partially transparent to allow the laser beam toe scap e (F ig .20.11). A s the photons m ove back and forthbetween the reflecting mirrors they continue to stimulateother excited atom s to em it photons. A s the processcontinues the num ber of photons multiply, and the resultingradiation is. therefore, m uch m ore intense and coherent thanlight from ordinary sources.H e liu m •N e o n L a s e rIt is a m ost com m on type of lasers used in physicslaboratories. Its discharge tube is filled with 8 5 % helium and1 5 % neon g a s . T h e neon is the lasing o r active m edium in thistube. B y chance, helium and neon h avo nearly identicalmetastable states, respectively located 20.61 eV and20.66 e V level. T h o high voltage electric discharge excitesthe electrons in som e of the helium atoms to the 20.61 oVstate. In this laser, population inversion in neon is achievedb y direct collisions with sam e energy electrons of heliumatoms. Th u s excited helium atoms collide with neon atoms,e a c h transferring its o w n 20.61 e V of en ergy to an electron inthe neon atom along with 0.05 eV of K .E . from the movingatom . A s a result, tho electrons in neon atom s are raised tothe 2 0 .66 e V state. In this w ay. a population inversion issustained in the neon g a s relative to an e n e rg y level of18.70 eV. S pontaneous em ission from neon a tom s initiatelaser action and stimulated em ission ca u s e s electrons in theneon to dro p from 2 0 .66 e V to the 18.70 e V level an d redlaser light of w avelength 6 32 .8 nm corresponding to 1.96 eVe n e rg y is generated (Fig, 20.12).U s e s of L a s e r in M e d icin e an d In d u s try1. La s e r b e am s are used as surgical tool for ’ welding’ detached retinas.2. T h e narrow intense beam of lasor c a n be used to destroy tissue in a localized area. Tin y organelles with a living cell have been destroyed b y using laser to study how the a b se n ce of that organelle affects the behavior of the cell.3. Finely focused be am of laser has be e n used to destroy cancerous and pre-cancerous cell.

Do You Know? T h e heat 0/ the laser seals off capillaries a n d lymphTh # h e U jn w w o n laaar boam » b o n g vessels to prevent spread of the disease.u M d to O a g n o M i t m m o» * » ey*.Th * u m oI tosor te c h n o lo g y in the T h e intense heat produced in small area by a laserf* M ofophthafrnctogy n wO*% p<**j boam is also used for welding and m achining metals and for drilling tiny holes in hard materials. T h e precise straightness of a laser be am is also useful to surveyors for lining up equipm ent especially in inaccessible locations. It is potential e n e rg y source for inducing fusion reactions. It c a n be used for telecomm unication along optical fibres. Laser beam can bo used to generate three- dimensional im ages of objects in a process called holography.W h e n an atom ic g a s or vapours at less than atm ospheric pressure is suitablyexa te d . usually by passing electric current through it. the emitted radiation has aspectrum w hich contains certain specific w avelenghts only.Postulates of Bohr's model of hydrogen atom are:A n electron, bound to tho nucleus in an atom , c a n m ove around the nucleus incertain circular orbits without radiating. Th e s e orbits are caBed the discretostationary states of the atom.O n ly those stationary states are allowed for w hich orbital angular m om entum isequal to an integral multiple of h i.e., m v r =W henever an electron m akos a transition, i.e., jum ps from high energy state E , to alower en ergy state E ,. a photon of en ergy h /is emitted so that h/= E . - E „.T h e transition of electrons in the h ydrogen or other light elem ents result in theem ission of spectral lines in the infrared, visible o r ultraviolet region ofelectromagnetic spectrum due to sm all en ergy differences in the transition levels.T h e X-rays emitted in inner shell transitions are called characteristic X-rays,because their en ergy depends upon the type o f target material.T h e X -ra y s that are emitted in all directions and with a continuous range offrequencies aro known as continuous X-rays.Laser is the acronym for Light Amplification b y Stimulated Em ission of Radiation 216

• The incident photon absorbed by an atom in the ground state thereby leaving the atom in the excited state £, is called stimulated or induced absorption.• Spontaneous or induced emission is that in which the atom emits a photon of energy h f * E 3- E , in any arbitrary direction.• Stimulated or induced emission is that inwhich the incident photon ofenergy h f = £ ,- £F, induces the atom to decay by emitting a photon that travels in the direction of the incident photon. For each incident photon, we will have two photons going in the same direction giving rise toan amplified as well as a unidirectional coherent beam. M 'I* H M M20.1 Bohr's theory of hydrogen atom is based upon sevoral assumptions. Do any of these assumptions contradict classical physics?20.2 What is meant by a lino spectrum? Explain, how line spectrum can be used for the identification of elements?20 3 Can the electron in the ground state of hydrogen absorb a photon of energy 13.6 eV and greater than 13.6 oV?20.4 How can the spectrum of hydrogen contain so many lines when hydrogen contains one electron?20.5 Is energy conserved when an atom emits a photon of light?20.6 Explain why a glowing gas gives only certain wavelengths of light and why thatgas is capable of absorbing the same wavelengths? Give a reason why it is transparent to other wavelengths?20.7 What do we mean when we say that the atom is excited?20 8 Can X-rays be reflected, refracted, diffracted and polarized just like any other waves? Explain.20.9 What are the advantages of lasers over ordinary light?20.10 Explain why laser action could not occur without population inversion between atomic levels? on™20.1 A hydrogen atoms is in its ground state (n = 1). Using Bohr’s theory, calculate (a) the radius of the orbit, (b) the linear momentum of the electron, (c ) the angular momentum of the electron (d) the kinetic enorgy (e ) the potential energy, and (f)the total enorgy. [Ans: (a ) 0.529 x 1 0 '5m (b) 1.99 x 1 0 \" kg ms ’ (c ) 1.05 x 10 * kg mV (d ) 13.6 eV (e) - 27.2 eV (0 -13.6 eV) 217

20.2 W hat are the energies in e V of quanta of wavelength? /. = 400. 500 and 700 nm. (Ans: 3.10 eV. 2.49 eV. 1.77 eV)20.3 A n electron jumps from a level E = -3 .5 x 10 ” J to £, = -1.20 x 10 ” J . W hat is the wavelength of the emitted light? (Ans: 234 nm),20.4 Find the wavelength of the spectral Nne corresponding to the transition in hydrogen from n ■ 6 state to n = 3 state? (Asn: 1094 nm )20.5 Compute the shortest wavelength radiation in the Balmer series? W hat value of n must be used? (Ans: 364.5 n m .n = oo)20.6 Calculate the longest wavelength of radiation for the Paschen series. (Ans: 1875 nm)20.7 Electrons in an X-ray tube are accelerated through a potential difference of 3000 V. If these electrons were slowed down in a target, what will be the minimum wavelength of X-rays produced? (Ans: 4.14 x 10\"'m )20.8 Th e wavelength of K X -ray from copper is 1.377 x 10 ” m W hat is the energy difference between the two levels from which this transition results? (Ans: 9.03 keV)20 9 Atungsten target is struck by electrons that have been accelerated from rest through 40 kV potential difference. Find the shortest wavelength of the bremsstrahlung radiation emitted. (A ns: 0.31 x 10 m)20.10 Th e orbital electron of a hydrogen atom m oves with a speed of 5.456 x 10' m s ’. (a ) Find the value of the quantum number n associated with this electron. (b ) Calculate the radius of this orbit (c ) Find the energy of the electron in this orbit. (A ns: n = 4. r, • 0.846 nm: E , = -0.85 e V ) 218

C h a p te r 21 NUCLEAR PHYSICSLearning ObjectivesAt t he end of this chapter the students will be able to:1. Understand the qualitative treatment of Rutherford's scattering experiment and the e vid e n c e it provides for the existence and sm all size of nucleus.2. Distinguish between nucleon number (m ass num ber) and atomic number.3. U n derstand that an elem ent can exist in various isotopic form s each with a different num ber of neutrons.-4. Understand the use of mass spectrograph to demonstrate the existence of isotopes and to m easure their relative abundance.5. Understand m ass defect and calculate binding energy using Einstein's equation.6 . Illustrate graphically the variation of binding energy per nucleon with the mass num ber.7. Appreciate the spontaneous and random nature of nuclear decay.8. Explain the m eaning of half-life.9. R e co g n ize and use d e c a y law.10. U nderstand and describe the interaction of nuclear radiation w ith matter.11. U n derstand the use of W ilson cloud cham ber. G e ig e r Muller counter and solid state detectors to detect the radiations.12. Appreciate that atomic num ber and mass num ber conserve in nuclear process.13. Describe energy and m ass conservation in sim ple reactions and in radioactive decay.14. U n derstand and describe the phenom ena of nuclear fission and nuclear fusion.15. Explain the working principle of nuclear reactor.16. B e a w a re of various types of nuclear reactors.17. S h o w an aw areness about nuclear radiation expo sure a n d biological effects of radiation.18. D e scribe in sim ple term s the use of radiations for m edical diagnosis and therapy.19. U n derstand qualitatively the importance of limiting exposure to ionizing radiation.20. O utlin e the use of tracer technique to obtain diagnostic information about internal structures.21. D e scribe exam ples of the use of radioactive tracers in diagnosis.22. Describe basic forces of nature.23. D escribe the m odem view of the building blocks of matter based o n hadrons, leptons and quarks. 219

Soon after the discovery of electron and proton in an atom, the quest started to find the w a y in w hich these charged particles are present in a n atom . From his experim ents Ernest Rutherford developed a nuclear model of the atom. His model of the atom consisted of a small dense, positively charged nucleus with negative electrons orbiting about it. In 1920 Rutherford suggested that there is probably another particle within the nucleus, neutral one, to which he ga ve the nam e neutron. Ja m e s Chadw ick discovered neutron in 1932. Do You Know? 21.1 A TO M IC N U C L E U S Nooaftvo otoOron A t the centre of each and every atom there is an infinitesimally small nucleus. T h e entire positive charge of Pruov&tiMve the atom and about 99 .9 percent of its m ass is concentrated in the nucleus. T h e tininess of tho nucleus can be imaginedFrom a -p »rlic le » jca tie n n goxporim ontt Lord Ruthortord by com paring that the radius of the atom is 10' tim es theconcluded that most o ( the port of * «atom is empty and lhal m a w d radius of the nucleus.concentralod n a very sm al re »o oca*ednucleus A nucleus consists of nucleons comprising of protons and neutrons. A proton has a positive charge equal to 1.6 x 10 ” C and its m ass is 1 6 7 3 x 1 0 ; kg A neutron has no charge on it. but its m ass is 1.675 x 10 ‘ kg T h e m a s s of a neutron is almost equal to m ass of proton. To nd'cate the m ass of atomic particles, instead of kilogram unified m ass scale (u ) is generally used. B y definition 1u is exactly one twelveth the m ass of c a rb o n \" atom (1 u = 1.6606 x 1 0 \" kg). In this unit the m ass of a proton is 1.007276 u and that of a neutron is 1.008665 u while that of an electron is 0 .0 0 0 5 5 u. T h e charge on a proton is equal in m agnitude to the charge on an electron. T h e charge on the proton is positive while that of an electron is negative. A s an atom on the w h o le ts electrically neutral, therefore, w e can conclude that the num ber of protons inside the nucleus is equal to the num ber of electrons outside the nucleus. T h e num ber of protons inside a nucleus is called the atom ic num ber or the charge num ber of an atom. It is denoted b y Z. Th u s the total charge of an y nucleus is Zo. hero e indicates charge o n one proton. T h e c o m b in e d n u m b e r o f all th e p ro to n s a n d n e u tro n s in a n u c le u s is k n o w n as its m a s s n u m b e r a n d is d e n o te d byA. T h e num ber of neutrons N present in a nucleus is g iven by N =(A -Z ) (2 1 .1 ) 220

W e now consider different elements of tho periodic table.H ydrogen atom is simplest of all the atoms. Its nucleus iscom posed of only one proton; that is for hydrogen A = 1.Z = 1. Th a t is w hy hydrogen is represented by the symbol JH.Next in the periodic table after the hydrogen element is thehelium eloment. Its nucleus contains two protons and twoneutrons. Th is m eans for helium A - 4 and Z = 2. and hencehelium is represented as j H e . W e now take the example ofuranium - a heavy element of the periodic table. T h e chargenum ber Z of uranium is 92 while its m ass num bor A is 235. This (•> (ProOum)is represented as . It has 92 protons while the num ber ofneutrons N is given by the equation N - A - Z = 235 -'92 = 143. •yIn this w a y the num ber of protons and neutrons in atoms of allthe elements of the periodic table can be determined. It hasbeen observed that the num ber of neutrons and protons in theinitial light elements of the periodic table is almost equal but intho later heavy elements the num ber of neutrons is greaterthan the num ber of protons m the nucleus.21.2 IS O TO P E S <b) (Oeutonum)Isotopes are such nudet of an element that have the same rta 21.1charge num ber Z . but have different m ass num ber A. that isin the nucleus of such an elem ent the num ber of protons is Do You Know?the sam e , but the num ber of neutrons is different. H elium , for 8<Xh X *o o n a n d c m u k ii <exam ple has two isotopes. Th e se are symbolically 36oo«oc*4represented as }H e and } H e .A s the charge num ber of helrumis 2. therefore, there are two protons in the helium nudeus.T h e neutron num ber of the first isotope is. according to Eq.21.1 is 3 - 2 = 1 a n d that in the second isotope} H e the num berof neutron is 4 - 2 = 2. H ydro ge n has three isotopesrepresented by ;H . '. H . J,H Its first isotope is called ordinaryh ydrogen o r protium. Th e re is on ly o n e proton in its nud eus.T h o s econ d isotope of h ydrogen is called deuterium. It haso n e proton and one neutron m its n u d e u s . Its n u d e u s iscalled deutoron. T h e third isotope of hydrogen has twoneutrons a n d one proton in its n u d e u s and it is called tritium.Th e isotopes of hydrogen are show n in Figs. 21.1 (a.b.c).T h e chem ical properties of all the isotopes of an eloment arealike, a s the chem ical properties of an element depend onlyupon the num bor of electrons around the n u d e u s . that isupon the charge num ber Z. w hich for all the isotopes of anelem ent is the sam e. It is. therefore, not possible to separatethe isotopes of a n eloment b y chemical m ethods. Physicalm ethods are found to be successful for this purpose. A 221

For Your Information dovice with the help of which not only the isotopes of any element can be separated from o n e another but their masses Som# atomic n i i M t can also be determined quite accurately is called mass spectrograph.Pw tld* M ass Spectrograph • 000055 A simple m ass spectrograph is show n in Fig. 21.2 (a ). Th e n 1006665 atom s or molecules of the elem ent under investigation, in vapour form, are ionized in the ions source S . A s a result of ■H 1C0T276 ionization, one electron is rem oved from the particle, leaving with a net positive charge *e. T h e positive ions, escaping the * 2.014102 slit S ,. are accelerated through a potential difference V applied betw een the two slits S . and S ,. ■h 301605 ■H# 301603 •H* 4002603 ■u 7016004 \"Bo 10013534 “ N 140031 169991 '0 F ig 21.2 T h e ions pass through the slit S , in the form of a na rrow beam T h e K .E . of singly charged ion at the slit S , will be given by ltm / = Vo (2 1 .2 ) 2 Th e ions are then subjected to a perpendicular and uniform m agnetic field 8 in a v a c u u m cham ber, w he re th ey are deflected in semicircular paths towards a detector. Th e detector records the num ber of ions arriving per second. Th e centripetal force applied by the m agnetic field is g iven by Bev=— - (2 1 .3 ) r v 222

Substituting the values of vfrom Eq. 21.2. w o getm* ( ' £ ) 8’ t2,'4)T h e above equation shows that the mass of each ionreaching the detector is proportional to S '. B y adjusting thevalue of B and keeping the term in the parentheses constant,ions of different masses are allowed to enter the detector. Agraph of the detector output as a function of S ' then gives anindication of what masses are present and the abundance ofeach mass.Fig. 21.2 (b ) shows a record obtained for naturally occurring 20 21 22neon gas showing three isotopes whose atomic massnumbers are 2 0 .2 1 . and 22. Th e larger is the peak, the more ftabundant is the isotope. Th us most abundant isotope of neonis neon-20. (b ) ( Proportional to atomic m a n ) Fig 2 1 2 Tha m asa apactrum oI21.3 M ASS D E F E C T AND BINDING E N E R G Y naS^a»y occurring noon. srto«KngIt is usually assumed that the whole is always equal to the sum nonpar ara 20.21. and 22 Th o largorof its parts. This is not so in the nucleus. Th e results of toe- peak. tho moro abundantexperiments on the masses of different nuclei show that the rtC-Opomass of the nucleus is always less than the total mass of all thoprotons and neutrons making up the nucleus. In tho nucleusthe missing mass is called the mass defect m given by.tn n = Z m „ * (A -Z )m .-m rMM (21.5)A s Z is the total number of protons in the nucleus and m, is them ass of a proton, then Z/n, is the total m ass of all the protons.A s shown in Eq. 21.1. (A - Z ) is the total num ber of neutronsand as m„ is the m ass of a single neutron. (A - Z ) m. is the totalm ass of all the neutrons. Th e term m ^ J s the experimentallym easured mass of the entire nucleus. Hence. Eq. 21.5represents the difference in mass between the sum of themasses of its constituents and the mass of the nudeus itself.T h e missing m ass is converted to energy in the formation ofthe nucleus. Th is energy is found from Einstein's massenergy rolation£»(A /n)c’ (21.6)and is called the binding energy (8 .E .) of the nudeus. Fromequations 21.5 and 21.6. the binding energy of a nudeus is B .E .* (A m )< ?= Z m ,c: * ( A -Z ) m c ^ -m c 1 ...... (21.7) 223

F o r Y o u r Inform ation Let us consider the exam ple of the deuteron nucleus to make the concept of m ass defect and binding energy more clear. E x a m p l e 2 1 .1 : Find the m a s s defect and binding en erg y of the deuteron nucleus. T h e experimental m ass of deuteron is 3.3435 x 10\"kg. S olu tion: Using equation 21.5. w e get the mass defect of deuteron as m = m .*m .-m 0 = 1.6726 x 10 \" k g + 1 .6 7 4 9 x 10 J ' kg - 3.3435 x 10 \" kg = 3.9754 x10°°kg Th e B .E . of deuteron as found from E q . 21.6 is A m e7 A m c ^ 3.9754 x 1 0 * k g x (3 x 10*ms ? = 6 .E . = 3.5729 x 1 0 '\J T o express the result in e V units, divide the B .E . obtained in joules by 1 .6 x 1 0 '*J. T h u s '-Ji.vnV.: r.xWo'-A Therefore, the bound constituents have less energy than (gfMttrmn*) w h e n they are free. Th a t is the B.E. co m e s from the m ass that is lost in the process of formation. Conversely, the bindingEn e rg y mu*< M « u p p M 10 brook lh »n o c M v j « p o n into o conotcuont en ergy is the am ount of en ergy that m ust be supplied to aprctont and neutrom nucleus if the nucleus is to be broken u p into protons and neutrons. Experim ents have revealed that such m ass defects exist in other elem ents as w ell. S h o w n in Fig. 21.3 is a graph between the mass defect per nucleon and charge CniKgo cx m bw Z Fig 21.3 224

num ber Z is obtained by finding the difference of m ass For Your Informationbetw een the total m a s s of all the protons and neutrons thatform the nucleus and the experimental mass of the nucleus $and dividing this difference by mass n u m b e r^ , i.e.. >( <*) 9Mass defect per nucleon Zmp + Q t Z ) m n) Am AAw here A m is the m a s s defect. Fro m the definition o f m assdefect it is quite obvious that for hydrogen, m ass defect iszero. T h e m ass defect is m ade d e a r with Einstein's equationE =Am e1. T h is equation show s that if for an y reason a m assA m is lost, them it is converted into energy.Let us n o w calculate the B E of helium . Fo r j H e A/n = 2 m , = 2.01519 u + 2.01796 u - 4.00281 u = 0.03034 usince 1 u = 1 .6 6 x 10*’kgThus A m = 0.03034 u x 1.66 x 1 0 '' kgu ’ = 5.03 x 10 ’’kg B E . = A/nC = 5.03x 10” k g x 9 x 1 0 '* m V'• « * * 10\"J \" 2 8 2 x 10' e V = 28.2 MOVTh is m eans that w hen two protons and two neutrons fusetogether to m ake helium nucleus, if an amount of 28 .2 M e Vonergy is given to the helium n u d e u s then it breaks u p intotwo protons and tw o neutrons. From this, w e c o n d u d e that 1 u = 1.6606 x 10J' kg = 931 MeVIn this w a y w o can calculate binding en ergy of every elem ent.S h o w n in Fig. 21 .4 is a graph betw een binding e n e rg y pernud e o n and the m ass num ber of different elements. Th isgraph show s that the binding en ergy per n u d e o n increases Fifl. 2%A 225

Fig. 21.5 with the m ass number till it reaches a maximum value of 8.8 M eV at mass number 58 and then it gradually decreases to a value of 7.6 M eV at m ass number 238. T h e binding energy per nucleon is maximum for iron. This shows that of all the elements iron is the most stable element. Later in this chapter it will be shown with the help of graph of Fig. 21.4 that when heavy element breaks into lighter elements or the lighter elements are fused to form heavier element then a large amount of energy can be obtained. 21.4 R AD IO ACTIVITY It h a s b e e n ob se rve d that those ele m e n ts w h o se ch arge n u m ber Z is greater than 82 are unstable. S o m e invisible radiations, that can affect the photographic plates em anate out of these elem ents. Such elements are called radioactive and the phen om en on is called radioactivity. T h e radiations com ing out of the radioactive elem ents are called alpha (a ), beta (3 ). and ga m m a (y ) radiation. Radioactivity w a s discovered by Henri Becquerel in 1896. He found that an ore containing uranium (Z = 9 2 ) em its an invisible radiation that penetrates through a black paper wrapping a photographic plate and affects the plate. After Becquerel's discovery Marie Curie and Pierre Curie discovered tw o new radioactive elem ents that they called polonium and radium. T h e analysis of the radiations emanating out of a radioactive material can be carried out by a simple experiment. Th e radioactive material is placed at the centre of a block of load by drilling a hole in tho block. Radioactive radiations enter a vacuum chamber after emerging out of this hole. After passing between the two parallel plates the radiations strike a photographic plate. These radiations, instead of impinging at one point, fall at three different points due to the potential difference between the plates (Fig. 21.5). From this experiment it can be concluded that all radiations from the radioactive material are not alike. T h e radiation that bends towards the negative plate is made up of positively charged particles. These are called ci-partides. Those radiations that bend towards the positive plate are composed of negatively charged particles. These are called (l-particles. Those radiations that go straight without bending have no 226

charge on them. Th e se are called y-rays.Further experiments reveal that u-partides are helium nuclei.Th e charge on them is *2e while their mass is 4u (atomicmass unit) that is every u-partide has two protons and twoneutrons in it. p-partides are in fa d fast moving electronswhich come out of the nudeus of a radioactive element, y-rayslike X-rays, are eledromagnetic waves which issue out of thenudeus of a radioactive element. Th e wavelength of theserays is much shorter, compared with the wavelength of X-rays.Nuclear Transm utationRadioadivity is purely a nudear phenomenon. Th is is not Fo r Y o u r Inform ationaffeded by any physical or chemical reaction. W henever anypartide I radiation is emitted out of any radioadive element, it Urjrnjm Thornjri (hotemis always accompanied by som e changes in the nudeus of daughter nucleus)the element. Therefore, this element changes into a newelement. Th is phenomenon is called radioadive decay. Th e melement formed due to this change is called daughterelement. Th e original element is called the paront element. n • decay o c o jh wtien an m ttebteDuring the nudear changes the laws of conservation of C o ro d n u cleus em its a n a • perttctemass, energy, momentum and charge remain applicable. e n d r u n e process « ■ converted ■mo < drtecent (or dm ghter) mjdeueW e know that three types of radiations u-partide. p-partideand y-rays are emitted by the naturally occurring radioadiveelements. W hen u-partide is emitted out of any nudeus thendue to law of conservation of matter, the mass number of thenudeus decreases by 4, and due to law of conservation ofcharge, the charge of the nucleus decreases by a magnitudeof 2e i.e.. the charge number of the nudeus decreases by 2. Itis due to the fa d that the m ass number and charge number ofthe emitted particle a is 4 and 2 respectively. Th e emission ofthe u-partide is represented by the following equationJ X ---------► *z *2Y + \ HeHere X represents the parent and Y the daughter elementTo explain the emission of a-partides w e take the example ofr a d iu m ^ R a T h e emission of an a-partide from radium 226.results in the formation of radon g a s ^ R n . Th is change isrepresented by the following equation “ «R a --------- ►^ R n . *HeIt m ay be remembered that the sum of the mass numbersand the charge num bers on both sides of the 227

For Your Information equation are equal. W h e n a (5-particle is emitted out of an y nucleus, then its m ass num ber does not undergo a n y changeParent nudeus a ■purtd* but its charge num ber increases by one. T h e em ission of a 0 'Me (5-particle from a n y elem ent X is represented by the following equation 5xTh e e rm tio n o4 r a p * n d * from Negative (5-particle is an electron and its em ission from theR o * u n -2 2 6 roscfis in the lo n ra w n nucleus becom es an incom prehensible enigm a, a s there isO lR *d on -2 229lr» no electron present in the nucleus. H ow ever, the em ission of electron from the nucleus can be thought of as a neutron emitting an electron and becom ing a proton, although the m odern explanation >snot that simple. p-paibde Th is m eans that the |5-particle is form ed at the time of !*> emission. T h a t is w h y at the time of em ission of aTh e emission or p-psrtlelo trom (5-particle the charge num berof the nucleus increases by oneP o lo n lu m -2 1 8 results m thefermeaoooT Astsbno-218. but n o ch ange in its m a s s num ber takes place as the m a s s of electron is exceedingly sm all as com p are d to the m a s s of a proton or a neutron. T h e transformation of an electron at the m om ent of its em ission is given below by an equation :h °e It has been observed that thorium *4^0T' h is transfo rm e d into protactinium ” *Pa after the em ission of 0 -p a rticle * h e following equation represents this reaction »< -. J3«_ 0 40 * 91P a * - i e Do You Know ? W hen a y-radiat>on issues out of nucleus then neither the charge number Z not the mass number A of the nucleus undergoes anyThorium ProUcbmum change. It is due to the fact that a y-radfation is simply a photon daughttr (electron) that has neither any charge nor an y mass. Its emission from the nucleus has som e resemblance with the emission of a photon of light from an atom. W e know that when an y electron of an kJ atom absorbs energy it jum ps from the ground state to a higherBOB energy state and the atom becomes excited. W hen the electronp -d o o iy occurs when a noutron n a nunsbibJo purer* nucleus decays •mo of this excited atom returns to its ground state then it emits the■ preton and en electron. Oio electronbo n o em tted as the p •perbde. In the absorbed energy in the form of a photon. In much the sam e wayprocess.INs parent nucleus Isvanstarmod intodaughter n udou* the nucleus is sometimes exdted to a higher state following the emission of a or |5-partide. Th is excited state of the nucleus is unstable state, in coming back to its ground state from the excited state, y-rarfcation is emitted. T h e emission of y-radiabon from a nucleus is generally represented by this equation 228

J x ’ --------- ► $ X . y radiationHere * X represents a n excited nucleus whilo *X showsground state of the nucleus.21.5 H ALF LIFEW e have seen that w h e n e ve r an a o r (V partide is emittedfrom a radioactive elem ent, it is transformed into som e otherelement. Th is radioactive de cay process is quite random andis not subjected to a n y sym m etry. T h is m e a ns that w o cannotforetell about an y particular atom a s to w h e n will it decay. Itcould de cay imm ediately o r it m ay rem ain unchanged formillions of year. T h u s w e cannot say anything about the life ofa ny particular atom of a radioactive element.Let us take the exam ple of a city with a population of onemillion and w e know that on the average ten person die everyday. Even with this know ledge w e cannot say with certaintythat which particular person will die on w hich particular day.W e can only s a y that on the w hole ten person will die. T h egreater the population o f the city, the greater the accuracy ofsuch predictions Like the population of a city, it is notpossible to talk about a n atom of a radioactive element. Form ore accurate result w e alw ays talk about large groups ofatom s and laws of statistics are applied upon them. Let ussuppose that w e bring a gro u p of 100.000 atom s underconsideration and w ait till such time that half of these i.e..50.00 0 de cay into their daughter element. T h is time is calledthe half-life 7 ,. of this elem ent. If the half-life of the saidelement bo one day. then after one day only 25.000 atomswill rem ain behind a n d after two da ys 12.500 atoms willremain behind. Th a t is with the passage of every one day. then u m b e r of atom s rem aining behind becom es half of thenum ber already present. T h is exam ple provides us thedefinition of half-life of a radioactive elem ent i.e.. \" T h e half-life 7,„ o f a radioactive elem ent is that period in w h ich half of the atom s decay” .Besides getting tho definition of half-life w e can deduce twoother conclusions from this exam ple. T h e s e are. firstly noradioactive elem ent c a n com pletely decay. It is due to thereason that in an y half-life period only half of tho nuclei decaya n d in this w a y an infinite time is required for all the atom s todecay. 229

Secondly, the number of atoms decaying in a particular period is proportional to the number of atoms present in the beginning of the period. If the number of atoms to start with is large then a large number of atoms will decay in this period and if the number of atoms present in the beginning is small then less atoms will decay. W e can represent these results with an equation. If at any particular time the number of radioactive atoms be N. then in an interval A/, the number of decaying atom. A N is proportional to the feme interval Af and the number of atoms N. i.e.. AN x -N A f or AN ■ - a N Af .............. (21.8) where >. is the constant of proportionality and is called decay constant. Eq. 21.8 shows that if the decay constant of any element is large then in a particular interval more of its atoms will decay and if the constant >. is small then in that very interval less number of atoms will decay. From Eq. 21.8 we can define decay constant /. as given below A here A N /N is the fraction of the decaying atoms. Th u s decay constant of an y element is equal to the fraction of the decaying atoms per unit time Th e unit of the decay constant is s . Th e negative sign in the Eq. 21. 8 indicates the decrease m the numbor of atoms N. Th e decay ability of any radioactive element can be shown by a graphic method also.o r., 2r„,3r„4rw sr., W e know that every radioactive element decay at a particular rate with time. If w e draw a graph between number of atoms inTm e(f) > the sample of the radioactive element present at different times and tho time then a curve as shown in Fig. 21.6 win beFr8la0.«*c2f1».v6odT4hc*<iyhi»*IheUw*n*T n„ woifW•i obtained. This graph shows that in the beginning the number of atoms present in the sample of the radioactive element was N„one-ha* o( V * radioactive nudo. with the passage of time the number of these atoms decreased due to their decay. This graph is called decay curve. After a period of one half-life N„ / 2 number of atoms of this radioactive element are left behind. If wo wait further for another half-period then half of tho remaining N ; / 2 atoms decay, and 1 /2 x N . /2 = (1 / 2)' N , atoms remain behind. After 230

the expiry of further period of a hatf-life. half of the rem aining( 1 12 ? N„ atom s decay. T h e num ber of atom s that rem ain u n ­decayed is 1 / 2 x (1 / 2 )' N , = (1 / 2)* N v W e can con clude fromthis exam ple that if w e have N„ num be r of a n y radioactiveelem ent then after a period of n half-fives the num be r ofatom s left behind is (1 /2 )' N„.It has been found that the estimate of d e c a y o f everyradioactive olem ont is according to the graph of F ig .21 .6 butthe half-life of e v e ry radioactive elem ent is different. Forexam ple the half-life of uranium -238 is 4 .5 x 10* years whiletho half-life of radium -226 is 1620 years. T h e half-life of som eradio active elem ents is very small, for exam ple, the half-lifeof radon gas is 3.8 days and that of uranium -239 is 23 .5minutes.From the a b o ve discussion it is found that the estimate of anyradioactive elem ent can be m ade from its half-life or bydetermining its de cay constant It can be proved with thehelp of calculus that the following relations exist betw een thed e ca y constant >. a n d the half-life >.T„= 0.693 (2 1 .9 )E q . 21 .9 s h o w s that if the de cay constant >. of an y radioactiveelement is know n, its half-life can be found.A n y stable element, besides the naturally occurringradioactive element, can be made radioactive. For this veryhigh energy particles are bom barded on the stable elem ent.T h is bo m ba rdm ent excites the nuclei and tho nuclei a/terbecom ing unstable becom e radioactive element, Suchradioactive elem ents are called artificial radioactiveelements.E xa m p le 21.2: Iodine-131 is an artificial radioactiveisotope. It is used for the treatment o f hum an thyroid gland.Its half-life is 8 days. In the drug store of a hospital 20 m g ofiodine-131 is present. It w as received from the laboratory 48da ys ago. Find the quantity of iodine-131 in the hospital afterthis period.S olu tion:A s the half-life o f iodine is 8 days, therefore in 8 d a y s half ofthe iodine d e ca ys. G ive n below in the table is the am ount ofiodine present after every 8 days. 231

Interval in Q uantity of In te rv a l In Quantity of days Io d in e days Iodine 0 20 mg 32 1.25 m g. .8 10 mg 40 0.625 mg 16 48 0.3125 mg 5m g24 2.5 mgTh u s 48 days after the receipt, the amount of iodine-131 leftbehind is only 0.3125 mg.21.6 IN TER A C TIO N O F RADIATION WITH MATTERAn a-particte travels a well defined distance in a mediumbefore coming to rest. Th is distance is called the range of theparticle. A s the particle passes through a solid, liquid or gas. itloses energy due to excitation and ionization of atoms andmolecules in the matter. Th e ionization m ay be due to directelastic collisions or through electrostatic attraction. Ionizationis the main interaction with matter to detect the particle or tomeasure its energy. Th e range depends on thei. charge, m ass and energy of the particle andii. the density of the medium and ionization potentials of the atoms of the medium.Since a-particle is about 7000 times more massive than anelectron, so it does not suffer any appreciable deflection fromits straight path, provided it does not approach too closely tothe nucleus of the atom. Th u s a-particle continues producingintense ionization along its straight path tinit loses all its energyand comes almost to rest It. then, captures two electrons fromthe medium and becomes a neutral helium atom.(l-particles also lose energy by producing ionization, However,its ionizing ability is about 100 times less than that of u-partides. A s a result its range « about 100 times more than ex-particles. (5-particles are more easily deflected by collisionsthan heavy u-particies. Th us the path of |5-partides in matter isnot straight but shows much straggling or scattering. Therange of (Pparticles is measured by the effective depth ofpenetration into the medium not by the length of erratic path.Th e more dense the material through which the particlemoves, the shorter its range will be. 232

a a nd |V-parttdes b oth radiate o n erg y a s X -ra y p ho ton s w h enth ey a re slow ed b y the electric field of the ch arged particles inasoltd m aterial.P h o to n s of y -ra y s . b e in g u n c h a rg e d , c a u s e v e r y littleionization. P hoton s a re rem oved from a b ea m b y eithersca tte rin g o r a b s o rp tio n in the m e d iu m . T h e y interact w ithm atter in three distinct w a y s, d opo nding m a in ly on theirenergy. (i) A t low e nergies (less than about 0 .5 M e V ). the dom inant process that rem oves photons from a beam•( i i ) is the photoelectric effect. (iii) A t inte rm e d ia te e n e rg ie s , the d o m in a n t p ro c e s s is C o m pto n scattering. A t higher e nergies (m o re than 1.02 M e V ). the dom inant p ro ce ss is pair production.In air y-ra ys intensity faUs off as the inverse s q u a re of thed istance from the so urce, in m u ch the sa m e m a n n e r a s lightfrom a lam p. In solids, the intensity decreases exponentiallywith increasing dep th of penetration into the m aterial. T h eintonsity /. of a b ea m after passing through a distance x in them edium is reduced to intensity/given b y tho relationw h ere p is the linear absorption coefficient of the m edium .T h is coefficient d e p e n d s on the e n e rgy of the photon a s wellas on the properties of the m edium .C h a rg e d particles a o r ft a n d y-radiabon p ro du ce fluore scen ceo r g lo w o n stnking s o m e su bs ta n ce like z in c su lp h id e, so diumiodide or barium platinocyanide coated screens. \"Fluorescence is the property of absorbing radiant energy of high frequency and re- oinlttlng energy of low frequency in the visible region of electromagnetic spectrum\".N eutro ns, being neutral particles, are extrem ely penetratingparticles. To be sto pped o r slow ed, a neutron m ust undergo adirect collision w ith a n ucleu s or so m e other particle that hasa m ass com parable to that of the neutron. Materials such asw ate r or plastic, w h ic h contain m ore lo w -m a s s nuclei p e r unitv o lu m e , a re u s e d to sto p n eu tro ns. N e u tro n s p ro d u c e a littleindirect ionization w hen they interact with materialscontaining hydrogen atom s and knock out protons. 233

Ta b le 21.1 Th e sum m ary of the nature of a . p & y radiation Characteristics a-particles ((-particles 7-r a y s1. Naturo H ekum nuclei of Electrons o r positrons from E M w a ve s from2. Typical charge 2e tho nucleus of charge i© excited n u d e i w ith no sources charge R adon-222 Stronbum -94 C o b a it-6 03. Ionization(Io n pairs mrrfin a * ) A b o u t 10* A b o u t 10* About 14 R a n g o in air S e v e ra l S everal m etres O b e y s inverse square5. Absorbed by centim etres law6. Energy 1 -5 m m of A l sheet A paper Variable energy 1 -1 0 cm of lead sheet spectrum7. Speed Em itted w ith tho - 1 x 1 ( f ms'1 Variable energy sam e energy - 3 x 1 Cf ms'1 - 10r m s'1C a rv* r» 21.7 R AD IATIO N D E TE C TO R S F ig . 21.7 Nuclear radiations cannot be detected by our senses, hence, w e use som e observable detecting methods employing the interaction of radiation with matter. Most detectors of radiation m ake use of the fact that ionization is produced along the path of the particle. Th e s e detectors include W ilson cloud chamber. G eiger counter and solid state detectors, W ilso n C lo u d C h a m b e r It is a device which shows the visible path of an ionizing particle. It m akes use of the fact that supersaturated vapours condense preferentially on ions. If an ionizing particle passes through a region in which d o u d droptots are about to form, the droplets will form first along the particle's path, showing the path as a trad of droplets. T h e apparatus consists of a cylindrical glass cham ber closed at the upper end by a glass window and at the lower end by a movable piston (Fig. 21.7). A black felt pad soaked in alcohol is placed on a metal plate inside the chamber. T h e air soon becom es saturated with alcohol vapours. A rapid expansion is produced by pulling quickly the piston of the bicycle pump having the leather w asher reversed so that it rem oves air. T h e sudden cooling resulted from adiabatic expansion helps to form supersaturated vapours. A s radiation passes through the chamber, ions are produced along the path. T h e tiny droplets 234

of rrxxsture condense about these o n s a nd form vapour tracksshow ing the path of the radiation. T h e s e a ro the atomic versionso f the ice crystate left in the s k y b y a jet plane w h e n suitableconditions exist. T h e fog tracks are illuminated with a la m p andm a y be seen o r photographed through the glass window.T h e u -p a rtid e s le a ve thick, straight a n d continuous tracks (*) a -P a r t id ed u e to in te n s e io n iz a tio n p ro d u c e d b y th e m a s s h o w n in (b) ^ Par1,oeFig . 2 1 .8 (a ), |5-particles form thin a n d d isco n tinu o us tracks wexte nd ing in erratic m a n n e r s h o w in g froquont deflections ^(F ig . 2 1 .8 b ) a n d y -ra y s le ave n o definite tracks a long theirpath (F ig . 2 1 .8 c ). T h e length of the cloud tracks has beenfound proportional to tho e n e rg y o f the in o d e n t particfe. Ahigh potential difference of the o rd e r of 1 k V betw ee n the topa n d bottom o f tho c h a m b e r p ro v id e s a n e lectnc field w h ic hd e a rs a w a y all th o u n w a n te d ions fro m th e ch a m b e r to m a keit re a d y for u s e . T h e tra c k s s e e n a r e . th e re fo ro . th o s e o f ra y sthat p a ss tho c h a m b e r a s the e x p a n sio n occu rs.T h e c h a m b e r m a y b e placed in a strong m agnetic fieldw hich will be nd the paths providing information about thecharge, m ass an d en ergy of the radiating partide. In thisway. it h as helped in the discovery o f m a n y new particles.Geiger-M uller C o unterGeiger-M uller tube is a w ell-know n radiation detector - T ' '*'(Fig. 21 .9 a ). T h e discharge in the tube results from theionization produ ced by the incident radiation. It consists Qof a stiff central w ire acting a s a n a n o d e in a hollow metal (c) r-Rayscylinder acting a s a cathode filled with a suitable mixtureof g a s at about 0.1 atm ospheric pressure. O n e end of the f ■ ctoud dumber tr*cutube h as a thin m ica w in dow to allow the entry of u or <#«. P. t i*<mhoo»jl-particles an d other end is sealed by non-conductingmaterial a n d carries the connecting pins for the twoelectrodes. A high potential difference, (about 400 V forn e o n -b ro m in e filled tu b e s ) but slightfy less than that 400VT h n nwce mndow - Argonge* Cathode Anode Geiger - Muller TubeF ig 21.9 (a) 235

(b ) G M Tube w it\ v M t r on* n e c e s s a ry to p ro d u ce d isch a rg e through the g a s is Fig 21J m aintained betw een the electrodes. W h e n radiation enters the tube, ionization is p ro d u ce d . T h e free electrons are attracted tow ards the positively charged central w ire. A s they are accelerated tow ards the w ire b y a strong electric field, th e y collide with other m olecules of the gas and knock out m ore electrons w hich in turn d o the sa m e a n d p ro du ce a ca s ca d e of electrons that m o ve tow ards the central w ire. Th is m a k e s a short pulse of electric current to pass through a n e x te rn a l resistor. It is a m p lifie d a n d re g is te re d electronically. T h e counter, w hich also provides the pow er, is called a scaler. Th e cascade of electrons produced by the entry of an ionizing particle is counted as a single pulse of approximately of the sam e size w hatever the energy or path of the particle m aybe. It cannot, thus, discriminate betw een the energies of the incident particle as output pulses are sam e. T h e entire electron pulse takes less than 1p s. However, positive ions, being very massive than the electrons, take several hundred times a s long to reach the outer cathode. D uring this time, called the dead time ( “10“*s ) o f the counter, further incoming particles cannot be counted. W h e n positive ions striko the cathode, secondary electrons are emitted from the surface. Th e s e electrons would be accelerated to give further spurious counts. T h is is prevented by m ixing a sm all am ount of quenching gas with the principal gas. T h e quenching gas must have an ionization potential lower than that of inert o r pnncipal gas. T h u s , the ions of quenching ga s reach the cathode before principal ga s ions. W h en they reach near the cathode, they capture electrons and become neutral molecules. Following neutralization, the excess energy of the quenching molecules is dissipated in dissociation of the molecules rather than in the release of electrons from the cathode. For exam ple, brom ine gas is added to neon gas. T h e brom ine m olecules ab so rb enorgy from the ions or secondary electrons and dissociato into brom ine atoms. T h e atom s then readily recom bine into m olecules again for the next pulse. T h o gas quenching is called self quenching Although all com m ercial G eiger tubes are self quenched, it is c o m m o n practice to use electronic quenching in addition. F o r this purposo. a large negative voltage is applied to the anode immediately after recording 236

the output pulse. Th is reduces the electric field bleow thecritical value for ionization by collision. T h e negative voltagerem ains until all the positive ions are collected at cathodethus preventing secondary pulses.G eiger counter can be used to determine the range orpenetration pow er of ionizing particles T h e reduction in thecount rate by inserting metal plates of varying thicknessbetw een the source and the tube helps to estimate thepenetration power of the incident radiation.G eiger counter is not suitable for fast counting. It is becauseof its relatively long 'd e a d tim e ' of the order of m ore than amillisecond w hich limits the counting rate to a few hundredcounts per second. If particles a re incident o n the tube at afaster rate, not all of them will b e counted since som e willarrive during the de ad time. Solid state detectors are fasten ough , m ore efficient and accurateS o lid State D etectorA solid state detector is a specially designed p -n Junction(Fig. 2 1.10 ) operating under a reversed bias in whichelectron-hole pairs are produced by the incident radiation tocause a current pulse to flow through the external circuit. Th edetector is m ade from a p-type silicon or germ anium . A n n -type thin layer is produced by doping the top surface withdonor type impurity. T h e lop and bottom surfaces are coatedwith a thin layer of gold to m ake g o o d conducting contact withe x te rn a l c irc u it. Th e co m b in e d th ick n e ss ofn-type and gold layer absorbs so less energy of the incidentparticle that the junction m ay be assum e d to be situated atthe front surface. Th is is know n a s the surface barrier typedetector. A reverse bias is applied through the twoconducting layers of gold. Th is enlarges the charge freeregion around the junction called depletion region. Normallyno current flows through the circuit. W h e n a n incident particlepenetrates through the depletion region, it produceselectron-hole pairs. Th e se mobile charge carriers movetowards the respective sides d u e to applied electric field. T h isgives rise to a current in the external circuit duo to w hich apulse of voltage is generated across the resistance R . Th ispulse is amplified a n d registered b y a scaler unit. T h e size ofthe pulse is found proportional to the en ergy absorbed of theincident particle. T h e energy needed to produce an electron-hole pair is about 3 e V to 4 e V w hich m akes the device usefulfor detecting low energy particles. Th e collection time ofelectrons and holes is m uch less than gas filled counters andhence a solid state detector can count very fast. It is m uchsmaller in size than a ny other detector and operates at lowvoltage. T h e a b o ve m entioned type detector is used fordetecting « or fl-partidos but a specially designed device

can be used for y-rays.21.8 N U C L E A R R E A C TIO N SW hile studying radioactivity, w e have seen that an aparticle is emitted from radium -226 and radon-222 isobtained. Th is nuclear ch ange is represented by thefollowing equation^R a ►Such an equation represents a nuclear reaction. Abovem entioned nuclear reaction takes place on its o w n accord.H ow ever, it w a s Rutherford w ho . first of all, expressed hisopinion that besides natural radioactive d e c a y processes,other nuclear reactions can also occur. A particle x isbombarded on any nucleus X and this process yield anucleus Y and a light object y as g iven belowX + x - » Y «-yRutherford performed an experiment on tho nuclear reactionIn 1918. H e bom barded a -p a rtid e s o n nitrogon. H e observedthat as a result of this reaction, oxygen is obtained and aproton is emitted. Th a t is\"N r- jH e ► ’J O + JHT h is reaction indicated that w h e n u-particle enters thenucleus of \" N then an exortation is produced in it. A nd as aresult of it \ 0 and a proton are produced. Since theoxpehment of Rutherford, innumerable nu d e a r reactionshave been observed. Fo r nu d e a r reactions to take place,thefulfillment of certain conditions is a must.Before and after an y n u d e a r reaction the num ber of protonsand neutrons must remain the sam e because protons andneutrons can neither be destroyed nor can they be createdW e elaborate this point from the exam ple of Rutherford'snud ear reaction of ',N and‘ H e. here’| N * j H e ►’J O + JHNum ber of protons = 7 ♦ 2 = 8 ♦ 1Num bor of neutrons = 7 + 2 = 9 + 0A n u d e a r reaction can tako place only when the total energy ofthe reactants mdudiog the rest m ass energy is equal to thetotal energy of the products. For its explanations w e again takethe example of the nud e ar reaction of Rutherford involving '‘ N and ‘ H e . in this reaction the m ass of the reactants is

Mass o f = 14.0031 uM assofjH e=4.0026uTotal m ass of the reactants = 18.0057 uIn the sam e w a y the m ass of the products isM ass of ’J O = 16.9991 uM ass of JH = 1.0078 uTotal m ass of the products after the reaction = 18.0069 uT h is s h ow s that the total m a s s after the reaction is greaterthan the total m ass before the reaction by 0.0012 u. W e knowthat a 1u m ass = 931 M e V energy, therefore, a m assdifference of 0 .0012u is equivalent to an en ergy of931 M e V x 0.001 2 u = 1.13 MeV. H e n c e this reaction ispossible only w hen a n additional m ass of 0.001 2 u is addedinto the reactants or the m inimum kinetic energy of the u -partide is 1.13 M e V such a s obtained from 2JJ P o . T h e en ergyof these u-partides is oqual to 7.7 M oV which is greater than1.13 MeV. Had these u-partides been obtained from asource that give out u -p a rtid e s w hose en ergy w a s less than1.13 M e V then this reaction w ould not h a v e taken place.Fro m the conditions described above w o c a n tell w hether an yn u d e a r roaction is possible o r not. T h e re is an interestingaspect in a n u d e a r roaction that it can take place in theopposite direction also. W e know th a t1J O is obtained by theinteraction’ with an u -p a rtid e of appropriate onergy. If w eaccelerate protons, with the help of a m achine like cydotron .and increase their velo dty and then bom bard these highve lo d ty protons on ’JO . Rutherford's n u d e a r reaction o f \" Na n d u -p a rtid e will proceed in the backw ard direction as’JO + JH » 1J N + j H eB y bom barding different elements with u-p artides. protonsand neutrons, m any n ud ear reactions have been produced.N o w w e describe one such n u d e a r reaction with the help ofwhich Ja m e s Chadw ick discovered neutron in 1932. W henjB e w as bombarded with u-partides emitted out of *JJPo,then as a result of a n u d e a r reaction.'JC and a neutron wereobtained. T h is reaction is s h ow n below with an equation»Be+ JH o ►’| C + <JnA s neutron carries n o charge, thereforo. it presenteda m e a t am ount of difficulty for its identification. A n y h o w 239

when neutrons w ere passed through a block of paraffin, fast moving protons w ere ejected out and these were easily identified. It m ay be rem em bered that a large am ount of hydrogen is present in paraffin and the nuclei of hydrogen Proton* atom s are protons. T h e em ission of protons is the ' n t > - * consequence of elastic collisions between the neutrons and 0 -*1© -* the protons. T h is indicates that the m a s s of neutron is equal a— » -* O - * to the m ass of the proton. It m a y be rem em bered that w henPKox r t * an object of certain m ass collides with another object of equal O - * m ass at rost. then a s a result of elastic collision, the moving object com es to rest and the stationary object begins to move 80 P«r»ffln with the velocity of the colliding object. T h e discovery of Fig. J1.11 neutron has brought in a revolution in nuclear reactions, as the neutrons carry no charge so they can easily enter the nucleus. Fig. 21.11 show s the arra ngem ent of Chadw ick’s experiment for the discovery of neutron. 21.9 N U C L E A R FISSION Otto Hahn and Fritz Strassm ann of G erm any while working upon the nuclear reactions m ade a startling discovery. T h e y observed that w h e n slow m oving neutrons are bom barded o n ^ U . then as a result of the nuclear reaction’.\"B a.” Kr and an avera ge of three neutrons are obtained. It m ay be rem em bered that the m ass of both krypton and barium is less than that of the m a s s of uranium . T h is nuclear reaction w as different from hither to studied other nuclear reactions, in two w ays. First as a result of the breakage of the uranium nucleus, two nuclei of almost equal size are obtained, w hereas in the other nuclear reactions the difference between the m asses of the reactants and the products was not large. Seco n dly a very large am ount of en ergy is given out in this reaction \" S u c h a reaction in w h ic h a h eavy nucleus like that of uranium splits up into tw o nuclei of roughly equal size along with the om ission of energy during the reaction is called fission reaction\". Fission reaction of ” j U can be represented by the equation * g U + ^ n ---------► ’£ B a - £ K r ♦ 3gn ♦ Q here Q is th e e n e rg y g ive n out in this reaction. B y c o m p a rin g th e total e n e r g y o n th e left s id e o f th e equation w ith total e n e rg y o n the right side, w e find that in the fission of o n e u ra n iu m nu cle u s about 2 0 0 M e V 240

energy is given out. It m ay be kept in mind that there is nodifference between the sum of the m ass and the chargenumbers on both sides of the equation. Fission reactionis shown in Fig. 21.12 (a ) and (b). Fission reaction canbe easily explained with the help of graph of Fig. 21.4.Th is graph show s that the binding energy per nucleon isgreatest for the middle elements of the periodic table andthis binding energy per nucleon is a little less for the lightor very heavy elements i.e., the nucleons in the light orvery heavy elements are not so rigidly bound. Forexample the binding energy per nucleon for uranium isFig 21.12 Proem* o4 FlM lon r*»cH©nabout 7.7 M e V and the products of the fission reaction ofuranium, namely barium and krypton, have bindingenergy of about 8.5 MeV per nucleon. Thus when auranium nucleus breaks up. as a result of fission reaction,into barium and krypton, then an energy at the rate of(8 5-7.6) = 0.9 M e V per nucleon is given out. This meansthat an energy 235 x 0.9 = 211.5 M e V is given out in thefission of one uranium nucleus.T h e fission process of uranium does not ahvays produce thesa m e fragm ents (B a . K r). In fact a n y of the hvo nuclei presentin the u p p e r horizontal part o f binding e n e rg y could beproduced T w o possible fission reactions of uranium are g ivenbelow as an example: Jo -------- ►’^ S n + '^ M o + S in + Q 241

*»U *on * ’« X e + & Sr + 2 in + QH ence in the uranium fission reaction several products maybe produced. All of these products (fragm ents) areradioactive. Fission reaction is not confined to uranium alone;it is possible in m any other heavy elements. H ow ever, it hasbeen observed that fission takes place v e ry easily with thesiow neutrons in uranium -235 and plutonium -239, and mostlythese two are used for fission purposes.Fissio n C h a in R eactionW e have observed that during fission reaction, a nucleus ofuranium -235 absorbs a neutron and breaks into two nuclei ofalmost equal m asses besides emitting two or three neutrons.B y properly using these neutrons fission reaction can beproduced in m ore uranium atoms such that a fission reactionc a n continuously maintain itself. T h is process is calledfission chain reaction. S uppose that w e have a definiteamount of and a slow neutron originating from anysource produces fission reaction in one atom of uranium. Outo f this reaction about three neutrons are emitted. Ifconditions are appropriate these neutrons produce fission insom e m ore atoms of uranium. In this w ay this processrapidly proceeds and in an infinitesimal small time a largeamount of energy along with huge explosion is produced.Fig.2 1 .13 is tho representation of fission chain reaction.It is possible to produce such conditions in which only oneneutron, out of all the neutrons created in o n e fission reaction,becom es the cause of further fission reaction. T h e otherneutrons either escape out or are absorbed in any otherm edium except uranium. In this case the fission chainreaction proceeds with its initial speed. To understand theseconditions carefully look at Fig. 21.14. In Fig. 21.14 (a ) afission reaction in a thin sheet of is show n to be inprogress. T h e resulting neutrons scatter in the air and so theycannot produce an y fission chain reaction. Fig. 21.14 (b )shows som e favourable conditions for chain reaction. Som eof the neutrons produced in the first fission reaction produceonly one m ore fission reaction but here also no chain reactionisproduced. ln F ig .2 1 .1 4 (c)a s p h e re o f issh ow n. Ifthesphere is sufficiently big. then most of the neutrons producedby the fission reaction get absorbed in before theyescape out of the sphere and produce chain reaction. Such a

m ass of uranium in which o n e neutron, out of all the neutrons For Your Informationproduced in o n e fission reaction, produces further fission iscalled critical m ass. T h e volum e of this m ass of uranium is Neutroncalled critical volum e. In a oonlroBaO tftam raacaon. on*yIf the m a s s of uranium is m u ch greater than the critical on* n a jy o n , o n a v r a g a , Irom oac/im ass, then the chain reaction p rocee ds at a rapid speed fiaaion * v* n t c a u » « « a no the ran d a h u g e explosion is p ro d u ce d . A to m b o m b w o rk s at nudaus to flaaton As a raauft.this p rin c ip le . If th e m a s s o f u ra n iu m is le s s th a n the energy a raieaead at a steady orcritical m a s s , th e c h a in re a c tio n d o e s not p ro c e e d . Ifthe m a ss o f uranium is eq ual to the critical m a ss, the ccrtnMdrmch ain reaction p ro c e e d s at its initial sp e e d a n d in thisw a y w e g e t a sou rce o f e n e rgy. E n e rg y , in a n atom icreactor, is obtained acco rd in g to this principle. T h ech ain reaction is not allow ed to run w ild, a s in a n atom icbo m b but is controlled b y a series of rods, usually m adeof ca d m iu m , that are inserted into the reactor. C a d m iu mis a n e le m e n t that is ca p a b le of ab so rb in g a largenum ber of neutrons w ithout becom ing unstable orradioactive. H ence, w he n the cadm ium control rods areinserted into the reactor, th ey absorb ne utrons to cutd o w n o n the n um ber of ne utrons that are available forthe fission pro cess. In this w a y the fission reaction iscontrolled.In a nuclear pow er station the reactor plays the sam e parta s do e s furnace in a therm al pow er station. In a furnace,coal or oil is burnt to produce heat, while in a reactor fissionreaction produces heat. W h e n fission takes place in theatom of u ranium or a n y othe r h e a v y atom , then an e n e rg y atthe rate of 200 M e V per nucleus ts produ ced T h is en ergyappears in the form of kinetic energy of the fissionfragments. Th e s e fast m oving fragm ents besides collidingwith one another also collide with the uranium atom s. In thisw a y their kinetic en ergy gets transform ed in heat energy.Th is heat is used to produce steam which in turn rotates theturbine. Turbine rotates the generator which produceselectricity. A sketch of a nuclear pow er station is show n inFig. 21.15. 243

Hoat exehoogo out Fig. 21.ISA reactor usually has four important parts. Th e se are:1 • T h e m ost important and vital part of a reactor is called core. Here the fuel is kept in the shape of cylindrical tubes. Reactor fuels are of various types. Uranium w as used as fuel in the olem entary reactors. In this fuel the quantity of 2^ U is increased from 2 to 4 percent. It m ay be rem em bered that the quantity of 2» U in the naturally occurring uranium is only 0.7 percent. N ow -a-days plutonium-239 and uranium- 233 are also being used as fuel.2 Th e fuel rods are placed in a substance of small atomic weight, such as water, heavy water, carbon or hydrocarbon otc. Th e se substances are called moderators. T h e function of these m oderators is to slow dow n the speed of the neutrons produced during the fission process and to direct them towards the fuel. H e avy water, it m a y be rem em bered, is m ade of 2H . a h eavy isotope of hydrogen instead of J H . T h e neutrons produced in the fission reaction are very fast and energetic and are not suitable for producing fission in reactor fuel like 2£ u o r ^ J P u e tc . For this purpose slow neutrons are m ore useful. To achieve this, moderators are used. B esides m oderator there is a n arra ngem ent for the control of num ber of neutrons, s o that of all the 244

neutrons produced in fission, only one neutron produces further fission reaction. T h e purpose is achieved either by cadmium or by boron because they have the property of absorbing fast neutrons. Th e control rods made of cadmium or boron are m oved in or out of the reactor core to control the neutrons that can initiate further fission reaction. In this w a y the speed of the chain reaction is kopt under control. In case of em ergency or for repair purposes control rods are allow ed to fall back into the reactor and thus stop the chain reaction and shut dow n the reactor.4 Heat is produced d u e to chain roaction taking place in the core of the reactor. T h e temperature of the core, therefore, rises to about 500 ’C . T o produco steam from this heat, it is transported to heat exchanger with the help of water, h eavy w ater o r a n y other liquid under high pressure. In the heat oxchanger this heat is used to produce high temperature steam from ordinary water T h o stoam is then used to run the turbine which in turn rotates the generator to produce electricity. T h o temperature of the steam com ing out of the turbine is about 300 *C. Th is is further cooled to convert it into w ater agam . To cool this steam, w ater from som e river or sea is. generally, used..In Karachi nuclear pow er plant (K A N U P ). h e a v y w ater is being used as a m oderator and for the transportation of heat also from the reactor core to heat exchanger, heavy water is used. To cool steam com ing out of the turbine, sea water is being used.T h e nuclear fuel o n ce used for charging the reactor can keepon operation continuously for a few months. Th e re after thefissile material begins to decrease. N o w the used fuel isrem oved and fresh fuel is fed instead. In the used up fuelintensely radioactive substances rem ain. T h e half-life ofthese radioactive remnant materials is m any thousand years.T h e radiations and the particles emitted out of this nuclearw aste is very injurious and harmful to the living things.Unfortunately there is no proper arrangement of the disposalof the nuclear waste. Th is cannot be dum ped into oceans orleft in a n y place w h e re they will contam inate theenvironm ent, such as through the soil o r the air. T h e y m ustnot be allowed to get into the drinking water. Th e best placeso far found to store those w astes is in the bottom of old salt 245

mines, which are very dry and are thousands of metres below the surface of tho Earth. Here they can remain and decay without polluting the environment. There are two main typos of nuclear reactors. Th o se are: (i) Therm al reactors (ii) Fast reactorsO r-K *y T h e thermal reactors are called th erm al* because the ^ 235 mm neutrons m ust be slowed do w n to ’ thermal energ»es* to produce further fission. T h e y use natural uranium or slightly oNpl M * 0 * 1 enriched uranium as fuel. Enriched uranium contains a greater percentage of U -2 3 5 than natural uranium does There are several designs of thermal reactors. Pressurized w ater reactors (P W R ) are the m ost w idely used reactors in the world. In this type of reactors, the w ater is prevented from boiling, being kept und er high pressure. Th is hot w ater is used to bo*l another circuit of w ater w hich produces steam for turbine rotation of electricity generators. n . Fast reactors are designed to m ake use of U -2 3 8 . w hich is v about 9 9 % content of natural uranium. E ach U -2 3 8 nucleus absorbs a fast neutron and changes to plutonium-239. ’X j - O + r ’> ♦ > » N p --------- » * £ P u ♦** nvck’*' r*#tt>on Plutonium can be fissioned by fast neutrons, hence.•Men tr*n«nwM mto is ^ need e d in fast reactors T h e core of fastijetfeneurenium • react0fS consists 0f a mixture of plutonium and uraniumpMon*w •* dioxide surrounded by a blanket of uranium-238. Neutrons that escape from the core interact with uranium- 238 in the blanket, producing thereby plutonium-239. Th u s m ore plutonium fuel is bred in this w a y and natural uranium is used m ore effectively. 21.10 FU S IO N R E A C TIO N W o know that the energy given out per nucleon per fission of heavy elem ent like that of uranium is 0 .9 MeV. It is due to the fact that the binding en ergy p e r n u d e o n of the fission fragments is greater than uranium In fact en e rg y is obtained from any nuclear reaction in which the binding energy per nucleon of the products increases. Is there a n y other reaction besides the fission reaction from w hich energy could be obtained? In order to answ er this question w e must ponder 246

over Fig-21.4 again. Th is graph shows that the binding energyper nucleon increases upto A = 50. H e n ce w h e n two lightnude* m erge together to form a heavy nucleus w hose m assnum ber A is less than 50. then en ergy is given o u t In sectionon 'M a s s Defect and Binding Energy” w e have observed thatw hen two protons and two neutrons merge to form a heliumnu d o u s . then about 28 M e V energy is given out.\"Such a nuclear reaction in w hich tw o light nuclei mergeto form a h e a v y nucleus is called fusion reaction\".During a fusion reaction so m e m ass is lost and its equivalenten ergy is given out. In a fusion reaction, m ore en ergy pernud e o n can be obtained as compared to the fission reaction.But unfortunately it is com paratively m ore difficult to producefusion. T w o positively charged light n u d e i m ust be broughtvery d o s e to o n e another. To d o s o work has to be doneagainst tho olcctrostatic force of repulsion between thepositively charged nudei. T h u s a very large am ount ofen ergy is required to produce fusion re adion . It is true that agreater am ount of energy can be obtained during a fusionreaction com p are d to that produced during a fission reaction,but in order to start this readion a very large am ount ofen ergy is spent. O n the contrary no difficulty is faced to startthe fission reaction because neutron has no charge on it andit has to face no repulsive force while reaching the n u d e u s .Let us now take the exam ple of a fusion reaction w hen twodeuterons are m erged to form a helium nucleus. 24 M e Ven ergy is released during this process i.e..2H ♦ 2H ►j H o ♦ 24 M e VBut there is a v e ry little chance of the formation of j H enudeus by the m erger of two deuterons. T h e probability ofoccurring such a readion is great w here one proton or oneneutron is produ ced as given below:2H + 2H ¥ *,H ♦ JH + 4 .0 M e VOf ^ H + 2H -------------► j H e + ,Jn + 3 .3 M e VIn both of these reactions about 1.0 M e V onergy per n u d e o nis produced w hich is equal to the on o rgy produced duringfission. If 2H a n d 3,H are forced to fuse then 17.6 M e V en ergyis obtained i.e.. 2, H * *H --------- ► j H e + o n + 17.6 M e VW e know that for fusion of tw o light n u d e i the work has to be 247

Oo You Know? done to overcom e the repulsive force w hich exists betweenOzon* on t w M rftM of Elrtti it • them . Fo r this the tw o n u d e i are hurled towards one anotherc o rro w * and poMcnou* g a t but « at a very high speed. O n e method to d o so is to give theset w holflN o<2 0 -5 0 b n fro n lh « E w f » nuclei a very large velocity with the help of an accelerator.hA m b K o n w v M l o A n l Th is method has been used in the research study of nucleara b tcrta rfmost a l u v rwfctton* fusion of *H and j,h . But this method o f nuclear fusion forvM d ia ra h a rm U lo lv'n g M n g * getting energy cannot be used on a largo scale. F o r Yo u r Inform ation T h e re is another m ethod to produce fusion reaction .It is Ufc-a x o M raduito-o cauta based upon the principle that the speed of atoms of a (I) Sunburn. M n d rw M and tkin substance increases with the increase in the temperature of that substance. To start a fusion re adion the temperature at cmom w hich the required speed of the light n u d e i can be obtained is about 10 million d e g ree s Celsius. A t such extraordinarily high (1 ) Savara crop damage temperature the reaction that takes place is called thermo­ (» ) dacay o ( m ieroorganM ina nuclear reaction. Ordinarily such a high tomperature cannot (k>) d brupr *»• oeaan aooayMem be achieved. H ow ever during tho explosion of am atom bom b this temperature can be had for a ve ry short time. Until now the fusion re a d io n is taking place only in a h ydrogen bom b. Th a t extraordinary high temperature is obtained d u rirg the explosion of an atom bo m b, due to this high temperature the fusion re a d io n betw een 2H and J,H sets in. In this w a y a ve ry large am ount of en ergy is given out with the explosion. A very large am ount of energy can be had from a fusion reaction, but till n o w this re a d io n has not been brought under control like a fission re a d io n and s o is not being used to produce eledrid ty. Efforts are in full swing in this field and it is hoped that in near future som e method w ould be found to control this readion a s w ell. T h e S u n is com p osed primarily of hydrogen. It has a little am ount of helium and a slight am ount of other heavy elements. A tremendous amount of energy keeps issuing out of it continuously at all times. Th o temperature of its core is about 20 million de g roe s Celsius and its surface temperature is about 6000 de groe s Celsius. Its e n e rg y is d u e to fusion readion called p -p readion. During this process two hydrogen nuclei or two protons fuse to from deuteron This reaction takes place as ]H + J H > 2H + ° e * E n e rg y With the fusion reaction of deuteron with proton, s ^ e an isotope of helium is formed i.e.. 248

* H + ! H ---------> * H e + y + E n e r g yIn the last stage the two nuclei of jH e react in the following For Your Informationmanner^He « jH c > j H e ♦ ]H ♦ 1,H * E n e rg yIn this reaction six protons take part and finally a heliumnucleus and tw o protons are formed. That is. the result ofdifferent stages of this reaction is that four protons haveformed o n e helium nucleus. It has been estimated that in thisp -p chain reaction, 25 .7 M e V energy is given out i.e.. 6.4 M eVper nucleon en ergy is obtained which is m uch greater thanthe e n e rg y given out per nucleon (1 M e V ) during a fissionreaction.21.11 R A D IA TIO N E X P O S U R EW h e n a G e ig e r tube is used in an y experiment, it records fho-chart show ing proportion ©<radiation even w h e n a radioactive source is nowhere near it. radiation from diflerenl sourcesTh is is caused by radiation called background radiation. It is absorbed b y avorage personpartly d u e to cosm ic radiation which com e s to us from outerspace and partly from naturally occurring radioactive T H » symbol * u rw o rsa ty used tosubstance in tho Earth’s crust. T h e cosm ic radiation consists mdiM le an area whore raOoochntyof high en ergy charged particles and electromagnetic is b o n g h*nd*d or artdoal radabonsradiation. T h e atm osphere acts as a shield to a bsorb so m e of are bemg producedthese radiations a s well a s ultraviolet rays. In recent past, thedepletion of ozone layer in the upper atmosphere has beendetected w hich particularly filters ultraviolet rays reaching us.Th is m ay result in increased eye and skin diseases. T h edepletion of o zone layer is suspected to be caused due toexcessive release of som e chemicals in the atmosphere suchas chloroflourocarbons (C F C ) used in refrigeration, aerosolspray a n d plastic foam industry. Its use is now being replacedby environmentally friendly chemicals. M any buildingmaterials contain small amounts of radioactive isotopes.Radioactive radon gas enters buildings from the ground. Itgets trapped inside the building which makes radiation levelsm uch higher from radon inside than outside. A goodventilation can reduce radon level inside the building. Alltypes of food also contain a little radioactive substance. Th emost com m on are potassium -40 and carbon-14 isotopes.S o m e radiation in the environm ent is added by humanactivities. Medical practices, mostly diagnostic X -rayprobably contribute the m ajor portion to it. It is an unfortunatefact that m any X -ra y exposures such as routine chest X -ra y 249