allen physics part 2

ALLENÒ CBSE 49 4. When a parallel beam of monochromatic source of light of wavelength l is incident on a single slit of width a, show how the diffraction pattern is formed at the screen by the interference of the wavelets from the slit. Show that, besides the central maximum at q = 0, secondary maxima are observed at q= æ n + 1 ö l and the minima at q= nl/a. (Refer to NCERT Q. No. 9) çè 2 ÷ø a Why do secondary maxima get weaker in intensity with increasing ‘n’ ? Explain. l Sol. As per dividation of slit, the first two third part is divided into two halves having path diff. 2 . The contribution due to these two halves is 180° out of phase & get cancelled. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Only the remaining one third part of the slit contributes to the intensity at a point between the www.notesdrive.com two minima which will be much weaker than the central maxima. Thus with increasing ‘n’ the intensity of maxima gets weaker. 5. (i) State the essential conditions for diffraction of light. (ii) Explain diffraction of light due to a narrow single slit and the formation of pattern of fringes on the screen. (iii) Find the relation for width of central maximum in terms of wavelength ‘l’, width of slit ‘a’, and separation between slit and screen ‘D’. (iv) If the width of the slit is made double the original width, how does it affect the size and intensity of the central band? Sol. (i) Wavelength of light should be comparable to the size of the obstacle. (ii) Suppose a plane wave front is incident on a slit AB (of width b). Each and every part of the exposed part of the plane wave front (i.e. every part of the slit) acts as a source of secondary wavelets spreading in all directions. The diffraction is obtained on a screen placed at a large distance. (In practice, this condition is achieved by placing the screen at the focal plane of a converging lens placed just after the slit). Aq P bq y O Plane q Lens wave front BD Screen Slit E

50 Physics ALLENÒ • The diffraction pattern consists of a central bright fringe (central maxima) surrounded by dark and bright lines (called secondary minima and maxima). • At point O on the screen, the central maxima is obtained. The wavelets originating from points A and B meets in the same phase at this point, hence at O, intensity is maximum. Secondary minima For obtaining nth secondary minima at P on the screen, path difference between the diffracted waves D = b sinq = nl (n = 0, 1, 2 ......) • Angular position of nth secondary minima sinq ;q= nl b Distance of nth secondary minima from central maximawww.notesdrive.com node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxxn = D×q =nlD=nlf b b where D = distance between slit and screen. f ; D = Focal length of converging lens. Secondary maxima : For nth secondary maxima at P on the screen. l Path difference, D = bsinq = (2n+1) 2 ; where n = 1,2,3......... • Angular position of nth secondary maxima (2n +1)l sinq ; q = 2b Distance of nth secondary maxima,from central maxima xn = D. q= (2n +1)lD = (2n +1)lf 2b 2b Secondary maxima I0 Central maximum Second order First order minimum Second First minimum I0/22 I0/61 – 3—bl – 2—bl – —lb O —lb 2—bl 3—bl Central maxima E

ALLENÒ CBSE 51 (iii) The central maximum : Width of the central maximum is simply the distance between the 1st order minima from the centre of the screen on both sides of the centre. The position of the minima given by ‘y’ (measured from centre of screen) is : y tanq ; q = D For small q, sinq ; q l = bsinq ; bq q = y = l D b node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxy=lD b www.notesdrive.com The width of the central maximum is simply twice this value width of central maximum = 2 lD b angular width of central maximum = 2q = 2 l b (iv) If the width of the slit is made double then the size of the central maxima reduces to half and intensity increases upto four times. 6. Wavefront is a locus of points which vibrates in same phase. A ray of light is perpendicular to the wavefront. According to Huygens principle, each point of the wavefront is the source of a secondary disturbance and the wavelets connecting from these points spread out in all directions with the speed of wave. The figure shows a surface XY separating two transparent media, medium- 1 and medium-2. The lines ab and cd represent wavefronts of a light wave travelling in medium- 1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium -2 after refraction. bd Xa c medium-1 f hY medium-2 eg (i) Light travels as a (A) parallel beam in each medium (B) convergent beam in each medium (C) divergent beam in each medium (D) divergent beam in one medium and convergent beam in the other medium Ans. (A) E

52 Physics ALLENÒ (ii) The phases of the light wave at c, d, e and f are fc, fd, fe, and ff respectively. It is given that fc ¹ ff (A) fc can not be equal to fd (B) fd can be equal to fe (C) (fd – ff ) is equal to (fc – fe) (D) (fd – fc) is not equal to (ff – fe) Ans. (C) (iii) Wavefront is the locus of all points, where the particles of the medium vibrate with the same (A) phase (B) amplitude (C) frequency (D) period Ans. (A) (iv) A point source that emits waves uniformly in all directions, produces wavefronts that are (A) spherical (B) elliptical (C) cylindrical (D) planar Ans. (A) (v) What are the types of wavefronts?www.notesdrive.com (A) Spherical node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx(B) Cylindrical(C) Plane(D) All of these Ans. (D) EXERCISE - V (RACE) CHAPTER-9 : RAY OPTICS & OPTICAL INSTRUMENTS For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : Critical angle is minimum for violet colour. Reason : Because critical angle qc = sin-1 æ 1 ö and m µ 1 . ç m ÷ è ø l (i) a (ii) b (iii) c (iv) d 2. Assertion : Within a glass slab a double convex air bubble is formed. This air bubble behaves like a converging lens. Reason : Refractive index of air is more than the refeactive index of glass. (i) a (ii) b (iii) c (iv) d 3. Assertion : A beam of white light gives a spectrum on passing through a hollow prism. Reason : Speed of light outside the prism is different from the speed of light inside the prism. (i) a (ii) b (iii) c (iv) d 4. Assertion : Microscope magnifies the image. Reason : Angular magnification for image is more than object in microscope. (i) a (ii) b (iii) c (iv) d 5. Assertion : Sperical aberration occur in lenses of larger aperture. Reason : The two rays, paraxial and marginal rays focus at different points. (i) a (ii) b (iii) c (iv) d E

ALLENÒ CBSE 53 6. Write the working principle of optical fibre. 7. State the two advantage of reflecting telescope over refracting telescope. 8. What is dispersion of light. 9. What is the SI unit of power of lens ? 10. Give one use of optical fibres. 11. What happens to the focal length of convex lens, when it is immersed in water? 12. Out of red and blue lights, for which colour is the refractive index of glass greater? 13. How does magnifying power of a telescope change on decreasing the aperture of its objective lens? 14. How does the focal length of a convex lens change if monochromatic red light is used instead of monochromatic blue light ? 15. Two thin lenses of power +7D and -3D are in contact.What is the focal length of the combination ? node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx RACE SOLUTIONS www.notesdrive.com CHAPTER-9 : RAY OPTICS & OPTICAL INSTRUMENTS 1. (i) 2. (iv) 3. (i) 4. (i) 5. (i) 6. Total internal reflection. 7. (i) Reflecting telescope does not suffer from chromatic aberration. (ii) Reflecting telescope's mirror are easier to mount. 8. The spliting of white light into band of seven different colours is called dispersion of white light. 9. Dioptre (D) 10. Optical fibres used in medical investigation. 11. The focal length of convex lens increases, when it is immersed in water. 12. The refractive index of glass for blue light is greater than that for red light i.e. mb > mr. 13. The magnifying power of a telescope is independent of the aperture of the objective lens. 14. We know that from the lens maker's formula 1 = (n - 1) æ 1 - 1 ö f ç R1 R2 ÷ è ø Q lR > lB, nR < nB \\ fB > fA 15. P=P1+ P2 = 7 – 3 P = 4D 100cm f = 4 = 25 cm. E

54 Physics ALLENÒ EXERCISE - V (RACE) CHAPTER-10 : WAVE OPTICS For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : If white light is used in YDSE, then the central bright fringe will be white. Reason : Because all the wavelengths produce their zero order maxima at the same position. (i) a (ii) b (iii) c (iv) d www.notesdrive.com 2. Assertion : Diffraction of sound wave are more easily observed as compare to light waves. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx Reason : Wavelength of sound waves is more as compared to light. (i) a (ii) b (iii) c (iv) d 3. Assertion : In interference light energy is redistributed. Reason : There is no gain or loss of energy, which is consistent with the principle of energy conservation. (i) a (ii) b (iii) c (iv) d 4. Assertion : No interference pattern is detected when two coherent sources are infinitely close to each other. Reason : The fringe width is inversely proportional to the distance between the two slits. (i) a (ii) b (iii) c (iv) d 5. Assertion : In Young's experiment, the fringe width for dark fringes in different from that for white fringes. Reason : In Young's double slit experiment the fringes are performed with a source of white light then only black and bright fringes are obeserved. (i) a (ii) b (iii) c (iv) d 6. In a YDSE with identical slits, the intensity of the central bright fringe is I0 . What will be intensity at the same point, if one of the slits is covered ? 7. A light wave enters from air to glass. How will the frequency be affected? 8. In YDSE how is the fringe width change when distance between the slits is decreased? 9. Which type of wave front will emerge from a point source. 10. Define the term wavefront. 11. What is the condition for first minimum in case of diffraction due to a single slit? 12. What is the geometrical shape of the wavefront in each of the following cases? (a) Light diverging from a point source. (b) Light emerging out of a convex lens, when a point source is placed at its focus. 13. State two conditions which must be satisfied for two light sources to be coherent. 14. State the condition for destructive interference. 15. How is a wavefront to the direction of corresponding rays? E

ALLENÒ CBSE 55 RACE SOLUTIONS CHAPTER-10 : WAVE OPTICS 1. (i) 2. (i) 3. (i) 4. (ii) 5. (iv) 6. I0 4 Þ I = I1 + I2 + 2 I1I2 cos f node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxI = 4I' cos2 f[I1 = I2 = I'] 2 www.notesdrive.com According to Questions Io = 4I' I'' = Io 4 Now one slit is covered, I2 = 0 So Io = I' + 0 + 0 I' = Io 4 7. Frequency will not be changed. 8. Q Bµ 1 , So d Fringe, width will increase. 9. Spherical wave front. 10. Wave front is defined as the locus of all such particles of the medium which are vibrating in the same phase at any instant. l 11. The angular position of first minimum, q1 = a 12. (a) Spherical in shape (b) Plane wavefront 13. 1. The two light waves should be of same wavelength. 2. The two light waves should either be in phase or should have a constant phase difference. l 14. Path difference, x = (2n + 1) 2 , where n = 0, 1, 2, 3…………. 15. The wavefront is perpendicular to the direction of rays. E

56 Physics ALLENÒ EXERCISE - VI (MOCK TEST) CHAPTER-9 : RAY OPTICS & OPTICAL INSTRUMENTS 1. How does the focal length of a convex lens change if monochromatic red light is used instead of monochromatic blue light? [1] 2. The refractive index of the meterial of an equilaterial prism is 3 . What is the angle of minimum deviation? [1] 3. When a ray of light enters from one medium to another, does the frequency changes? [1] 4. What are optical fibres? Give their one use? [2] www.notesdrive.com 5. A convex lens made up of refractive index n1 is kept in a medium of refractive index n2. Parallel node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx rays of light are incident on the lens. Complete the path of rays of light emerging from the convex lens if (i) n1 > n2 (ii) n1 = n2 [2] [2] 6. A substance has critical angle of 45° for yellow light what is its refractive index? 7. A thin converging lens has a focal length f in air. If it is completely immersed in a liquid, briefly explain how the focal length of the lens will vary? [2] 8. Draw a ray diagram to illustrate image formation by a cassegrain type reflecting telescope? Hence state two advantages of it over refracting type telescopes? [3] 9. The focal lengths of an objective lens and eyepiece are 192 cm and 8 cm respectively in a small telescope. Calculate it’s magnifying power and the separation between the two lenses. [3] 10. Show that the limiting value of the angle of prism is twice its critical angle? Hence define critical angle? [3] 11. Derive the mathematical relation between refractive indices n1 and n2 of two radii and radius of curvature R for refraction at a convex spherical surface. Consider the object to be a point source lying on the principle axis in rarer medium of refractive index n1 and a real image formed in the denser medium of refractive index n2. Hence derive lens maker’s formula. [5] E

ALLENÒ CBSE 57 12. The total internal reflection of the light is used in polishing diamonds to create a sparking brilliance. By polishing the diamond with specific cuts, it is adjusted the most of the light rays approaching the surface are incident with an angle of incidence more than critical angle. Hence, they suffer multiple reflections and ultimately come out of diamond from the top. This gives the diamond a sparking brilliance. Critical angle node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docxTotal AirDiamond reflection www.notesdrive.com (i) The refractive index for a diamond is (A) 1.41 (B) Same as glass (C) 2.42 (D) 1 (ii) The basic reason for the extraordinary sparkle of suitably cut diamond is that (A) It has low refractive index (B) It has high transparency (C) It has high refractive index (D) It is very hard (iii) The extraordinary sparkling of diamond (A) Does not depend on its shape (B) Depends on its shape (C) Has no fixed reason (D) None (iv) A diamond is immersed in a liquid with a refractive index greater than water. Then the critical angle for total internal reflection will (A) Increase (B) Decrease (C) Depend on the nature of the liquid (D) Remains the same (v) Optical Fibre Cables work on the principle of (A) Dispersion of light (B) Refraction of light (C) Total internal reflection (D) Interference of light E

58 Physics ALLENÒ (MOCK TEST) CHAPTER-10 : WAVE OPTICS CHAPTER 1. In YDSE interference pattern if the slit width are in the ratio 1 : 4, then find out the ratio of minimum and maximum intensity ? [1] 2. What happens, if the monochromatic light used in Yong's double slit experiment is replaced by white light? [1] 3. What happen to the fringe width if entire arrangement of YDSE is dipped in water? [1] 4. Why no interference is observed, when two coherent sources are [2] www.notesdrive.com (i) infinitely close to each other node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 01_Unit-6.docx (ii) far apart from each other? 5. Give two differences fringes formed in single slit diffraction and Young’s double slit experiment. [2] 6. State Huygens’s principle for constructing wave fronts? [2] 7. A light wave enters from air to glass. How will the following be affected: [2] (i) Energy of the wave (ii) Frequency of the wave: 8. (i) In Young’s double slit experiment, two slits are 1 mm apart and the screen is placed 1 m away from the slits. Calculate the fringe width when light of wavelength 500 nm is used. [3] (ii) What should be the width of each slit in order to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern? 9. Derive Snell’s law on the basis of Huygen’s Wave theory when light is travelling from a denser to a rarer medium. [3] 10. Write any two necessary conditions for interference of light. Draw curve for intensity distribution in Young’s doble slit experiment. [3] 11. What is meant by diffraction of light ? Draw a graph to show the relative intensity distribution for a single slit diffraction pattern. Obtain an expression for the first minima of diffraction. [5] E

ALLENÒ CBSE 59 UNIT-VII : DUAL NATURE OF RADIATION AND MATTER CHAPTER- 11 : DUAL NATURE OF RADIATION AND MATTER 1. PHOTO ELECTRIC EFFECT (PEE) The phenomenon of the emission of electrons, when metals are exposed to light (of a certain minimum frequency) is called photo electric effect. Work Function (f0) : The minimum energy required to eject a electron out from a metal surface. The work function (f0) depends on the properties of the metal and the nature of the surface. f0 ® Cs = 2.14 V, Pt = 5.65 eV Threshold frequency (n0) : The minimum value of frequency below which photo emission is not possible. www.notesdrive.com 2. EXPERIMENTAL SETUP FOR STUDY OF PEE : Quartz *window S Evacuated glass tube Photosensitive plate Electrons CA Commutator V node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx +– Various experiments : Experiment-I : Effect of intensity of light on photo electric current By keeping the frequency of incident radiation & the accelerating potential fixed, the intensity of light is varied and the resulting photo electric current is measured each time. It is found that the number of photo electrons emitted per second is directly proportional to the intensity of incident radiation. Photoelectric current Intensity of light Variation of photoelectric current with intensity of light. E

60 Physics ALLENÒ Experiment-II : Effect of potential (anode) on photo electric current By keeping the intensity & frequency of incident light constant, variation of photo electric current with the potential difference between cathode anode studied. It is found that photo current is found increased up to the level of saturation current by increasing the anode potential. Also at a given frequency of incident radiation the stoping potential is independent of its intensity. www.notesdrive.com PhotocurrentI3 > I2 > I1 I3 Stopping potential II12 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx –V0 0 Collector plate Retarding potential potential Saturation current : The maximum value of photo electric current is called the saturation current. Stopping potential / cut-off potential (V0) : The minimum negative (retrading) potential (given to collector) for which the photo electric current becomes zero, is called stopping potential. 1 mVm2ax = eV0 2 Þ K max = eV0 Experiment-III : Effect of frequency of incident light on stopping potential By keeping intensity of light constant, variation of photoelectric current with potential difference between emitter & collector plate, studied. We obtain different values of stopping potential but the value of saturation is same for incident radiation of different frequencies. It is found that greater the frequency, greater is the maximum kinetic energy of the photoelectron. Photoelectric current v3 > v2 > v1 –V03 –V02 –V01 0 Collector plate potential Retarding potential Variation stopping potential with frequency : Stopping Metal A potential Metal B (V0) ' 0 Frequency of incident radiation (v) E

ALLENÒ CBSE 61 3. SUMMARY OF EXPERIMENTS RELATED TO PHOTOELECTRIC EFFECT We now summarise the experimental features and observations described in this section. (i) For a given photosensitive material and frequency of incident radiation (above the threshold frequency), the photoelectric current is directly proportional to the intensity of incident light. (ii) For a given photosensitive material and frequency of incident radiation, saturation current is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity. (iii) For a given photosensitive material, there exists a certain minimum cut-off frequency of the incident radiation, called the threshold frequency, below which no emission of photoelectrons takes place, no matter how intense the incident light is. Above the threshold frequency, the stopping potential or equivalently the maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of the incident radiation, but is independent of its intensity. (iv) The photoelectric emission is an instantaneous process without any apparent time lag (~10–9s or less), even when the incident radiation is made exceedingly dim. 4. FAILURE OF WAVE THEORY Wave picture is unable to explain the most basic features of photoelectric emission. 5. EINSTEIN'S PHOTOELECTRIC EQUATION (ENERGY QUANTUM OF RADIATION) Albert Einstein explained the various laws of photoelectric emission on the basis of Plank's quantum theory. According to Plank's quantum theory, light radiations consist of tiny packets of energy called photon. One quantum of light radiation is called a photon which travels with the speed of light. The energy of a photon is given by E = hv where, h is Planck's constant and v is the frequency of light radiation. When a photon of energy hv falls on a metal surface, the energy divides in following two ways : node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx www.notesdrive.com (i) A part of energy is used to overcome the surface barrier to come out as photoelectron from metal surface. This part of energy is called work function. It is expressed as f0 = hn0. E

62 Physics ALLENÒ (ii) The remaining part of the energy is used in giving a velocity v to the emitted photoelectron. This is equal to the maximum kinetic energy of the photoelectrons æ 1 m v 2m ax ö , where m is çè 2 ÷ø the mass of the photoelectrons. According to the law of conservation of energy, hv = f0 + 1 mvm2 ax = hv0 + 1 mvm2 ax 2 2 \\ 1 mv2max = Kmax = hv - f0 2 This equation is called Einstein's photoelectric equation. Graph between (Kmax) and frequency Kmax Metal A B (K)max = hn – f0 fA Metal Comparing with equation, Y = mx – c slope = m = tanq = h (same for all metals) fB (f0)B > (f0)A www.noteMetalsBMetaldArive.com qq n node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx Graph between stopping potential (V0) and frequency (n) V0 (fe0)B (fe0)A q q Frequencyn Q eV0 = hn – f0 ; V0 = æ h ö n - æ f0 ö èç e ÷ø èç e ø÷ slope = m = tanq = h (same for all metals) e 6. PARTICLE NATURE OF LIGHT : THE PHOTON • In interaction of radiation with matter, radiation behaves as if it is made of particles called photons. • Each photon has energy E = hn and momentum P = hn/c = h l • All photon of light at particular frequency, have the same energy & momentum. • Photons are electrically neutral. E

ALLENÒ CBSE 63 7. WAVE NATURE OF MATTER De-Broglie proposed that the wave length l associated with a particle of momentum p is given as hh l = p = mv • De Broglie wavelength associated with moving particles If a particle of mass m moving with velocity v. Kinetic energy of the particle E = 1 mv 2 = p2 2 2m node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docxmomentum of particle p = mv = 2mE www.notesdrive.comthe wave length associated with the particles isl=h=h= h p mv 2mE De Broglie wavelength associated with the charged particles :- • For an Electron le = 12.27 ´10-10 m = 12.27 Å so lµ 1 V V V • For Proton lp = 0.286´10-10 m = = 0.286 Å V V • For Deuteron ld = 0.202 Å V • For a Particles \\ la = 0.101 Ao V • For uncharged particle [at temperature T] l = h = h= h= h p 2mk 3 3mk BT 2m ´ 2 k BT Here, m = Mass of particle kB = Boltzman constant (1.38 × 10–23 J/K) T = Temperature E

64 Physics ALLENÒ NCERT IMPORTANT QUESTIONS CHAPTER - 11 : DUAL NATURE OF RADIATION AND MATTER 1. The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium and (b) the wavelength of the incident light if the photo current is brought to zero by a stopping potential of 0.60V. Given h = 6.63 x 10-34 Js. Sol. (a) v0 = f0 = 2.14eV = 2.14 ´1.6 ´10-19 J = 5.16 × 1014 Hz h 6.63 ´ 10 -34 6.63´10-34 Js Js (b) eV0 = hc - f0 or hc l l = (eV0 + f0 ) Q (0.6 + 2.14) eV = 2.74 ev www.notesdrive.com l =21.794.8´91´.610´-1260J-m19 J = 454 nm. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx 2. Monochromatic light of frequency 6.0 x 1014 Hz is produced by a laser. The power emitted is 2.0 × 10-3 W. (a) what is the energy of a photon in the light beam (b) How many photon per second on the average are emitted by the source ? Given h = 6.63 x 10-34 Js. Sol. (a) E = hv = 6.63 x 10-34 x 6.0 x 1014 = 3.98 x 10-19 J = 3.98´10-19 ev = 2.49 ev. 1.6 ´10-19 (b) No. of photons emitted per second (n) = P = 2.0´10-3 W E 3.98´10-19 J = 5.0 x 1015 photons per second. 3. What is de-Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volt. Sol. V = 100 volt De-Broglie wavelength, l = 12.27 Å = 12.27 = 1.227 Å V 100 4. The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission. Sol. Kmax = hn - hno eVo = h(n - no ) Vo = h(n - no ) e Vo = 6.62 ´10-34 ´ (4.9 ´ 1014 ) = 2.0 V 1.6 ´10 -19 E

ALLENÒ CBSE 65 5. Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made. Sol. K max = hc -f Þ eVo = ( ) ( )6.63´10-34 3´108 -f l 488 ´ 10 -9 eVo = 1240eV - f 488 [here Vo = 0.38 V ] f = (2.54 – 0.38) eV or f =2.16 eV 6. Write de Broglie hypothesis. Obtain the formula for de Broglie wavelength of an electron which is accelerated from rest through a potential V volt. Sol. De-Broglie postulate – According to de-Broglie postulate if a light wave can behave like a node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx particle then any moving particle of matter should also behave like a wave of definite www.notesdrive.com wavelength. If electron is accelerated from rest through V volt, then kinetic energy (k) = eV wavelength related to electron – le = h h Þ le = 12.27 Å 2mek Þ le = 2m eeV V 7. What is photo electric effect ? On which two factors the photo electric current depends. Sol. Photoelectric effect :- The phenomenon of emission of electron by metal plate on falling radiations of suitable frequency at it's surface is known as photoelectric effect. Factors affecting photoelectric current - (1) Intensity of incident radiation. (2) Potential of anode with respect to cathode. 8. Derive Einstein’s Photoelectric equation. Explain photoelectric effect with help of this equation. (Refer to Theory topic-5) 9. What do you understand from rest mass and kinetic mass of a photon P. Derive formula for momentum of photon P =h / l, where h is Plank’s constant and l is wavelength of photon. Sol. Rest mass of photon is zero According to Einstein energy mass relation, E = mc2 The inertial mass of a photon is m = E = hn c2 c2 Momentum of photon P = mc = hn ×c = hn Q n= c c2 c l P = h × c Þ ëéêP = h ù c l l ûú 10. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation. Sol. (i) The instantaneous ejection of photoelectrons. (ii) Existence of threshold freq. for a metal surface. (iii) The fact that K.E. of the emitted electrons is independent of the intensity of the light but depends on its frequency. E

66 Physics ALLENÒ CHAPTER - 11 : DUAL NATURE OF RADIATION AND MATTER EXERCISE-I ONE MARK QUESTIONS : 1. Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity. Photo electric current Sol. i Higher intensity I2 I1 I2 > I1 www.notesdrive.com V0 O V node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx The graph I2 corresponds to radiation of higher intensity (Also accept if the student just puts some indicative marks, or words, (like tick, cross, higher intensity) on the graph itself. 2. Name the phenomenon which shows the quantum nature of electromagnetic radiation. Sol. Photoelectric effect (PEE) 3. Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit. Sol. Intensity of radiation is defined as the energy associated with number of photons incident/emitted from a unit surface area in unit time. i.e. Intensity =ArEenae´rgtiyme SI unit :- Joule or Watt m–2 m2 -s 4. Draw a plot showing the variation of de Broglie wavelength of electron as a function of its K.E. Sol. l 0 E Ql = h 2mE E

ALLENÒ CBSE 67 EXERCISE-II TWO MARK QUESTIONS : 1. A photon and a proton have the same de-Broglie wavelength l. Prove that the energy of the photon is (2mlc/h) times the kinetic energy of the proton. Sol. Energy of photon, E1 = hc ...(1) l Energy of proton, E2 = h2 æ h ö ...(2) 2 ml 2 èççQ l = 2mE 2 ÷÷ø from (1) & (2), E1 = æ 2mlc ö E2 èç h ÷ø node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx2. If light wavelength 412.5 nm is incident on each of the metals given below, which ones will www.notesdrive.comshow photoelectric emission and why ? Metal Work Function Na (eV) 1.92 K 2.15 Ca 3.20 Mo 4.17 Sol. Given l = 412.5 nm Using, E = 1240 eV l nm E = 1240 = 3eV 412.5 So, Na(f = 1.92 eV) and K (f = 2.15 eV) both the metals will show photoelectric emission as incident energy is more than work function. 3. The wavelength l of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of a photon is (2lmc/h) times the kinetic energy of electron; where m, c and h have their usual meaning. Sol. Ephoton = hc ....(1) l Eelectron = h2 ....(2) æ hö 2ml2 ççèQl = 2mEelectron ÷÷ø Dividing (1) by (2), E photon = 2mcl Eelectron h E

68 Physics ALLENÒ 4. Calculate the de-Broglie wavelength of the electron orbiting in the n = 2 state of hydrogen atom. Sol. Velocity of an e– in first orbit (n = 1) of hydrogen atom, v = 2.18 × 106 m/s Now, velocity of e– in second orbit (n = 2) of hydrogen atom will be given as, v¢ = v/n \\ v¢ = 1.09 × 106 m/s So mv¢ = 9.9 × 10–25 Putting the values in l = h/mv¢ = 6.6 ´10–34 = 6.68Å 9.9 ´10–25 5. A proton and an a particle are accelerated through the same potential difference. Which one of the two has (i) greater de-Broglie wavelength, and (ii) less kinetic energy ? Justify your answer. Sol. l= h = h p 2qVm www.notesdrive.com (i) a–particle : h node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx2qaVm a = la proton : h 2qpVm p = lp Clearly, lp > la as ma > mp & qa > qp. So, proton has greater de-Broglie wavelength. (ii) As 1 mv2 = qV ; (K.E.)p < (K.E.)a as qp < qa. 2 So, proton has less K.E. 6. Plot a graph showing variation of de-Broglie wavelength l versus 1 , where V is accelerating V potential for two particles A and B carrying same charge but of masses m1, m2 (m1 > m2). Which one of the two represent a particle of smaller mass and why ? Sol. As, l = h Graph :- 2mqV Slope of l versus 1 graph will be inversely prop. to the square root of the mass of the V particles. Now, slope of B is greater & it shows that its mass is smaller. B l ­A O –1— ® ÖV E

ALLENÒ CBSE 69 EXERCISE-III THREE MARK QUESTION : 1. Plot a graph showing the variation of photoelectric current with intensity of light. The work function for the following metals is given : Na : 2.75 eV and Mo : 4.17 eV Which of these will not give photoelectron emission from a radiation of wavelength 3300 Å from a laser beam ? What happens if the source of laser beam is brought closer ? Sol. Photocurrent node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx O www.notesdrive.com Intensity For, l = 3300Å, E = hc = 6.6 ´10–34 ´ 3´108 = 6 ´10–19 J = 6 ´10–19 eV = 3.75 eV l 3300 ´10–10 1.6 ´10–19 Clearly, incident energy 3.75 eV < 4.17 eV, so photoelectric effect will not take place in Mo (4.17 eV). Photo current will increase in Na if the source of laser beam is brought closer. 2. (a) Define the term ‘intensity of radiation’ in photon picture. (b) Plot a graph showing the variation of photo current v/s collector potential for three different intensities I1 < I2 < I3, two of which (I1 and I2) have the same frequency n and the third has frequency u1 > u. (c) Explain the nature of the curves on the basis of Einstein’s equation. Sol. (a) Refer One Marks Q.3 I3 I2 (Intensity) I1 (b) V VO2 o1 O Potential (V) (c) From, eV0 = hn – hn0 [Clearly stopping potential (V0) depends on incident freq.(n)] or V0 = h n– h n0 e e So, for I1 & I2 intensities, freq. (n) is same, hence for them stopping potential (V0) is also equal. Further, for I3 intensity having freq (n1) stopping potential is more. (Q n1 > n) E

70 Physics ALLENÒ 3. The given graph show the variation of photo-electric current I (I) with the applied voltage (V) for two different materials 1 and for two different intensities of the incident radiation. Identify and explain using, Einstein’s photo electric equation 2 the pair of curves that correspond to (i) different material but 3 same intensity of incident radiation, (ii) different intensities 4 but same materials. V Sol. I 1 2 3 4 www.notesdrive.com V node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx (i) (1,2) & (3,4) are diff. materials of same intensity. As saturation current is same but stopping potential is different. (ii) (1,3) & (2,4) have diff. intensities but same materials. As saturation currents are different but stopping potential is same. 4. State two important properties of photon which are used to write Einstein’s photoelectric equation. Define (i) stopping potential and (ii) threshold frequency, using Einstein’s equation and drawing necessary plot between relevant quantities. Sol. (i) Energy of a photon is proportional to the freq. of light. (E = hn) (ii) Photons are quanta or discrete carriers of energy. Stopping potential :- In experiment of photoelectric effect, AB (nOA <nOB) the value of negative potential of anode at which (V0) stopping frequency (n) ® photoelectric current reduces to zero is called stopping potential potential for the given freq. of incident radiation. O Threshold freq. :- For a given material, there exist a certain min. frequency below which no photoelectron can come out from the metal surface. This is called threshold ferquency. 5. Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies uA > uB. (i) In which case is the stopping potential more and why ? (ii) Does the slope of the graph depend on the nature of the material used ? Explain. E

ALLENÒ CBSE 71 Sol. (V0) B stopping A potential (nA >nB) 0 (n) ® frequency From eV0 = Kmax = hn - f eV0 = hn – hn0 V0 = hn - hn0 e e node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx (i) From the above equation, we can conclude, more threshold freq., less would be the stopping www.notesdrive.com potential (V0). Here, nOA > nOB, hence stopping potential (V0) is greater for B than A. (ii) V0 = æh ö n - h (y = mx ± c) èç e ÷ø e n0 Slope of the above equation is æ h ö which is ‘independent’ of the nature of material. çè e ø÷ 6. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation. Sol. (i) The instantaneous ejection of photoelectrons. (ii) Existence of threshold freq. for a metal surface. (iii) The fact that K.E. of the emitted electrons is independent of the intensity of the light but depends on its frequency. 7. Using photon picture of light, show how Einstein's photoelectric equation can be established. Write two features of photoelectric effect which cannot be explained by wave theory. Sol. Einstein's states that electromagnetic radiation energy is built up of discrete units-called quanta of energy & has energy hv where 'h' is planck's constant & 'v' is the frequecy of light. In this phenomenon, an e– absorbs a quantum of energy (hv), if the energy absorbed exceeds the minimum energy needed to escape from metal (f0), e– is emitted with maximum K.E. Kmax = hv - f0 (Einstein's photo electric equation) Two features of PEE which can't be explained by wave theory- (a) A min. frequency (threshold frequency) exists for diff. metals below which PEE is not possible (b) PEE is an instantaneous phenomenon E

72 Physics ALLENÒ 8. Define the term ‘cut off frequency” in photoelectric emission. The threshold frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photo-electrons is v1. When the frequency of the incident radiation is increased to 5f, the maximum velocity of photo-electrons is v2. Find the ratio v1 : v2. Sol. Cut off frequency – The min. freq. of light which can emit photoelectrons from a material is known as cut-off frequency. From eq. KEmax = hn – hn0 1 mv12 = hf .........(1) 2 1 mv 2 = 4hf .........(2) 2 2 \\ v12www.notesdrive.com=1Þv1=1 v22 4 v2 2 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx EXERCISE-IV FIVE MARKS QUESTIONS :- 1. The photoelectric emission is possible only if the incident light is in the form of packets of energy, each having a definite value, more than the work function of the metal. This shows that light is not of wave nature but of particle nature. It is due to this reason that photoelectric emission was accounted by quantum theory of light. (i) Packet of energy are called (A) electron (B) quanta (C) frequency (D) neutron Ans. (B) (ii) One quantum of radiation is called (A) meter (B) meson (C) photon (D) quark Ans. (C) (iii) Energy associated with each photon (A) hc (B) mc (C) hn (D) hk Ans. (C) (iv) Which of the following waves can produce photoelectric effect (A) UV radiation (B) Infrared radiation (C) Radio waves (D) Microwaves Ans. (A) (v) Work function of alkali metals is (A) less than zero (B) just equal to other metals (C) greater than other metals (D) quite less than other metals Ans. (D) E

ALLENÒ CBSE 73 EXERCISE - V (RACE) CHAPTER - 11 : DUAL NATURE OF RADIATION AND MATTER For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : Photo electric effect demonstrate the wave nature of light. Reason : The number of photo electrons is proportional to the frequency of light node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx (i) a (ii) b (iii) c (iv) d www.notesdrive.com 2. Assertion : At saturation, photo electric current is maximum. Reason : At saturation, all the electrons emitted from cathode are able to reach at anode. (i) a (ii) b (iii) c (iv) d 3. Assertion : A photon is not a material particle. It is a quanta of energy. Reason : Photoelectric effect demonstrate wave nature of radiation. (i) a (ii) b (iii) c (iv) d 4. Assertion : Photo electric current increases if the distance between cathode and anode is increased. Reason : The momentum of photon is directly proportional to its wavelength. (i) a (ii) b (iii) c (iv) d 5. Assertion : Photo sensitivity of a metal is high if its work function is small. Reason : Work function = hn0 , where n0 is the threshold frequency. (i) a (ii) b (iii) c (iv) d 6. Define intensity of radiation on the basis of photon picture of light. Write its SI unit. 7. Define the term stopping potential. 8. Show on a plot the nature of variation of photoelectric current with the intensity of radiation incident on a photo sensitive surface? 9. The maximum kinetic energy of a photo electron is 3eV, what is its stopping potential? 10. Write Einstein's photo electric equation. 11. In photoelectric effect, why should the photoelectric current increases as the intensity of monochromatic radiation incident on a photosensitive surface is increased? Explain. 12. Show the variation of photocurrent with collector plate potential for different frequencies but same intensity of incident radiation. 13. Name the phenomenon which shows the quantum nature of electromagnetic radiation. E

74 Physics ALLENÒ 14. The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work function? Justify your answer. 15. How will the photoelectric current change on decreasing the wavelength of incident radiation for a given photosensitive material ? RACE SOLUTIONS CHAPTER - 11 : DUAL NATURE OF RADIATION AND MATTER www.notesdrive.com 1. (iv) 2. (i) 3. (iii) 4. (iv) 5. (ii) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx 6. Intensity of radiation is defined as the energy associated with number of photons incident / emitted from a unit surface area in unit time Intensity = Energy Area ´ time SI unit : Watt m–2 7. It is the value of negative potential of anode at which photo electric current reduces to zero is called stopping potential for the given frequency of incideent radiation. 8. Photo current Intensity 9. 3V 10. eV0 = Kmax = hu – f 11. An increase in intensity means increase in number of photons and thus, increase in photoelectric current. 12. For same intensity but different frequencies n1 > n2 > n3 of incident radiation. 13. Photoelectric effect. 14. As threshold frequency of metal A is greater, its work function, F = h n0, will also be greater than that of B. 15. The photoelectric current does not depend on the wavelength of incident radiation E

ALLENÒ CBSE 75 EXERCISE - VI (MOCK TEST) CHAPTER - 11 : DUAL NATURE OF RADIATION AND MATTER 1. Calculate the energy associated in eV with a photon of wave length 4000Å. [1] 2. Draw a plot showing the variation of de Broglie wavelength of electron as a function of its K.E.[1] 3. What is de-Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volt. [1] 4. Define the term threshold frequency and stopping potential for photoelectric effect. Show graphically, how the stopping potential for a given metal, varies with frequency of incident radiation. Mark threshold frequency on the graph. [2] 5. A particle is moving three times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.813 × 10–4. Calculate the particle mass and identify the node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx particle. [2] www.notesdrive.com 6. Calculate the maximum kinetic energy of electrons emitted from a photosensitive surface of work function 3.2eV, for the incident radiation of wavelength 300nm. [2] 7. If the intensity of light falling on the emitting substance of a photoelectric cell be increased then what will be the effect on (i) Current flowing from the cell (ii) potential difference required to stop the current completely? [2] 8. What is photoelectric effect? Why it can't be explained on the basis of wave nature of light? [3] 9. Write de Broglie hypothesis. Obtain the formula for de Broglie wavelength of an electron which is accelerated from rest through a potential V volt. [3] 10. Plot a graph showing the variation of photoelectric current with intensity of light. The work function for the following metals is given : Na : 2.75 eV and Mo : 4.17 eV Which of these will not give photoelectron emission from a radiation of wavelength 3300 Å from a laser beam ? What happens if the source of laser beam is brought closer ? [3] 11.(a) Write Einstein photoelectric equation and use it to explain: (i) Independence of maximum energy of emitted photoelectrons from intensity of incident light, (ii) Existence of threshold frequency for emission of photoelectrons. (b) Radiations of frequencies n1 and n2 are made to fall in turn on a photosensitive surface. The stopping potential required for stopping the most energetic photoelectrons in the two cases are V1 and V2 respectively. Obtain a formula for determining the Plank's constant and threshold frequency in terms of these parameters. (c) An electromagnetic wave of wavelength l is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength l1, prove that l = 2mc l12 . h E

76 Physics ALLENÒ 12. According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle or a wave is associated with moving material particle which controls the particle in every respect. The wave associated with moving material particle is called matter wave or de-Broglie wave whose wavelength called de-Broglie wavelength, is given by l = h/mv (i) The dual nature of light is exhibited by :- (A) diffraction and photo electric effect (B) photoelectric effect (C) refraction and interference (D) diffraction and reflection. (ii) If the momentum of a particle is doubled, then its de-Broglie wavelength will :- www.notesdrive.com (A) remain unchanged node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 02_Unit-7.docx (B) become four times (C) become two times (D) become half (iii) If an electron and proton are propagating in the form of waves having the same l, it implies that they have the same :- (A) energy (B) momentum (C) velocity (D) angular momentum (iv) Velocity of a body of mass m, having de-Broglie wavelength l , is given by relation :- (A) v = l h/m (B) v = l m/h (C) v = l/hm (D) v = h/ lm (v) Moving with the same velocity, which of the following has the longest de-Broglie wavelength? (A) b-particle (B) a-particle (C) proton (D) neutron E

ALLENÒ CBSE 77 UNIT-VIII : ATOMS & NUCLEI CHAPTER-12 : ATOMS 1. RUTHERFORD ATOMIC MODEL (Nuclear Model) 1.1 Experimental Set up : lead vacuum ZnS screen s cre e n lead box gold foil micros cope 10–7 m beam of q most a pass a-particle through source of a-particle node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docxabout 1 in 8000 issome are deviated repelled back through large angle q www.not Ne(q)sdrive.com A beam of 5.5 MeV a-particles emitted from a 214 Bi radioactive directed at a thin metal foil 83 made of gold. The scattered alpha-particles were observed through a rotatable detector consisting of zinc sulphide screen and a microscope. 1.2 Graph between number of scattered particle and scattering angle : N( q)µ cosec4 q 2 90° 180° q 1.3 Summary of Rutherford Model : • The whole positive charge of the atom & almost its entire mass is concentrated at the small region of the atom (nucleus) • The central core (nucleus) is surrounded by clouds of electron which makes the atom as electrically neutral. • Using high speed a-particle, the nucleuar diameter has been found to be in order of 10–14 m. • Distance of closest approach (r0) : Estimation of nuclear size At the distance r0 the entire kinetic energy of the a-particle is get converted into potential energy of the system. \\ K.E.(K) = P.E. = kq1q2 = (K) (2e)·Ze r0 r0 Þ r0 Ze2 = 2pe0K E

78 Physics ALLENÒ Impact parameter (b) : The trajectory traced by an a-particle depends upon the impact parameter of collision. Impact parameter defined as the perpendicular distance of the velocity vector of a-particle from the centre of the nucleus. q target b nucleus 1.4 Draw backs of Rutherford Model : (i) An accelerating charge radiates the nucleus spiralling inward and finally fall into the nucleus, which does not happen in an atom. This could not be explained by this model. (ii) It could not explain the characteristic line spectra of different elements. 2. BOHR ATOMIC MODEL Bohr adopted Rutherford model of the atom & added some arbitrary conditions. 2.1 Bohr postulates : (i) The electron in a stable orbit does not radiate energy. (ii) A stable orbit is that in which the angular momentum of the electron about nucleus is an www.notesdrive.com integral (n) multiple of h i.e. mvr = n h ; n=1, 2, 3, ..(n ¹ 0). node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx2p2p (iii) The electron can absorb or radiate energy only if the electron jumps from a lower to a higher orbit or falls from a higher to a lower orbit. The energy emitted or absorbed is a light photon of frequency n and of energy. E = hn 2.2 For Hydrogen atom (Z = atomic no.= 1) : Radius of orbit : rn = n2h2e0 pme2 rn =(0.53) n2 Å rn µ n2 Velocity of electron in nth orbit : v = e2 2nhe0 1 c v = 137 n Orbital frequency of electron (n) : me4 n = 4e02h3n3 n µ n-3 E

ALLENÒ CBSE 79 Electron energy : (i) Kinetic energy = me4 8n2h2e20 (ii) Potential energy = me4 - 4n2h2e20 Total energy = KE + PE En = me4 8n2h2e20 -13.6 En = n2 eV Frequency of emitted radiation (n) : node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docxn=c=-Rcé1-1ù l ê n2f n2i ú www.notesdrive.com ë û n =1 = R é 1 - 1ù = R é 1 - 1ù l ê ni2 ú êë n12 úû ë nf2 û n 2 2 2.3 line spectra of the hydrogen atoms n=5 Ionised atom n=4 O – 0.85 Brackett Excited – 1.5 Paschen series states series – 3.40 Balmer series – 13.6 n=1 Ground state Lyman series Line spectra originate in transitions between energy levels • Lyman Series : (Landing orbit n = 1) . Ultraviolet region n = R é1 - 1ù ; n2 > 1 êêë12 ú n22 ûú • Balmer Series: (Landing orbit n = 2) Visible region ; n = R é1 - 1 ù n2 > 2 êëê22 n22 ú ûú E

80 Physics ALLENÒ • Paschan Series : (Landing orbit n = 3) In the near infrared region n = R é1 - 1 ù ; n2 > 3 êêë32 n22 ú ûú • Brackett Series : (Landing orbit n = 4) In the mid infrared region n = R é 1 - 1 ù ; n2 > 4 ê 42 n22 ú ëê ûú • Pfund Series : (Landing orbit n = 5) In far www.notesdrive.cominfraredregionn = R é 1 - 1 ù ; n2 > 5 ê 52 n22 ú node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docxêëûú In all these series n2 = n1 + 1 is the a line = n1 + 2 is the b line = n1+3 is the g line...etc. where n1 = Landing orbit Total emission spectral lines (i) From n1 = n to n2 = 1 state = n(n - 1) 2 (ii) From n1 = n to n2 = m state = æ (n - m) (n - m +1) ö çèç 2 ÷÷ø Ionization energy of hydrogen atom The energy required to remove an electron from an atom. The energy required to ionize hydrogen atom is = 0 – ( – 13.6) = 13.6 eV. Ionization Potential Potential difference through which a free electron is moved to gain ionization energy = - En electronic charge 2.4 Draw back of Bohr model : (i) It is applicable only for Hydrogenic (single electron) atoms (ii) Model is unable to explain for the relative intensities of the frequencies emitted. E

ALLENÒ CBSE 81 CHAPTER- 13 : NUCLEI 1. NUCLEAR STRUCTURE Each nuclear species with a given Z and A is called a nuclide A nuclide is represented by ZAX Z = atomic number = p = number of proton n = number of neutrons A = mass number = number of nucleons A =n+p 2. PROPERTIES OF NUCLEI (i) Size of nucleus : (order is fermi) R µ A1/3 R = R0 A1/3 R0 ; 1.2 fm (ii) Volume : Volume µ A (iii) Mass : Measured in amu(u) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx (iv) Density (r) : r = A mp = 3 × 10–17 kg/m3 www.notesdrive.com 4 pR 3 3 3. EINSTEIN'S MASS ENERGY EQUIVALENCE E = mc2 4. MASS DEFECT Mass of a nucleus is always less than the sum of masses of its constituent nucleons. This difference is called mass defect. If observed mass of nucleus 2XA be M, mass of proton mP and mass of neutron is mN then mass defect = Dm = [ZmP + (A – Z) mn] – M or mass defect = mcalculated – mobserved 5. BINDING ENERGY (Eb) It it the energy required to separate a nucleus into its constituent nucleons Eb = Dmc2 D m = mass defect Binding energy is a measure of how well a nucleus is held together. 6. NUCLEAR FISSION Nuclear fission occues when a heavy nucleus (such as U235) splits into two smaller nuclei. In such reaction total mass of the product is less than the original mass of the heavy nucleus. 0 n1 +92 U 235 ® 56 Ba141 + 36 Kr92 + 30 n1 Measurement showed that about (Q value) 200 MeV of energy is released in each fusion event. 7. NUCLEAR FUSION When two light nuclei combine to form a heavier nuclei, the process called nuclear fussion. Total mass of final nuclei is less than the masses of original nuclei, this mass loss compensated by a release of energy. [41 H1 + 4e¯ ]® [2 He4 + 2e¯ ] + 2n + 6 g + 26.7 MeV E

82 Physics ALLENÒ NCERT IMPORTANT QUESTIONS CHAPTER-12 ATOMS 1. In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.) Sol. From Bohr’s second postulate mvr = nh 2p n = 2pmvr Þ n = 2 ´ 3.14´ 6 ´1024 ´ 3´104 ´1.5 ´ 1011 h 6.62 ´10-34 n = 2.6´1074 2. Calculate the longest and shortest wavelength in the Balmer series of hydrogen atom. Givenwww.notesdrive.com Rydberg constant = 1.0987 × 107m-1. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx Sol. The wavelength (l) of different spectral lines of Balmer series is given by 1 =R é1 - 1 ù l ê ú ë 22 n 2 û 2 Longest wave length is of Ha line or Ist line of the series. For which n2 = 3. 1 = 1.097×107 é 1 - 1 ù = 1.097× 107 × 5 l ëê 22 32 ûú 36 36 m = 36 ´1010 Å = 6563 Å l =5´1.097´107 5 ´1.097 ´ 10 7 For Shortest Wavelength , n2 = ¥ 1 =1.097 × 107 é 1 - 1 ù = 1.097´107 l êë 22 ¥2 ûú 4 l 4 = 4 ´1010 Å = 3646 Å =1.097´107 1.097 ´ 10 7 3 Calculate the radius of the third Bohr orbit of hydrogen atom and the energy of electron in the orbit. Sol. r = ?, n =3, E = ? As r= n2h2 4p2mke2 r= 32 (6.6 ´10-34 )2 = 392.04 ´ 10-68 8268.81´10-60 4 ´ 22 ´ 22 ´ 9.1´10-31 ´ 9 ´ 109 (1.6 ´10 -19 )2 7 7 r = 4.775 × 10-10 m = 4.775 Å E= -13.6 eV = -13.6 eV = –2.42 × 10–19 J n2 n2 E

ALLENÒ CBSE 83 4. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? Sol. Wave length of radiation emitted due to bombardment of electron beam is ( )( )hc 6.63´10-34 3´108 =994 Å l =E = 12.5´1.6 ´10-19 This wave length falls in the range of lyman series (912 Å to 1216 Å) thus we conclude that lyman series of wave length 994Å is emitted. 5. Write Bohr’s any two postulates for hydrogen atom. Sol. Bohr's postulate :- (i) An electron in an atom can revolve in certain stable orbits without the emission of radiant energy. The neccessary centripetal force for circular motion is provided by electrostatic force of attraction between the nucleus and electron. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx (ii) The electron revolves around the nucleus only in those orbits for which the angular www.notesdrive.com momentum is equal to an integral multiple of h (where h = plank's constant) or angular 2p momentum of revolving e– is quantized. 6. Write two drawbacks of Rutherford’s atomic model. Sol. (i) It failed to explain about stability of atom. (ii) If e– radiates energy continuously then observed spetrum should be continues but actually lines of certain frequencies are present. So it's spectrum is a line spectrum. Rutherford's model could not explain line spectrum of an atom. 7. Write first and second postulates of Bohr’s model of atom. Derive the expression for radius of stationary orbit of an electron and its orbital velocity. Sol. Bohr's first postulate : An electron in an atom can revolve in certain stable orbits without the emission of radiant energy. The neccessary centripetal force for circular motion is provided by electrostatic force of attraction between the nucleus and electron. Bohr's second postulate : The electron revolves around the nucleus only in those orbits for which the angular momentum is equal to an integral multiple of h (where h = plank's 2p constant) or angular momentum of revolving e– is quantized. From Bohr's second postulate mvnrn =n2hp vn nh ....(1) =2pmrn For Radius % Coulomb force : The electrostatic attraction force which is acting between electron and nucleus is called coulamb force. According to this - Fe = kze2 íîìFe = kq1q 2 rn2 r2 E

84 Physics ALLENÒ Centripetal force = mv 2 n rn In equilibrium = mv 2 n rn v 2 = kze2 ……(2) n mrn From equation (1) and (2) n2h2 kze2 4p2m 2rn2 =mrn rn =æèç 4p2he 2 ö n2 ……(3) www.notesdrive.com2km÷ z ø node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx Put the value p, k, h, m, e rn =0.529 ´10-8 ´ n2 cm z rn =0.529 ´ n2 Å z For velocity kze2 = mv2n rn2 rn kze2 =(mvn rn )(vn ) ……(4) ……(5) angular momentum mvn rn = nh - 2p From equation (3) and (4) kze2 =2nhp vn Þ vn = kze2 ´ 2p nh on putting the value of p, k, e, h vn =2.188 ´108 ´ z cm / s n 8. Using Rutherford model of the atom, derive the expression for the total energy of an electron in hydrogen atom. What is the significance of total negative energy possessed by the electron ? Sol. Acc. to Rutherford nuclear model of the atom, the electrostatic force of attraction Fe between the revolving electron & the nucleus provides the requisite centripetal force (FC). Thus, FC = Fe mv2 1 e2 (Q z = 1) r =4pe0 × r2 E

ALLENÒ CBSE 85 Thus the relation between the orbit radius and the e– velocity is, r = e2 4pe 0 mv2 The K.E. (K) & electrostatic potential energy (U) of electron in hydrogen atom are – K = 1 mv2 = e2 v 2 8pe0 r & U = -e2 (–ve sign shows, that nature of force is attractive) r e- 4pe0r +Ze Thus, the total mech. energy E of an e– is, E = K+ U = e2 - e2 = -e2 8pe0r 4pe0r 8pe0r Total energy (E) is –ve which shows that e– is bound to the nucleus, if (E) were +ve then e– would leave the the atom. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx NCERT IMPORTANT QUESTIONS www.notesdrive.com CHAPTER-13 NUCLEI 1. Obtain approximately the ratio of the nuclear radii of the gold isotope 79Au197 and silver isotope 47Ag107 Sol. A1 = 197 and A2 = 107 R=RoA1/3 R1 é A1 ù1 / 3 é197 ù1/3 R2 ê A2 ú êë107 úû = ë = =1.225 û 2. Obtain the binding energy of the nuclei 26Fe56 and 83Bi209 in units of MeV from the following data : m(26Fe56) = 55.934939 a.m.ui m(83Bi209) = 208.980388 a.m.u. (Take 1 a.m.u. = 931.5 MeV) Sol. (i) F56 nucleus contains 26 protons and (56 - 26) = 30 Neutrons 26 Mass of 26 protons = 26 × 1.007825 = 26.20345 a.m.u. Mass of 30 neutrons = 30 × 1.008665 = 30.25995 a.m.u Total mass of 56 nucleons = 56.46340 a.m.u. Mass of 26Fe56 nucleus = 55.934939 a.m.u. Mass defect, Dm = 56.46340 - 55.934939 = 0.528461 a.m.u. Total binding energy = 0.528461 × 931.5 MeV = 492.26 MeV (ii) Same as part-(i) 3. From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant. (i.e. independent of A). Sol. Density of nucleus matter = Mass of nucleus Volume of nucleus Þ r= mA , [ R = R0A1/3 4 pR 3 3 r = 3m Þ A×1.66×1027kg » 1017 kg/ m3 4pR ( )r = 3 4 ×3.14 × 1.2×10-15 3 A 0 3 It is clear that the density of nucleus is constant. E

86 Physics ALLENÒ 4. The fission properties of 94Pu239 are very similar to those of 92U235. The average energy released per fission is 180 MeV. How much energy,in MeV, is released if all the atoms in 1 kg of pure 94Pu239 undergo fission? Sol. The number of atoms in 1kg of 94Pu239 = 6.023 ´ 1023 ´ 1000 239 Energy released per fission = 180 MeV Energy released by 1kg of 94Pu239 = 6.023´ 1023 ´ 1000 ´180MeV = 4.53 × 1026 MeV 239 5. What will be the value of neutron multiplication factor for controlled chain reactions ? Sol. Neutron multiplication factor (k) = rate of production of neutrons rate of loss of neutrons Value of neutron multiplication factor for controlled chain reactions will be 1. 6. Differentiate between nuclear fission & fusion. Give an example of each. Which of the above reactions take place in nuclear reactor ? Sol. (i) Nuclear fission : Nuclear fission is the phenomenon of splitting of a heavy nucleus (usually A > 230) into two or more lighter nuclei. www.notesdrive.com Ex. U235 + 0n1 ®92U236®56Ba141 + 36Kr92 + 30n1 + Q node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx 92 (ii) Nuclear fusion : The process of combining of two lighter nuclei to form one heavy nucleus is called nuclear fusion. Ex. 41H1 ® 2He4 + 21e0 + 2n + 26.7 MeV * Nuclear reactor is based on controled chain reaction. 7. What is nuclear force ? Write it’s specific properties ? Sol. Nuclear Force:- The nuclear force is a force that acts between the protons and neutrons of an atom. Properties :- (i) Nuclear forces act between a pair of neutrons, a pair of protons and also between a neutron- proton pair with the same strength. (ii) The nuclear forces are very short range forces. (iii) The nuclear forces are depended on spin angular momentum of nuclei. (iv) Nuclear forces are non-central forces. 8. Draw a graph showing variation of potential energy of a pair of nucleon as a function of their separation indicate the region in which the nuclear force is (a) Attractive (b) Repulsive. Sol. The potential energy is a minimum at a distance r0 of about 0.8 fm. This means that the force is attractive for distances larger than 0.8 fm and repulsive if they are separated by distances less than 0.8 fm. E

ALLENÒ CBSE 87 CHAPTER - 12 : ATOMS EXERCISE-I TWO MARK QUESTIONS : 1. A photon emitted during the de-excitation of electron from a state n to the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2 eV, in a photo cell, with a stopping potential of 0.55 V. Obtain the value of the quantum number of the state n. OR A hydrogen atom in the ground state is excited by an electron beam of 12.5 eV energy. Find out the maximum number of lines emitted by the atom from its excited state. Sol. Stopping potential V0 = 0.55 V Work function W0 = 2eV node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docxMaximum kinetic energy, kmax = 0.55 eV www.notesdrive.comEnergy of emitted photon , E = hv = kmax + W0 E = 2.55 eV En2 - En1 = 2.55 eV, En1 = – 3.4 eV En2 = 2.55 – 3.4 En2 = -0.85eV From, En2 = - 13.6 eV n2 -13.6 = n2 Þ n2 = 4 -0.85 OR En = En1 + 12.5 eV En2 = –13.6 eV + 12.5 eV = – 1.1 eV En2 =– 13.6 eV Þ n2 = -13.6 = 3.51 Þ n2 ; 4 n22 -1.1 Maximum spectral lines = n(n -1) ; = 4(4 -1) = 6 22 2. A hydrogen atom initially in the ground state absorbs a photon with energy 12.5 eV. Calculate the longest wavelength of the radiation emitted and identify the series to which it belongs. [Rydberg constant R = 1.1 × 107 m–1] Sol. We know, 1 = R æ 1 - 1 ö lmax ç nf2 ÷ è n 2 ø i The energy of the incident photon = 12.5 eV Energy of ground state = –13.6eV \\ Energy after absorption of photon can be –1.1eV E

88 Physics ALLENÒ This means that electron can go the excited state ni = 3. Now it emits photons of maximum wavelength on going to nf = 2 1 = R ì 1 - 1 ü lmax í 22 32 ý î þ 36 36 lmax = 5R = 5´1.1´107 = 6.545 × 10–7m = 6545 A° It belongs to Balmer Series. 3. Calculate the shortest wavelength of the spectral lines emitted in Balmer series. (Refer to NCERT Q. No. 2) [Given Rydberg constant, R = 107 m–1] 4. A 12·5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. www.notesdrive.com Determine the wavelengths and the corresponding series of the lines emitted. node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx Sol. When electron beam of 12.5ev is used to excite hydrogen, then the probable transition is from n=1 to n=3 (i.e. E3–E1 = –1.51–(–13.6) = 12.09eV). On de excitation the electron may jump by following ways 3 ® 1, 3 ® 2, 2 ® 1 Possible wavelength & their corresponding series of lines emitted are - (a) From n=3 to n=1, 1 = R é1 - 1 ù l êë12 32 úû 1 = R êéë1 - 1 ù = R æèç 8 ö l 9 ûú 9 ø÷ or l= 9 = 1.09×10–7m = 109 nm 8R It belongs to Balmer series. (b) From n = 3 to n= 2, 1 = R é 1 - 1 ù = R é 1 - 1 ù l êë 22 32 úû êë 4 9 úû 1 = R é 5 ù l êë 36 ûú l= 36 = 6.545 × 10–7m = 6545Å 5R It belongs to Lyman series. (c) From n=2 to n=1, 1 = R é1 - 1 ù = R êëé1 - 1 ù l ëê12 22 ûú 4 ûú 1 3R = l 4 l = 1.29×10–7m = 129nm E

ALLENÒ CBSE 89 5. When is Ha line in the emission spectrum of hydrogen atom for balmer series obtained ? Calculate the frequency of the photon emitted during this transition. Sol. Ha is a specific deep-red visible spectral line in the Balmer series with a wavelength of 656.28 nm, it occurs when a hydrogen electron transits from its 3rd to 2nd lowest energy level. This transition produces H-alpha photon & the 1st line of Balmer series. 1 = R é 1 - 1 ù = R é 1 - 1 ù l ê nf2 ú ëê 22 32 úû ë n 2 û i 1 = 1.097 ´107 é5 ù l ëê 36 úû n = c = 3´108 ´1.097´107 ´ 5 l 36 node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx = 4.57 ´1014 Hz www.notesdrive.com 6. Calculate the wavelength of radiation emitted when electron in a hydrogen atom jumps from n = ¥ to n = 1. Sol. Using, 1 = R é 1 - 1 ù = R é1 - 1ù = R l ê n2f ú ëê12 ¥ ûú ë n 2 û i 11 = 0.9115 × 10–7 m = 911 nm l = R = 1.097´107 m–1 7. Given the value of the ground state energy of hydrogen atom as –13.6 eV, find out its kinetic and potential energy in the ground and second excited states. Sol. Ground state energy of hydrogen atom as –13.6 eV, En = Total energy = 13.6 eV - n2 Rhc P.E. = = -2Rhc , T.E. = – Rhc = -Rhc = -13.6 Þ Rhc = 13.6 eV K.E. = n2 , n2 n2 (1)2 R = 1.097 ´107 m-1(Rydberg's constant) h = 6.6 ´10-34 J - s (Planck's constant) c = 3´108 m / s Þ In ground state, (n = 1) K.E. = Rhc = Rhc n2 & P.E. = –2Rhc Þ In second excited state, (n = 3) K.E. = Rhc Þ 13.6 = 1.51eV & P.E. = -2Rhc Þ -2 ´13.6 = -3.02eV 9 9 9 9 E

90 Physics ALLENÒ 8. Show that the radius of the orbit in hydrogen atom varies as n2, where n is the principal quantum number of the atom. (Refer to NCERT Q. No. 7) 9. Given the ground state energy E0 = –13.6 eV and Bohr radius r0 = 0.53 Å. Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state. Sol. Given ground state energy E0 = –13.6 eV Energy in the first excited state = E1 = -13.6 eV = –3.4 eV (2)2 \\ l= h = 6.63 ´ 10 –34 ( )( )2mE1 2 ´ 9.1´10–31 3.4 ´1.6 ´10–19 6.63 ´10 –34 9.95 ´ 10 –25 l = www.notesdrive.com=6.6´10-10 m 10. Using Rutherford model of the atom, derive the expression for the total energy of an electron in node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx hydrogen atom. What is the significance of total negative energy possessed by the electron ? (Refer to NCERT Q. No. 8) OR Using Bohr’s postulates of the atomic model, derive the expression for radius of nth electron orbit. Hence obtain the expression for Bohr’s radius. (Refer to NCERT Q. No. 7) 11. Define ionization energy. How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge ? Sol. Ionisation energy is the energy that an isolated, gaseous atom in the ground state must absorb to discharge an electron. Ionisation energy of hydrogen atom is, EH µ me So, due to this replacement of e–, E¢H = 200 me i.e. E¢H = 200 EH = 200 × (–13.6) = –2720 eV 12. In the ground state of hydrogen atom, its bohr radius is given as 5.3 × 10–11 m. The atom is excited such that the radius becomes 21.2 × 10–11 m. Find (i) the value of the principal quantum number and (ii) the total energy of the atom in this excited state Sol. Given, r1 = 5.3 × 10–11 m, r2 = 21.2 × 10–11 m, n1 = 1 We know that, r µ n2 r1 = n12 r2 n 2 2 1 = 5.3 ´10 –11 Þ n22 = 4 Þ n2 = 2 Þ n22 21.2 ´10–11 Now, from E = - 13.6 = -13.6 = –3.4 eV n2 4 E

ALLENÒ CBSE 91 13. Use Bohr model of hydrogen atom to calculate the speed of the electron in the first excited state. Sol. From Bohr’s first postulate, mv2n = kZe2 rn rn2 From Bohr’s second postulate, mvnrn = nh/2p On solving we get rn = n2h2 = 0.53 n2 Å 4p2kme2Z Z (radius) Here, k = 1 = 9 ´109 Nm 2 - C2 4pe0 2p2 kme2 Z æ Z ö h = Plank 's constant vn = nh = çè 2.165 ´106 n ø÷ m / s (velocity) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx6.6 ×10–34J - s www.notesdrive.comNow for 1st excited state (n = 2), v2 = 2.165 × 106 × 1 = 1.08 × 106 m/s 2 14. When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de Broglie wavelength associated with the electron change ? Justify your answer. Sol. We know, l = h = h p 2mk So, l1 = k4 l4 k1 but kn = (En ) µ -1 n2 Hence, l1 = 1 l4 16 \\ l1 = l4 4 15. The energy levels of an atom are given below in the diagram. 0 eV –1 eV A B C D E –3 eV –10 eV Which of the transitions belong to Lyman and Balmer series ? Calculate the ratio of the shortest wavelengths of the Lyman and the Balmer series of the spectra. E

92 Physics ALLENÒ Sol. 0 eV A B CDE –1 eV –3 eV –10 eV Lyman Series – C & E Balmer Series – B & D using 1 = R é 1 - 1ù lwww.notesdrive.comên12 ú ë n 2 û node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx2 Ratio of shortest wavelength lL = n12 = 1 lB 4 n 2 2 16. The energy levels of a hypothetical atom are shown below. Which of the shown transitions will result in the emission of photon of wavelength 275 nm ? 0 eV AB –2 eV C –4.5 eV D –10 eV Sol. Using relation, E = 1242 eV l(in nm) Here, l = 275 nm, E = 1242 = 4.5eV 275 This energy of photon exist corresponding to ‘B’. 17. State Bohr’s postulate of hydrogen atom which successfully explains the emission lines in the spectrum of hydrogen atom. Use Rydberg formula to determine the wavelength of Ha line. [Given : Rydberg constant R = 1.03 × 107 m–1] Sol. Acc. to Bohr’s postulate when an e– jumps from one orbit to another, the energy diff. between them is emitted in the form of energy i.e. as photon. Ei – Ef = hn = hc l E

ALLENÒ CBSE 93 Now, Rydberg formula for the Balmer series is 1 = R æ 1 - 1 ö where R = Rydberg constant l ç ÷ è n 2 n 2 ø f i = 1.09 × 107 m–1 The Ha – line of the Balmer series is obtained when an e– jumps to the second orbit (nf = 2) from the third orbit (ni = 3). Further, 1 = 1.03 ´107 æ 1 - 1 ö l èç 22 32 ÷ø = 6.99 × 10–7 = 699 nm ‘l’ lies in the visible region. 18. How does one explain, Bohr’s second postulate of quantization of orbital angular momentum node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx using de Broglie hypothesis ? www.notesdrive.com Sol. Behavior of particle waves can be viewed analogous to the waves travelling on a string. Particle waves can lead to standing waves held under resonant conditions. When a stationary string is plucked, a number of wavelengths are excited but only those wavelengths survive which form a standing wave in the string. Thus, in a string standing waves are formed only when the total distance travelled by a wave is an integral number of wavelengths. Hence, for any electron moving in nth circular orbit of radius rn, the total distance is equal to the circumference of the orbit, 2prn. l Nucleus 2prn = nl .......(1) here, l is de Broglie wavelength. We know, l = h/p or l = h/mvn ........(2) Where, mvn is the momentum of electron revolving in the nth orbit. From equation (2) & equation (1) we get, 2prn = nh/mvn L = mvnrn = nh/2p Hence, de Broglie hypothesis successfully proves Bohr’s second postulate. E

94 Physics ALLENÒ 19. (i) State Bohr postulate of hydrogen atom that gives the relationship for the frequency of emitted photon in a transition. (ii) An electron jumps from fourth to first orbit in an atom. How many maximum number of spectral lines can be emitted by the atom ? To which series these lines correspond ? Sol. (i) Acc. to Bohr’s theory, energy is quantised i.e. for each orbital the corresponding energy is given as – DE = hn (ii) There will be 6 spectral lines & they correspond to . A- Lymanserieswww.notesdrive.com B- Balmer series node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx 20. Calculate the energy in fusion reaction : 2 H + 2 H ® 3 He + n where BE of 2 H = 2.23 MeV and of 3 H e = 7.73 MeV. 1 1 2 1 2 Sol. B.E. of 2 H = 2.23 MeV 1 Hence total intial B.E. = 2.23 + 2.23 = 4.46 MeV Therefore, Energy released, DE = 7.73 – 4.46 = 3.27 MeV EXERCISE-II THREE MARK QUESTIONS : 1. Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius r with orbital speed v is equal to evr/2. Hence using Bohr's postulate of quantization of angular momentum, deduce the expression for the magnetic moment of hydrogen atom in the ground state. Sol. Here we consider an e– of mass me which is revolving with velocity v on a circular path of radius r. \\ I = e = e = ev ...(1) T 2pr 2pr v Q m = IA \\ m= ev ´ pr 2 2pr m= evr ...(2) 2 Now, m= evr ´ me = eL (Q mevr = L) 2 me 2m e or mr = - eLr ...(3) 2me E

ALLENÒ CBSE 95 From m= eL = e(nh) çèæQ L = m evr = nh ö 2me 2me (2p) 2p ÷ø m = n eh 4pm e for H-atom in n = 1 (ground state) m = eh = 9.27 × 10–24J/T 4pm e 2. (i) State Bohr’s quantization condition for defining stationary orbits. How does de Broglie hypothesis explain the stationary orbits ? (ii) Find the relation between the three wavelength l1, l2 and l3 from the energy level diagram shown below. C node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx l1 l3 www.notesdrive.com B Sol. (i) l2 (ii) A Bohr’s quantization condition : The electron can revolve round the nucleus only in those circular orbits in which angular momentum of an electron is integral multiple of h , where 2p h is Planck’s constant. mvr = n h [m = mass, v = velocity of electron] 2p There circular orbits are stationary orbits. de Broglie hypothesis : de-Broglie interpret Bohr’s 2nd postulate in terms of wave nature of electron. According to him – The electron can revolve in certain stable orbits for which the angular momentum is some integral multiple of h/2p. Mathematically, 2prn = nh mv therefore, mvrn = nh 2p ECB = hc Þ EBA = hc Þ ECA = hc l1 l2 l3 here ECB = Energy gap between level B and C ECA = Energy gap between level A and C EBA = Energy gap between level A and B ECA = ECB + EBA Þ hc hc hc 1 1 1 Þ l3 = l1l2 =+Þ =+ l1 + l2 l3 l1 l2 l3 l1 l2 E

96 Physics ALLENÒ 3. A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon. Sol. Energy in the n = 4 level = -E0 = - E0 42 16 \\ Energy required to take the electron from the ground state, to the n = 4 level = æ - E0 ö - ( -E0 ) = Eo æèç1 - 1 ö = 15 Eo çè 16 ø÷ 16 ø÷ 16 = 15 × 13.6 × 1.6 × 10–19 J 16 Let the frequency of the photon be u, we have hu = 15 × 13.6 × 1.6 × 10–19 16 www.notesdrive.com \\ u = 15´13.6 ´1.6 ´10-19 Hz ; 3.1´1015 Hz node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx16 ´ 6.63´10-34 (Also accept 3 × 1015 Hz) 4. (a) Write the basic nuclear process involved in the emission of b+ in a symbolic form, by a radioactive nucleus. (b) In the reactions given below : (i) 161C ® z B + x + u (ii) 12 C + 12 C ® 20 Ne + c He y 6 6 a b Find the value of x, y and z and a, b and c Sol. (a) ZXA ¾b¾+ ® Z–1YA + +1e0 + n + energy In b+, mass no. (A) remains same but atomic no. (Z) decreases by 1 unit. (b) x ¾® +1e0, y ¾® 5, z ¾® 11 a ¾® 10, b ¾® 2, c ¾® 4 5. According to the third postulates of Bohr's model, when an atom makes a transition from the higher energy state with quantum number ni to the lower energy state with quantum number nf(nf < ni), the difference of the energy is carried away by the photon of frequency such that hn = Eni - Enf Since both ni and nf are integers, this immediately shows that in transitions between the different atomic levels, light is radiated in various discrete frequencies. For hydrogen atom spectrum, the Balmer formula corresponds to nf = 2 and ni = 3, 4, 5 etc. This result of the Bohr's model suggested the presence of other series spectra for hydrogen atom - those corresponding to the transitions resulting from nf = 1 and ni = 2, 3 etc ; and nf = 3 and ni = 4, 5 etc. and so on. Such series were identified in the course of spectroscopic investigations and are known as Lyman, Balmer, Paschen, Brackett and Pfund-series. The electronic transitions corresponding to this series are shown E

ALLENÒ CBSE 97 n=¥ n=6 3.4 eV Humphrey (Infrared)1.0 eV1.5 eV n=5 Pfund 0.65 eV (Infrared) 1.9 eV n=4 Bracket (Infrared) n=3 Paschen (Infrared) n=2 Balmer (visible) n=1 102eVwww.notesdrive.com (i) The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are respectively (A) 3.4 eV, 3.4 eV (B) –3.4 eV, –3.4 eV (C) –3.4 eV, –6.8 eV (D) 3.4 eV, –6.8 eV Ans. (D) (ii) Given the value of Rydberg constant is 107 m–1, the wave number of the last line of the Balmer series in hydrogen spectrum will be (A) 0.5 × 107 m–1 (B) 0.25 × 107 m–1 (C) 2.5 × 107 m–1 (D) 0.025 × 104 m–1 Ans. (B) (iii) The ratio of wavelength of last line of Balmer series and the last line of Lyman series (A) 0.5 (B) 2 (C) 1 (D) 4 Ans. (D) (iv) The wavelength of Balmer series lies in node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx (A) ultraviolet region (B) infrared region (C) far infra-red region (D) visible region Ans. (D) (v) In the empirical formula for the observed wavelength (l) for hydrogen is 1 = R é 1 - 1 ù where l êë 42 n2 úû n is integral values higher than 4 then it represents ----------- series. (A) Balmer series (B) Brackett series (C) Pfund series (D) Lyman series Ans. (B) E

98 Physics ALLENÒ 6. Neutrons and protons are identical particle in the sense that their masses are nearly the same and the force called nuclear force, does not distinguish them. Nuclear force is the strongest force. Stability of nucleus is determined by the neutron Proton ratio or mass defect or packing fraction. Shape of nucleus is calculated by quadruple moment and the spin of nucleus depends on even or odd mass number. Volume of nucleus depends on the mass number. Whole mass of the atom (nearly 99%) is centered at the nucleus. (i) The correct statement about the nuclear force is are : (A) Charge independent (B) Short range force (C) Non conservative force (D) all of these Ans. (D) (ii) The range of nuclear force is the order of : (A) 2 × 10–10 m (B) 1.5 × 10–20 m (C) 1.2 × 10–4 m (D) 1.4 × 10–15 m www.notesdrive.com Ans. (D) node06\\B0BB-BC\\Kota\\Board Material\\Physics\\CBSE\\CBSE Booklet\\Part-2\\ 03_Unit-8.docx (iii) A force between two protons is same as the force between proton and neutron. The nature of the force is : (A) electrical force (B) weak nuclear force (C) gravitational force (D) strong nuclear force Ans. (D) (iv) All the nucleons in an atom are held by : (A) Nuclear forces (B) Vander Waal's forces (C) Tensor forces (D) Coulomb forces Ans. (A) EXERCISE - V (RACE) CHAPTER-12 : ATOMS For question numbers 1 to 5 two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is NOT the correct explanation of A (c) A is true but R is false (d) A is false and R is also false 1. Assertion : The mass of the atom is concentrated in the nucleus. Reason : The mass of a nucleus can be either less than or more than the sum of the masses of nucleons present in it. (i) a (ii) b (iii) c (iv) d 2. Assertion : Bohor's orbits are also called stationary states. Reason : In Bohr's orbits electron revolves in fixed path. (i) a (ii) b (iii) c (iv) d E