JEE-Physics From Bohr Model O n = 1, n = 2, 3, 4.......K series N4 12 N N N series n = 2, n = 3, 4, 5.......L series M MMM M series 3 12 L 2 n = 3, n = 4, 5, 6.......M series L L L L series 12 K 1 First line of series = KK K K Second line of series = K series Third line of series = Transit ion Wave– E n er g y E n er g y Wavelength length difference hc K = (E K E L ) LK K hK –(E –E ) K KL 3 12400 Characteristic X-ray = eVÅ (2 1) = hK (EK EL ) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 2 MK K hK –(EK–EM) hc K X-ray from a molybdenum Relative intensityK = (E K E M ) target at 35 kV 1 Bremsstrahlung continuum L L (3 1) = h 12400 K = eVÅ (EK EM ) ML hL –(E –E ) hc 0.02 0.04 0.06 0.08 0.10 0.12 L LM L = (E L E M ) wavelength (nm) (3 2) = hL 12400 = eVÅ (EL EM ) MOSELEY'S LAW Moseley studied the characteristic spectrum of number of many elements and observed that the square root of the frequency of a K– line is closely proportional to atomic number of the element. This is called Moseley's law. (Z – b) (Z – b)2 = a (Z – b)2 ...(i) Z = atomic number of target K = frequency of characteristic spectrum K b = screening constant (for K– series b=1, L series b=7.4) a = proportionality constant 1 1 From Bohr Model = RcZ2 ...(ii) n 2 n 2 1 2 1 1 Z Comparing (i) and (ii) a = Rc n 2 n 2 1 2 E 13
JEE-Physics • Thus proportionality constant 'a' does not depend on the nature of target but depend on transition. Bohr model Moseley's correction 1 . For single electron species 1 . For many electron species 1 1 1 1 2. E = 13. 6Z 2 eV 2. E = 13.6 (Z–1)2 eV n 2 n 2 n 2 n 2 1 2 1 2 1 1 1 1 3. = RcZ2 3. = Rc(Z–1)2 n 2 n 2 n 2 n 2 1 2 1 2 1 1 1 1 1 1 4. = RZ2 n 2 n 2 4. = R (Z – 1)2 n 2 n 2 1 2 1 2 • For X–ray production, Moseley formulae are used because heavy metal are used. When target is same 1 When transition is same 1 (Z b)2 11 n12 n 2 2 ABSORPTION OF X–R AY When X–ray passes through x thickness then its intensity I = I e–x 0 I = Intensity of incident X–ray 0 I = Intensity of X–ray after passing through x distance = absorption coefficient of material I0 • Intensity of X–ray decrease exponentially. I • Maximum absorption of X–ray Lead x • Minimum absorption of X–ray Air Half thickness (x ) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 1/2 The distance travelled by X–ray when intensity become half the original value x = n2 1/2 Example When X–rays of wavelength 0.5Å pass through 10 mm thick Al sheet then their intensity is reduced to one sixth. Find the absorption coefficient for Aluminium . Solution 2.303 lo g Io 2.303 log10 6 2.303 0.7781 0.1752 / mm x I 10 10 14 E
JEE-Physics DIFFR ACTION OF X–R AY Diffraction of X–ray is possible by crystals because the interatomic spacing in a crystal lattice is order of wavelength of X–rays it was first verified by Lauve. Diffraction of X–ray take place according to Bragg's law 2d sin = n d = spacing of crystal plane or lattice constant or distance between adjacent atomic plane = Bragg's angle or glancing angle = Diffracting angle n = 1, 2, 3 ....... d For Maximum Wavelength sin = 1, n = 1 max = 2d so if > 2d diffraction is not possible i.e. solution of Bragg's equation is not possible. PROPERTIES OF X–R AY • X–ray always travel with the velocity of light in straight line because X–rays are em waves • X–ray is electromagnetic radiation it show particle and wave both nature • In reflection, diffraction, interference, refraction X–ray shows wave nature while in photoelectric effect it shows particle nature. • There is no charge on X–ray thus these are not deflected by electric field and magnetic field. • X–ray are invisible. • X–ray affects the photographic plate • When X–ray incidents on the surface of substance it exerts force and pressure and transfer energy and momentum • Characteristic X–ray can not obtained from hydrogen because the difference of energy level in hydrogen is very small. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 Example Show that the frequency of K X–ray of a material is equal to the sum of frequencies of K and L X–rays of the same material. Solution The energy level diagram of an atom with one electron knocked out is shown above. L M K K L K Energy of K X–ray is E =E – E K and of K L and of thus K X–ray is E = E – E K M K E L X–rays is E = E – E L M L E = E + E or K = K + L K K L 15
JEE-Physics PHOTO ELECTRIC EFFECT PHOTOELECTRIC EFFECT It was discovered by Hertz in 1887. He found that when the negative plate of an electric discharge tube was illuminated with ultraviolet light, the electric discharge took place more readily. Further experiments carried out by Hallwachs confirmed that certain negatively charged particles are emitted, when a Zn plate is illuminated with ultraviolet light. These particles were identified as electrons. The phenomenon of emission of electrons from the surface of certain substances, when suitable radiations of certain frequency or wavelength are incident upon it is called photoelectric effect. EXPLANATION OF PHOTOELECTRIC EFFECT • On the basis of wave theory :According to wave theory, light is an electromagnetic radiation consisting of oscillating electric field vectors and magnetic field vectors. When electromagnetic radiations are incident on a metal surface, the free electrons [free electrons means the electrons which are loosely bound and free to move inside the metal] absorb energy from the radiation. This occurs by the oscillations of electron under the action of electric field vector of electromagnetic radiation. When an electron acquires sufficiently high energy so that it can overcome its binding energy, it comes out from the metal. • On the basis of photon theory: According to photon theory of light, light consists of particles (called photons). Each particle carries a certain amount of energy with it. This energy is given by E=h, where h is the Plank's constant and is the frequency. When the photons are incident on a metal surface, they collide with electrons. In some of the collisions, a photon is absorbed by an electron. Thus an electron gets energy h. If this energy is greater than the binding energy of the electron, it comes out of the metal surface. The extra energy given to the electron becomes its kinetic energy. EXPERIMENTS • Hertz Experiment : Hertz observed that when ultraviolet rays are incident on negative plate of electric discharge tube then conduction takes place easily in the tube. ultraviolet rays cathode anode Evacuated quartz tube NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 A • Hallwach experiment : Hallwach observed that if negatively charged Zn plate is illuminated by U.V. light, its negative charge decreases and it becomes neutral and after some time it gains positive charge. It means in the effect of light, some negative charged particles are emitted from the metal. • Lenard Explanation : He told that when ultraviolet rays are incident on cathode, electrons are ejected. These electrons are attracted by anode and circuit is completed due to flow of electrons and current flows. When U.V. rays incident on anode, electrons are ejected but current does not flow. For the photo electric effect the light of short wavelength (or high frequency) is more effective than the light of long wavelength (or low frequency) 16 E
JEE-Physics • Experimental study of photoelectric Effect : When light of frequency and intensity I falls on the cathode, electrons are emitted from it. The electrons are collected by the anode and a current flows in the circuit. This current is called photoelectric current. This experiment is used to study the variation of photoelectric current with different factors like intensity, frequency and the potential difference between the anode and cathode. Light of intensity I and frequency Photoelectrons Cathode Anode (Photosensitive metal) A i (Photoelectric current) V Potential Divider ( i ) Variation of photoelectric current with potential difference : With the help of the above experimental setup, a graph is obtained between current and potential difference. The potential difference is varied with the help of a potential divider. The graph obtained is shown below. i NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 V -V0 O Anode Potential The main points of observation are : (a) At zero anode potential, a current exists. It means that electrons are emitted from cathode with some kinetic energy. (b) As anode potential is increased, current increases. This implies that different electrons are emitted with different kinetic energies. (c) After a certain anode potential, current acquires a constant value called saturation current. Current acquires a saturation value because the number of electrons emitted 7 per second from the cathode are fixed. (d) At a certain negative potential, the photoelectric current becomes zero. This is called stopping potential (V ). Stopping potential is a measure of maximum kinetic energy of the emitted electrons. Let KE be 0 max the maximum kinetic energy of an emitted electron, then KE = eV . max 0 i (ii) Variation of current with intensity The photoelectric current is found to be directly proportional to intensity of Intensity incident radiation. (iii) Effect of intensity on saturation current and stopping potential i (a) Saturation current increases with increase in intensity. I2 (I2>I1) I1 (b) Stopping potential (and therefore maximum kinetic energy) is independent V0 Anode Potential V of intensity. 17 E
JEE-Physics (iv) Effect of frequency V0 (stopping potential) (a) Stopping potential is found to vary with frequency of incident light linearly. Greater the frequency of incident light, greater the stopping potential. (b) There exists a certain minimum frequency 0 below which no stopping 0 (frequency) potential is required as no emission of electrons takes place. This frequency is called threshold frequency. For photoelectric emission to take place, > 0. GOLDEN KEY POINTS • Photo electric effect is an instantaneous process, as soon as light is incident on the metal, photo electrons are emitted. • Stopping potential does not depend on the distance between cathode and anode. • The work function represented the energy needed to remove the least tightly bounded electrons from the surface. It depends only on nature of the metal and independent of any other factors. • Failure of wave theory of light (i) According to wave theory when light incident on a surface, energy is distributed continuously over the surface. So that electron has to wait to gain sufficient energy to come out. But in experiment there is no time lag. Emission of electrons takes place in less than 10–9 s. This means, electron does not absorb energy. They get all the energy once. (ii) When intensity is increased, more energetic electrons should be emitted. So that stopping potential should be intensity dependent. But it is not observed. (iii) According to wave theory, if intensity is sufficient then, at each frequency, electron emission is possible. It means there should not be existence of threshold frequency. • Einstein's Explanation of Photoelectric Effect Einstein explained photoelectric effect on the basis of photon–electron interaction. The energy transfer takes place due to collisions between an electrons and a photon. The electrons within the target material are held there by electric force. The electron needs a certain minimum energy to escape from this pull. This minimum energy is the property of target material and it is called the work function. When a photon of energy E=h collides with and transfers its energy to an electron, and this energy is greater than the work function, the electron can escape through the surface. • Einstein's Photoelectric Equation h KEmax NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 Here h is the energy transferred to the electron. Out of this, is the energy needed to escape. The remaining energy appears as kinetic energy of the electron. V0 Now KE = eV (where V is stopping potential) h max 0 0 e slope = h = + eV V = h 0 0 e e Thus, the stopping potential varies linearly with the frequency of incident radiation. h E Slope of the graph obtained is . This graph helps in determination of Planck's constant. e 18
JEE-Physics GOLDEN KEY POINTS • Einstein's Photo Electric equation is based on conservation of energy. • Einstein explained P.E.E. on the basis of quantum theory, for which he was awarded noble prize. • According to Einstein one photon can eject one e– only. But here the energy of incident photon should greater or equal to work function. • In photoelectric effect all photoelectrons do not have same kinetic energy. Their KE range from zero to Emax which depends on frequency of incident radiation and nature of cathode. • The photo electric effect takes place only when photons strike bound electrons because for free electrons energy and momentum conservations do not hold together. Example Calculate the possible velocity of a photoelectron if the work function of the target material is 1.24 eV and wavelength of light is 4.36 × 10–7 m. What retarding potential is necessary to stop the emission of electrons? Solution As KE = h – 1 m v 2 ax h hc max 2 m hc 6 .6 3 1034 3 10 8 4.36 10 7 2 2 1.24 1.6 1 0 19 vmax = = = 7.523 × 105 m/s m 9.11 1031 The speed of a photoelectron can be any value between 0 and 7.43 × 105 m/s If V is the stopping potential, then eV = 1 m v 2 ax 00 2 m V= 1 m v 2 ax hc = 12400 – 1.24 = 1.60 V hc 12400 10 10 V m 0 m e 2 e e e 4360 Example The surface of a metal of work function is illuminated by light whose electric field component varies with time NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 as E = E [ 1 + cos t] sin0t. Find the maximum kinetic energy of photoelectrons emitted from the surface. 0 Solution The given electric field component is E=E sin t + E sin 0t cos t = E sin 0t + E0 [ sin (0 + ) t + sin(0–)t] 0 0 0 2 The given light comprises three different frequencies viz. , 0 + 0 – The maximum kinetic energy will be due to most energetic photon. KE = h– = h 0 – 2 or = max 2 2 E 19
JEE-Physics Example When light of wavelength is incident on a metal surface, stopping potential is found to be x. When light of x wavelength n is incident on the same metal surface, stopping potential is found to be n 1 . Find the threshold wavelength of the metal. Solution Let 0 is the threshold wavelength. The work function is hc . 0 hc hc ex hc hc =– Now, by photoelectric equation ex = 0 ...(i) ...(ii) n 1 n 0 From (i) and (ii) hc hc n 1 hc n 1 hc nhc hc = n2 n 0 0 n 0 PHOTON THEORY OF LIGHT • A photon is a particle of light moving with speed 299792458 m/s in vacuum. • The speed of a photon is independent of frame of reference. This is the basic postulate of theory of relativity. • The rest mass of a photon is zero. i.e. photons do not exist at rest. E hc h 1 • Effective mass of photon m = = = i.e. m c2 c2 c So mass of violet light photon is greater than the mass of red light photon. (R > V) • According to Planck the energy of a photon is directly proportional to the frequency of the radiation. E or E = h E hc joule ( c = ) or hc 12400 eV – Å hc 12400( eV ) E= e = e A Here E = energy of photon, c = speed of light, h = Planck's constant, e = charge of electron h = 6.62 × 10–34 J–s, = frequency of photon, = wavelength of photon • Linear momentum of photon p = E h h NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 c c • A photon can collide with material particles like electron. During these collisions, the total energy and total momentum remain constant. • Energy of light passing through per unit area per unit time is known as intensity of light. Intensity of light I E P ...(i) At A Here P = power of source, A = Area, t = time taken N = number of photon incident in t time E = energy incident in t time = Nh Intensity N(h) n(h) nN no. of photon per sec. I = = ...(ii) t At A P n(h) P P From equation (i) and (ii), = n = h = hc = 5 × 1024 J–1 m–1 × P × A A • When photons fall on a surface, they exert a force and pressure on the surface. This pressure is called radiation pressure. 20 E
JEE-Physics • Force exerted on perfectly reflecting surface Let 'N' photons are there in time t, Nh incident photon Momentum before striking the surface (p )= p1 = h 1 Nh p= h Momentum after striking the surface (p )= 2 2 reflected photon Change in momentum of photons = p – p = 2 N h 21 But change in momentum of surface = p = 2Nh 2Nh n 2h but n= P ; So that force on surface F = t hc F = 2h P 2P and Pressure = F 2P 2I I P hc c A cA c A incident photon • Force exerted on perfectly absorbing surface p1= h Nh no reflected 0 Nh photon p2= 0 F = p1 p2 = t t h ; F= P n P and Pressure = F P I t = n c hc A Ac c • When a beam of light is incident at angle on perfectly reflector surface pihnoctiodnent refplehcotteodn F= 2P n 2h cos 2IA cos Pressure = F 2I cos c cos = c A c Example The intensity of sunlight on the surface of earth is 1400 W/m2. Assuming the mean wavelength of sunlight to be 6000 Å, calculate:– (a) The photon flux arriving at 1 m2 area on earth perpendicular to light radiations and (b) The number of photons emitted from the sun per second (Assuming the average radius of Earth's orbit to be 1.49 × 1011 m) Solution hc 12400 (a) Energy of a photon E = =2.06 eV = 3.3 × 10–19 J 6000 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 IA 1400 1 Photon flux = E = 3.3 1019 = 4.22 × 1021 photons/sec. (b) P IA 1400 4 (1.49 1011 )2 = 1.18 × 1045 Number of photons emitted per second n = = 3.3 10 19 EE Example In a photoelectric setup, a point source of light of power 3.2 × 10–3 W emits monochromatic photons of energy 5.0 eV. The source is located at a distance 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and radius 8 × 10–3 m. The efficiency of photoelectron emission is one for every 106 incident photons. Assuming that the sphere is isolated and initially neutral and that photoelectrons are instantly swept away after emission, Find (i) the number of photoelectrons emitted per second. (ii) the time t after light source is switched on, at which photoelectron emission stops. E 21
JEE-Physics Solution Energy of a single photon E=5.0 eV = 8 × 10–19 J Power of source P = 3.2 × 10–3 W P 3.2 10 3 number of photons emitted per second n = 8 10 19 = 4× 1015/s E The number of photons incident per second on metal surface is n0 = n r2 4R 2 4 1015 2 4 0.8 2 n = 0 8 10 3 1.0 1011 photon/s 1.0 1011 Source R=0.8m r=0.8× 10–3m Number of electrons emitted = 106 = 105 /s KEmax = h– = 5.0 –3.0 = 2.0 eV The photoelectron emission stops, when the metallic sphere acquires stopping potential. q As KE = 2.0 eV Stopping potential V = 2V 2 = 4 0 r q = 1.78 × 10–12 C max 0 1.78 10 12 Now charge q = (number of electrons/second) × t× e t = 105 1.6 1019 = 111s PHOTO CELL A photo cell is a practical application of the phenomenon of photo electric effect, with the help of photo cell light energy is converted into electrical energy. • Construction : A photo cell consists of an evacuated sealed glass tube containing anode and a concave cathode of suitable emitting material such as Cesium (Cs). • Working: When light of frequency greater than the threshold frequency of cathode material falls on the cathode, photoelectrons emitted are collected by the anode and an electric current starts flowing in the external circuit. The current increase with the increase in the intensity of light. The current would stop, if the light does not fall on the cathode. mA + anode light e– e– e– e– NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 glass tube e– cathode - key • Application (i) In television camera. (ii) In automatic door (iii) Burglar’s alarm (iv) Automatic switching of street light and traffic signals. 22 E
JEE-Physics MATTER WAVES THEORY DUAL NATURE OF LIGHT Experimental phenomena of light reflection, refraction, interference, diffraction are explained only on the basis of wave theory of light. These phenomena verify the wave nature of light. Experimental phenomena of light photoelectric effect and Crompton effect, pair production and positron inhalational can be explained only on the basis of the particle nature of light. These phenomena verify the particle nature of light. It is inferred that light does not have any definite nature, rather its nature depends on its experimental phenomenon. This is known as the dual nature of light. The wave nature and particle nature both can not be possible simultaneously. De-Broglie HYPOTHESIS De Broglie imagined that as light possess both wave and particle nature, similarly matter must also posses both nature, particle as well as wave. De Broglie imagined that despite particle nature of matter, waves must also be associated with material particles. Wave associated with material particles, are defined as matter waves. • De Broglie wavelength associated with moving particles If a particle of mass m moving with velocity v Kinetic energy of the particle E 1 mv2 p2 momentum of particle p = mv = 2mE the wave length 2 2m hh h 1 1 1 associated with the particles is p v E p mv 2mE The order of magnitude of wave lengths associated with macroscopic particles is 10–24 Å. The smallest wavelength whose measurement is possible is that of – rays ( 10–5 Å). This is the reason why the wave nature of macroscopic particles is not observable. The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron, – particle, atom, molecule etc. is of the order of 10–10 m, it is equal to the wavelength of X–rays, which is within the limit of measurement. Hence the wave nature of these particles is observable. • De Broglie wavelength associated with the charged particles Let a charged particle having charge q is accelerated by potential difference V. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 Kinetic energy of this particle E 1 mv2 qV Momentum of particle p mv 2mE = 2mqV 2 hh h The De Broglie wavelength associated with charged particle p 2mE 2mqV • For an Electron m = 9.1 × 10–31 kg, q = 1.6 × 10–19 C, h = 6.62 × 10–34 J–s e E De Broglie wavelength associated with electron 6.62 1034 2 9.1 1031 1.6 1019 V 12.27 1010 meter 12.27 so 1 A VV V Potential difference required to stop an electron of wavelength is V = 150.6 volt (Å)2 2 23
JEE-Physics mp = 1.67 x 10–27 kg De Broglie wavelength associated with proton • For Proton p = 6.62 1034 ; p = 0.286 10 10 0.286 Å 2 1.67 1027 1.6 10 19 V meter V V • For Deuteron m = 2 × 1.67 × 10–27 kg, q = 1.6 × 10–19 C dd d = 6.62 1034 = 0.202 A 2 2 1.67 1027 1.6 1019 V V • Fo r Parti cles q = 2 × 1.6 × 10–19 C, m = 4 × 1.67 × 10–27 kg 6.62 1034 0.101 = A 2 4 1.67 1027 2 1.6 1019 V V DE BROGLIE WAVELENGTH ASSOCIATED WITH UNCHARGED PARTICLES • Kinetic energy of uncharged particle E 1 mv2 p2 2 2m m = mass of particle, v = velocity of particle, p = momentum of particle. • Velocity of uncharged par ticle v 2E m • Momentum of particle p mv 2mE hh h wavelength associated with the particle p mv 2mE h2 h2 Kinetic energy of the particle in terms of its wavelength E eV 2m2 2m2 1.6 1019 For a neutron mn = 1.67 × 10–27 kg 6.62 1034 0.286 1010 meter eV 0.286 eV 2 1.67 1027 E A EE EXPLANATION OF BOHR QUANTIZATION CONDITION NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 According to De Broglie electron revolves round the nucleus in the form of stationary 6th Bohr orbit waves (i. e. wave packet) in the similar fashion as stationary waves in a vibrating string. Electron can stay in those circular orbits whose circumference is an integral multiple of De–Broglie wavelength associated with the electron, 2r = n h and 2r = n mvr nh mv 2 This is the Bohr quantizations condition. equivalent straightened orbit 24 E
JEE-Physics Example Find the initial momentum of electron if the momentum of electron is changed by pm and the De Broglie wavelength associated with it changes by 0.50 % Solution d 100 0.5 d 0.5 1 and p = p 100 200 m p h , differentiating dp h h 1 = p | dp| d pm 1 p = 200 p d = – 2 =– × p p 200 m Example An –particle moves in circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. Find the De Broglie wavelength associated with the particle. Solution hh 6.62 1034 mv 2 r qvB = p = qBr = 2 1.6 1019 0.25 83 104 meter = 0.01 Å Example A proton and an –particle are accelerated through same potential difference. Find the ratio of their de- Broglie wavelength. Solution hh h E qV mv 2mE 2mqV For proton mp = m, q = e For –particle m =4 m, q = 2e, mpqp 1 p mq 2 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 Example A particle of mass m is confined to a narrow tube of length L. (a) Find the wavelengths of the de–Broglie wave which will resonate in the tube. (b) Calculate the corresponding particle momenta, and (c) Calculate the corresponding energies. Solution (a) The de–Broglie waves will resonate with a node at each end of the tube. 2L n=1,2,3..... N A N Few of the possible resonance forms are as follows : n n , L=/2 (b) N A N AN Since de–Broglie wavelengths are n h L=2(/2) pn p= h nh n 1,2,3.... NN n n , NA A AN 2L L=3( /2) (c) The kinetic energy of the particles are K= p 2 n2h2 , n = 1, 2, 3,.... n n 8 L2 m 2m E 25
JEE-Physics NUCLEAR PHYSICS ATOMIC NUCLEUS The atomic nucleus consists of two types of elementary particles, viz. protons and neutrons. These particles are called nucleons. The proton (denoted by p) has a charge +e and a mass m = 1.6726 × 10–27 kg, which is p approximately 1840 times larger than the electron mass. The proton is the nucleus of the simplest atom with Z=1, viz the hydrogen atom. The neutron (denoted by n) is an electrically neutral particle (its charge is zero). The neutron mass is 1.6749 × 10–27 kg. The fact that the mass of a neutron exceeds the mass of a proton by about 2.5 times the electronic masses is of essential importance. It follows from this that the neutron in free state (outside the nucleus) is unstable (radioactive). With half life equal to 12 min, the neutron spontaneously transforms into a proton by emitting an electron (e–) and a particle called the antineutrino . This process can be schematically written as follows : n1 p1 + e0 + 0 1 –1 The most important characteristics of the nucleus are the charge number Z (coinciding with atomic number of the element) and mass number A. The charge number Z is equal to the number of protons in the nucleus, and hence it determines the nuclear charge equal to Ze. The mass number A is equal to the number of nucleons in the nucleus (i.e., to the total number of protons and neutrons). Nuclei are symbolically designated as X A or XA ZZ where X stands for the symbol of a chemical element. For example, the nucleus of the oxygen atom is symbolically written as O186 or 8O16 . The shape of nucleus is approximately spherical and its radius is approximately related to the mass number by R = 1.2 A1/3 × 10–15 m = 1.2 × 10–15 × A1/3 m Most of the chemical elements have several types of atoms differing in the number of neutrons in their nuclei. These varieties are called isotopes. For example carbon has three isotopes C12, C13, C14. In addition to stable 666 isotopes, there also exist unstable (radioactive) isotopes. Atomic masses are specified in terms of the atomic mass unit or unified mass unit (u). The mass of a neutral atom of the carbon C12 is defined to be exactly 12 u. 6 1u = 1.66056 × 10–27 kg = 931.5 MeV. BINDING ENERGY The rest mass of the nucleus is smaller than the sum of the rest masses of nucleons constituting it. This is due to the fact that when nucleons combine to form a nucleus, some energy (binding energy) is liberated. The binding energy is equal to the work that must be done to split the nucleus into the particles constituting it. The difference between the total mass of the nucleons and mass of the nucleus is called the mass defect of the nucleus represented by m = [Zm + (A–Z)m ] – m p n nuc Multiplying the mass defect by the square of the velocity of light, we can find the binding energy of the nucleus. BE = mc2 = [(Zm + (A–Z)m )–m ]c2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 p n nuc If the masses are taken in atomic mass unit, the binding energy is given by BE = [(Zm + (A–Z)m )–m ] 931.5 MeV p n nuc Let us take example of oxygen nucleus. It contains 8 protons and 8 neutrons. We can discuss concept of binding energy by following diagram. 8 protons mrealesassdeescerneearsgeys + 8 neutrons ambsaosrsbisnecrneearsgeys Nucleus of oxygen 8m + 8m > mass of nucleus of oxygen pn 26 E
JEE-Physics For nucleus we apply mass energy conservation, 8m + 8m = mass of nucleus + B.E. pn c2 For general nucleus A X , mass defect = difference between total mass of nucleons and mass of the nucleus Z m = [Zm + (A–Z)m ]–M pn B.E. = mc2 (joules) = (m)in amu × 931.5 MeV Binding Energy per Nucleon Stability of a nucleus does not depend upon binding energy of a nucleus but it depends upon binding energy per B.E. B.E. nucleon B.E./nucleon = Stability mass number A 8.8MeV NiB.E./nucleon (MeV) C N Fe He Li H A (mass number) (i) B.E./A is maximum for A =62 (Ni), It is 8.79460 ± 0.00003 MeV/nucleon, means most stable nuclei are in the region of A=62. (ii) Heavy nuclei achieve stability by breaking into two smaller nuclei and this reaction is called fission reaction. small mass numbers large mass number (iii) Nuclei achieve stability by combining and resulting into heavy nucleus and this reaction is called fusion reaction. small mass numbers large mass number (iv) In both reactions products are more stable in comparison to reactants and Q value is positive. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 NUCLEAR COLLISIONS We can represent a nuclear collision or reaction by the following notation, which means X (a,b) Y a X Y +b (bombarding particle) (at rest) We can apply : (i) Conservation of momentum (ii) Conservation of charge (iii) Conservation of mass–energy For any nuclear reaction a X Y+b K1 K2 K3 K4 By mass energy conservation (i) K + K + (m + m )c2 = K + K + (m + m )c2 12 x 34 Yb E 27
JEE-Physics (ii) Energy released in any nuclear reaction or collision is called Q value of the reaction (iii) Q = (K + K) – (K + K) = KP –KR = (mR – mP)c2 3 4 1 2 (iv) If Q is positive, energy is released and products are more stable in comparison to reactants. (v) If Q is negative, energy is absorbed and products are less stable in comparison to reactants. Q = (B.E.)product – (B.E.)reactants Example Let us find the Q value of fusion reaction B.E. B.E. 4He + 4He 8Be, if A of He = X and A of Be = Y Q = 8Y – 8X Q value for decay XA YA–4 + He4 Q = K+ K ....(i) Y ...(ii) Z Z–2 2 Momentum conservation, p =p Y p2 K= 2m p2 4 4K K= Y A 4 A 2m4 Q K 4K A K A 4 A 4 A 4 K A Q For decay A > 210 which means maximum part of released energy is associated with K.E. of . If Q is negative, the reaction is endoergic. The minimum amount of energy that a bombarding particle must have in order to initiate an endoergic reaction is called Threshold energy E , given by th E = –Q m1 where m = mass of the projectile. th m2 1 1 E = minimum kinetic energy of the projectile to initiate the nuclear reaction th m = mass of the target 2 Example How much energy must a bombarding proton possess to cause the reaction Li7 + H1 Be7 + n1 3 1 4 0 (Mass of Li7 atom is 7.01600, mass of H1 atom is 1.0783, mass of Be7 atom is 7.01693) 3 14 Solution Since the mass of an atom includes the masses of the atomic electrons, the appropriate number of electron masses must be subtracted from the given values. Reactants : Total mass = (7.01600 – 3 m ) + (1.0783 – 1 m ) = 8.0943 – 4m NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 e ee Products : Total mass = (7.01693 –4me)+ 1.0087 = 8.02563 – 4me The energy is supplied as kinetic energy of the bombarding proton. The incident proton must have more than this energy because the system must possess some kinetic energy even after the reaction, so that momentum is conserved with momentum conservation taken into account, the minimum kinetic energy that the incident particle must possess can be found with the formula. where, Q = – [(8.02563 – 4m ) – (8.0943 –4 m )] 931.5 MeV = – 63.96 MeV ee E =– 1 m Q=– 1 1 (–63.96) = 73.1 MeV th M 7 NUCLEAR FISSION In 1938 by Hahn and Strassmann. By attack of a particle splitting of a heavy nucleus (A > 230) into two or more lighter nuclei. In this process certain mass disappears which is obtained in the form of energy (enormous amount) A + p excited nucleus B + C + Q 28 E
JEE-Physics Hahn and Strassmann done the first fision of nucleus of U235). When U235 is bombarded by a neutron it splits into two fragments and 2 or 3 secondary neutrons and releases about 190 MeV ( 200 MeV) energy per fission (or from single nucleus) Fragments are uncertain but each time energy released is almost same. Possible reactions are U235 + n1 Ba + Kr + 3 n1 + 200 MeV or U235 + n1 Xe + Sr + 2 n1 + 200 MeV 0 0 0 0 and many other reactions are possible. • The average number of secondary neutrons is 2.5. • Nuclear fission can be explained by using \"liquid drop model\" also. • The mass defect m is about 0.1% of mass of fissioned nucleus • About 93% of released energy (Q) is appear in the form of kinetic energies of products and about 7% part in the form of – rays. NUCLEAR CHAIN REACTION : The equation of fission of U235 is U235 + n1 Ba + Kr + 3 n1 + Q. 0 0 These three secondary neutrons produced in the reaction may cause of fission of three more U235 and give 9 neutrons, which in turn, may cause of nine more fission of U235 and so on. Thus a continuous 'Nuclear Chain reaction' would start. If there is no control on chain reaction then in a short time (10–6 sec.) a huge amount of energy will be released. (This is the principle of 'Atom bomb'). If chain is controlled then produced energy can be used for peaceful purposes. For example nuclear reactor (Based on fission) are generating electricity. NATUR AL UR ANIUM : It is mixture of U235 (0.7%) and U238 (99.3%). U235 is easily fissionable, by slow neutron (or thermal neutrons) having K.E. of the order of 0.03 eV. But U238 is fissionable with fast neutrons. Note : Chain reaction in natural uranium can't occur. To improve the quality, percentage of U235 is increased to 3%. The proposed uranium is called 'Enriched Uranium' (97% U238 and 3% U235) LOSSES OF SECONDARY NEUTRONS : Leakage of neutrons from the system : Due to their maximum K.E. some neutrons escape from the system. Absorption of neutrons by U238 : Which is not fissionable by these secondary neutrons. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 CRITICAL SIZE (OR MASS) : In order to sustain chain reaction in a sample of enriched uranium, it is required that the number of lost neutrons should be much smaller than the number of neutrons produced in a fission process. For it the size of uranium block should be equal or greater than a certain size called critical size. REPRODUCTION FACTOR : (K) = rate of production of neutrons rate of loss of neutrons (i) If size of Uranium used is 'Critical' then K = 1 and the chain reaction will be steady or sustained (As in nuclear reaction) (ii) If size of Uranium used is 'Super critical' then K > 1 and chain reaction will accelerate resulting in a explosion (As in atom bomb) (iii) If size of Uranium used is 'Sub Critical' then K < 1 and chain reaction will retard and will stop. E 29
JEE-Physics NUCLEAR REACTOR (K = 1) : Credit To Enricho Fermi Construction : • Nuclear Fuel : Commonly used are U235 , Pu239. Pu239 is the best. Its critical size is less than critical size of U235. But Pu239 is not naturally available and U235 is used in most of the reactors. • Moderator : Its function is to slow down the fast secondary neutrons. Because only slow neutrons can bring the fission of U235. The moderator should be light and it should not absorb the neutrons. Commonly, Heavy water (D O, molecular weight 20 gm.) Graphite etc. are used. These are rich of protons. Neutrons collide with the 2 protons and interchange their energy. Thus neutrons get slow down. • Control rods :They have the ability to capture the slow neutrons and can control the chain reaction at any stage. Boron and Cadmium are best absorber of neutrons. • Coolant : A substance which absorb the produced heat and transfers it to water for further use. Generally coolant is water at high pressure FAST BREADER REACTORS The atomic reactor in which fresh fissionable fuel (Pu239) is produced along with energy. The amount of produced fuel (Pu239) is more than consumed fuel (U235) • Fuel : Natural Uranium. • P r o c e s s : During fission of U235, energy and secondary neutrons are produced. These secondary neutrons are absorbed by U238 and U239 is formed. This U239 converts into Pu239 after two beta decay. This Pu239 can be separated, its half life is 2400 years. U238 + n1 U239 2 Pu239 (best fuel of fission) 92 0 92 94 This Pu239 can be used in nuclear weapons because of its small critical size than U235. • Moderator : Are not used in these reactors. • Coolant : Liquid sodium NUCLEAR FUSION : It is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus. A + B C + Q (Energy) The product (C) is more stable then reactants (A and B) & m < (m + m ) C ab and mass defect m = [(m + m )– m] amu a b c Energy released is E = m 931 MeV The total binding energy and binding energy per nucleon C both are more than of A and B. E = E – (E + E) c a b Fusion of four hydrogen nuclei into helium nucleus : 4( H1) He4 + 2 + 0 + 2 + 26 MeV NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 1 2 • Energy released per fission >> Energy released per fusion. • Energy per nucleon in fission 200 0.8 5 MeV << energy per nucleon in fusion 24 6MeV 235 4 REQUIRED CONDITION FOR NUCLEAR FUSION • High temperature : Which provide kinetic energy to nuclei to overcome the repulsive electrostatic force between them. • High Pressure (or density) : Which ensure frequent collision and increases the probability of fusion. The required temperature and pressure at earth (lab) are not possible. These condition exist in the sun and in many other stars. The source of energy in the sun is nuclear fusion, where hydrogen is in plasma state and there protons fuse to form helium nuclei. 30 E
JEE-Physics HYDROGEN BOMB It is based on nuclear fusion and produces more energy than an atom bomb. Pair production Pair Annihilation A –photon of energy more than 1.02 MeV, when interact with a nucleus When electron and positron combines they produces pair of electron (e—) and positron (e+). The energy equivalent to annihilates to each other and only energy is rest massof e— (or e+)=0.51MeV. Theenergy equivalent torest massof pair released in the form of (e— + e+) = 1.02MeV. two gama photons. If the energy of electron and Forpair productionEnergyof photon 1.02MeV. positron are negligible then energy of each If energy of photon is more than 1.02 MeV, the extra energy (E–1.02) photonis0.5/MeV MeV divides approximately in equal amount to each particle as the kinetic energyor EPh1.02 MeV 2 (KE)e– or e+ = If E< 1.02MeV, pairwill not produce. Example In a nuclear reactor, fission is produced in 1 g for U235 (235.0439) in 24 hours by slow neutrons (1.0087 u). Assume that Kr92 (91.8973 u) and Ba141 (140.9139 amu) are produced in all reactions and no energy is lost. 35 56 (i) Write the complete reaction (ii) Calculate the total energy produced in kilowatt hour. Given 1u = 931 MeV. Solution The nuclear fission reaction is U235 + n1 Ba141 + Kr92 + 3 n1 92 0 56 36 0 Mass defect m= [(m + m) – (m + m + 3m )] = 256.0526 – 235.8373 =0.2153 u u n Ba Kr n Energy released Q = 0.2153 × 931 = 200 MeV. Number of atoms in 1 g = 6.02 1023 = 2.56 × 1021 235 Energy released in fission of 1 g of U235 is E = 200 × 2.56 × 1021 = 5.12 × 1023 MeV = 5.12 × 1023 × 1.6 × 10–13 = 8.2 × 1010 J 8.2 1010 = 3.6 106 kWh = 2.28 × 104 kWh Example It is proposed to use the nuclear fusion reaction : H2 + H2 He4 in a nuclear reactor of 200 MW rating. If 1 1 2 the energy from above reaction is used with at 25% efficiency in the reactor, how many grams of deuterium will be needed per day. (Mass of H2 is 2.0141 u and mass of He4 is 4.0026 u) 12 Solution Energy released in the nuclear fusion is Q = mc2 = m(931) MeV (where m is in amu) Q = (2 × 2.0141 – 4.0026) × 931 MeV = 23.834 MeV = 23.834 × 106 eV Since efficiency of reactor is 25% NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 So effective energy used = 25 23.834 106 1.6 1019 J = 9.534 × 10–13 J 100 Since the two deuterium nuclei are involved in a fusion reaction, therefore, energy released per deuterium is 9.534 1013 . 2 200 106 86400 = 3.624 × 1025 For 200 MW power per day, number of deuterium nuclei required = 9.534 1023 2 Since 2g of deuterium constitute 6 × 1023 nuclei, therefore amount of deuterium required is = 2 3.624 1025 =120.83 g/day E 31 6 1023
JEE-Physics RADIOACTIVITY The process of spontaneous disintegration shown by some unstable atomic nuclei is known as natural radioactivity. This property is associated with the emission of certain types of penetrating radiations, called radioactive rays, or Becquerel rays ( rays). The elements or compounds, whose atoms disintegrate and emit radiations are called radioactive elements. Radioactivity is a continuous, irreversible nuclear phenomenon. Radioactive Decays Generally, there are three types of radioactive decays (i) decay (ii) and decay (iii) decay • decay B.E. A In decay, the unstable nucleus emits an particle. By emitting particle, the nucleus decreases it's mass energy number and move towards stability. A A=210 Nucleus having A>210 shows decay.By releasing particle, it can attain higher stability and Q value is positive. • decay Region of stable nuclei In beta decay (N/Z) ratio of nucleus is changed. This decay is shown by unstable nuclei. In beta decay, either a neutron is converted into proton or N N=Z proton is converted into neutron. For better understanding we discuss N/Z graph. There are two type of unstable nuclides 45° Z • A type Move towards stability For A type nuclides (N/Z) > (N/Z) stable A A N To achieve stability, it increases Z by conversion of neutron into proton Z n1 p1 + e–1 + , XA YA + e 1 0 1 Z Z+1 particle This decay is called –1 decay. Kinetic energy available for e–1 and is, Q K K NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 K.E. of satisfies the condition 0 < K < Q • B type B For B type nuclides (N/Z) (N/Z) stable Move towards B N stability To achieve stability it decreases Z by the conversion of a proton into Z X A YA Z neutron. That is, p n e , Z 1 e positron neutrino particle 32 E
JEE-Physics • decay : when an or decay takes place, the daughter nucleus is usually in higher energy state, such a nucleus comes to ground state by emitting a photon or photons. 67Co 67Ni* E=2.5MeV E=1.17MeV E=1.33MeV Order of energy of photon is 100 KeV e.g. 67 C o 67 Ni * , 67 N i * 67 Ni photon 27 28 28 28 higher energy state Features –particles Properties of , and rays –rays –particles Identity Helium nucleus or doubly Fast moving electrons Electromagnetic wave Charge ( –0 or –) (photons) Mass ionised helium atom ( He4 ) Electronic (– e) Neutral Speed 2 rest mass = 0 = (rest mass of electron) Only c = 3 × 108 m/s K.E. Twice of proton (+2e) 4m 1% of c to 99% of c –photons come out with Energy p (All possible values same speed from all spectrum between this range) types of nucleus. Ionization (rest mass of –particles come out with So, can not be a power (>>) different speeds from the characteristic speed. m –mass of proton same type of nucleus. p So that it can not MeV be a characteristic speed. Line and discrete 1.4 × 107 m/s. to MeV 1 (or 1 times of ) 2.2 × 107 m/s. Continuous 100 (or linear) (Only certain value 100 times of –rays 1(100 times of ) (or 1 times of ) between this range). No deflection 100 Their speed depends on With the help of energy 1 times of –rays levels in nucleus nature of the nucleus. 100 So that it is a (100 times of ) Deflection (More than ) characteristic speed. By weak nuclear MeV interactions NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 Line and discrete (or linear) 10,000 times of –rays Penetration 1 times of –rays power (>>) 10000 Effect of electric Deflection or magnetic field Explanation By Tunnel effect of emission (or quantum mechanics) E 33
JEE-Physics Laws of Radioactive Decay 1 . The radioactive decay is a spontaneous process with the emission of , and rays. It is not influenced by external conditions such as temperature, pressure, electric and magnetic fields. 2 . The rate of disintegration is directly proportional to the number of radioactive atoms present at that time i.e., rate of decay number of nuclei. dN Rate of decay = (number of nuclei) i.e. N dt where is called the decay constant. This equation may be expressed in the form dN dt . N N dN t n N t N0 NN 0 dt 0 where N is the number of parent nuclei at t=0. The number that survives at time t is therefore 0 N=N e–t and t = 2.303 log10 N0 this function is plotted in figure. 0 Nt N0 N'=N0(1–e–t) Graph : Time versus N (or N') 0.63N0 N N0 2 0.37N0= N0 e N=N0e–t (0,0) Th Ta time • Half life (T ) : It is the time during which number of active nuclei reduce to half of initial value. h If at t = 0 no. of active nuclei N then at t = T number of active nuclei will be N0 0h 2 From decay equation N = N e–t 0 N 0 = N e–Tn T= n 2 0.693 0.7 20 h = • Mean or Average Life (T ) : It is the average of age of all active nuclei i.e. a sum of times of existance of all nuclei in a sample 1 T= = a initial number of active nuclei in that sample (i) At t = 0, number of active nuclei = N then number of active nuclei at NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 0 t = T is N N 0 e Ta N 0 e1 N0 0.37N0 37% of N0 a e (ii) Number nuclei which have been disintegrated within duration T is a N' = N – N = N – 0.37 N = 0.63 N = 63% of N 00 0 0 0 • T= 1 = Th = Th = 1.44 T a 0.693 h n 2 • Within duration T 50% of N decayed and 50% of N remains active h0 0 • Within duration T 63% of N decayed and 37% of N remains active a0 0 34 E
JEE-Physics ACTIVITY OF A SA MPLE (OR DECAY R ATE) It is the rate of decay of a radioactive sample R dN N or R = R e–t 0 dt • Activity of a sample at any instant depends upon number of active nuclei at that instant. R N (or active mass) , R m • R also decreases exponentially w.r.t. time same as the number of active nuclei decreases. • R is not a constant with N, m and time while , T and T are constant h a • At t = 0, R = R then at t = T R = R0 and at t = T R = R0 or 0.37 R 0h 2 a e 0 • Similarly active mass of radioactive sample decreases exponentially. m = m e–t 0 • Activity of m gm active sample (molecular weight M ) is R = N 0.693 N AV m w = MW Th here N = Avogadro number = 6.023 × 1023 AV SI UNIT of R : 1 becquerel (1 Bq)= 1 decay/sec Other Unit is curie : 1 Ci = 3.70 × 1010 decays/sec 1 Rutherford : (1 Rd) =106 decays/s Specific activity : Activity of 1 gm sample of radioactive substance. Its unit is Ci/gm e.g. specific activity of radium (226) is 1 Ci/gm. Example The half–life of cobalt–60 is 5.25 yrs. After how long does its activity reduce to about one eight of its original value? Solution The activity is proportional to the number of undecayed atoms: In each half–life, the remaining sample decays to half of its initial value. Since 1 1 1 1 therefore, three half–lives or 15.75 years are required 2 2 2 8, for the sample to decay to 1/8th its original strength. Example A count rate meter is used to measure the activity of a given sample. At one instant the meter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute. (i) Find the decay constant. (ii) Also, find the half–life of the sample. Solution NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 Initial activity A = dN N 0 4750 ...(i) Final activity A = dN N 2700 ...(ii) i dt f dt t 0 t5 4750 N0 Dividing (i) by (ii), we get 2700 = N t The decay constant is given by 2.303 log N0 2 .3 0 3 lo g 4 7 5 0 0 .1 1 3 m in -1 t Nt 5 2700 0.693 0 .6 9 3 Half–life of the sample is T= = 0 .1 1 3 =6.14 min E 35
JEE-Physics • Parallel radioactive disintegration B Let initial number of nuclei of A is N then at any time number of nuclei of 0 A A, B & C are given by N = N + N + N dN A d N B NC 0 A B C dt dt C A disintegrates into B and C by emitting particle. dN B dN C d dt dt 2N A dt Now, NA 1N A and NB NC 1 2 dN A NA teff t1 t2 dt 1 2 eff 1 2 t1 t2 Example The mean lives of a radioactive substances are 1620 and 405 years for –emission and –emission respectively. Find out the time during which three fourth of a sample will decay if it is decaying both by –emission and –emission simultaneously. Solution When a substance decays by and emission simultaneously, the average rate of disintegration av is given by av= + when = disintegration constant for –emission only = disintegration constant for –emission only 1 1 11 1 11 , av= + Mean life is given by T = Tm T T 1620 405 324 m N0 1 100 av × t = 2.303 log N t , 324 t = 2.303 log 25 t = 2.303 × 324 log 4 = 449 years. Example A radioactive decay is given by A t1/28 yrs B Only A is present at t=0. Find the time at which if we are able to pick one atom out of the sample, then probability of getting B is 15 times of getting A. Solution A B NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 at t 0 N0 0 at t t N N0 N N Probability of getting A, P = N0 A Probability of getting B, P = N0 N P = 15 P N0 N N N = 16N N= N 0 B N0 B A N0 15 0 16 N0 1 Remaining nuclei are th of initial nuclei, hence required time t=4 half lives =32 years 16 36 E
JEE-Physics Radioactive Disintegration with Successive Production rate ofproduction A B dN A N A ....(i) dt dNA 0 NA 0, rate of production R dt when N in maximum N max = = A A t dN A t t 0 N A By equation (i) dt, Number of nuclei is NA 1 e t 0 Example 1 021persec A 1 30 B A shows radioactive disintegration and it is continuously produced at the rate of 1021 per sec. Find maximum number of nuclei of A. Solution 1 At maximum, rproduction = rdecay 1021 = 30 N N=30 × 1021 Soddy and Fajan's Group Displacement Laws : (i) –decay : The emission of one –particle reduces the mass number by 4 units and atomic number by 2 units. If parent and daughter nuclei are represented by symbols X and Y respectively then, XA YA–4 + 2He4() Z Z–2 (ii) –decay : Beta particles are said to be fast moving electrons coming from the nucleus of a radioactive substance. Does it mean that a nucleus contains electrons? No, it is an established fact that nucleus does not contain any electrons. When a nucleus emits a beta particle, one of its neutrons breaks into a proton, an electron (i.e., –particle) and an antineutrino n p e where n= neutron p = proton e = –particle Thus emission of a beta particle is caused by the decay of a neutron into a proton. The daughter nucleus thus has an atomic number greater than one (due to one new proton in the nucleus) but same mass number as that of parent nucleus. Therefore, representing the parent and daughter nucleus by symbols X and Y respectively, we have XA YA + Z Z+1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 (iii) –decay :When parent atoms emit gamma rays, no charge is involved as these are neutral rays. Thus there is no effect on the atomic number and mass number of the parent nucleus. However the emission of –rays represents energy. Hence the emission of these rays changes the nucleus from an excited (high energy) state to a less excited (lower energy) state. ve potential (electromagnetic radiation) +ve potential Lead E 37
JEE-Physics SOME WORKED OUT EXAMPLES Example#1 A photon of energy 12.09 eV is completely absorbed by a hydrogen atom initially in the ground state. The quantum number of the excited state is (A) 4 (B) 5 (C) 3 (D) 2 Solution Ans. (C) 13.6 Energy difference in hydrogen atom = 13.6 – n2 = 12.09 n2 9 n=3 Example#2 The figure indicates the energy level diagram of an atom and the origin of five spectral lines in emission spectra. Which of the spectral lines will also occur in the absorption spectra? 3rd 2nd first excited state 123 45 ground state (A) 1, 2, 4 (B) 2, 3, 4, 5 (C) 1, 2, 3 (D) 1,2,3,4,5 Ans. C Solution For absorpiton spectra, atom must be in ground state Example#3 The ionization energy of Li++ is equal to (A) 6hcR (B) 2hcR (C) 9hcR (D)hcR Solution Ans. (C) RhcZ2 Ionization energy = (Rhc)Z2 = 9hcR Note :E n n2 Example#4 The electron in a hydrogen atom makes a transition from n=n to n=n state. The time period of the electron 12 in state n is eight times that in state n . The possible values of n and n are 1 2 12 (A) n = 8, n =1 (B) n =4, n =1 (C) n =4, n =2 (D) n =2, n =4 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 12 12 12 12 Solution Ans. (C) Time period T = 2 r n3 But T = 8T n 3 = 8n 3 n =2n v 12 12 12 Example#5 When a hydrogen atom is excited from ground state to first excited state, then (A) its kinetic energy increases by 10.2 eV (B) its kinetic energy decreases by 13.6 eV (C) its potential energy increases by 10.2 eV (D) its angular momentum increases by h/2 Solution Ans. (D) In ground state , kinetic energy = 13.6 eV, Potential energy = – 27.2 eV In first excited state, kinetic energy = 3.4 eV, Potential energy = – 6.8 eV nh h Angular momentum is ; Difference of angular momentum for consecutive orbit 2 2 38 E
JEE-Physics Example#6 The energy of a tungsten atom with a vacancy in L shell is 11.3 KeV. Wavelength of K photon for tungsten is 21.3 pm. If a potential difference of 62 kV is applied across the X–rays tube following characteristic X–rays will be produced . (A) K,L series (B) only K & L series (C) only L series (D) None of these Solution Ans. (C) hc 1240 1240 (nm ) 21.3 103 E = E = 58.21 KeV E = 11.3 KeV E = 69.51 KeV 21 E < 62 KeV Therefore only L series will be produced. Example#7 The figure shows a graph between n An and n|n|, where A is the area enclosed by the nth orbit in a A1 n hydrogen like atom. The correct curve is- An 43 2 A1 1 4 2 O1 |n| (A) 4 (B) 3 (C) 2 (D) 1 Solution Ans. (A) An rn 2 n 4 An A1 r1 1 A1 A = r2n log =4 log (n) n e e Example#8 Consider the radiation emitted by large number of singly charged positive ions of a certain element. The sample emit fifteen types of spectral lines, one of which is same as the first line of lyman series. What is the binding energy in the highest energy state of this configuration ? (A) 13.6 eV (B) 54.4 eV (C) 10.2eV (D) 1.6 eV Solution Ans. (D) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 n(n 1) 15 n = 6 Element should be hydrogen like atoms so ion will be He+ 2 (13.6 )(Z2 ) (13.6)(4) 13.6 Binding energy 1.6eV n2 36 9 Example#9 The probability that a certain radioactive atom would get disintegrated in a time equal to the mean life of the radioactive sample is- (A) 0.37 (B) 0.63 (C) 0.50 (D) 0.67 Solution Ans. (B) Required probability P(t) = N 0 (1 et ) 1 e 1 1 e 1 0.63 N0 E 39
JEE-Physics Example#10 Figure shows the graph of stopping potential versus the frequency of a photosensitive metal. The plank's constant and work function of the metal are V0 V2 V1 1 2 (A) V2 V1 V2 1 V1 2 e (B) V2 V1 V2 1 V1 2 e 2 2 e, 2 1 2 1 e, 2 1 (C) V2 V1 e, V2 1 V1 2 e (D) V2 V1 V2 1 V1 2 e 2 1 2 1 2 1 e, 2 1 Solution Ans. (D) Equation of straight line V V1 V2 V1 V V2 V1 V2 1 V1 2 1 2 1 2 1 2 1 But V h 0 h V2 V1 e and 0 V2 1 V1 2 e e e 2 1 2 1 Example#11 Consider a hydrogen like atom whose energy in nth excited state is given by E = 13 .6 Z 2 when this n n2 excited atom makes a transition from excited state to ground state, most energetic photons have energy E = 52.2224 eV and least energetic photons have energy E = 1.224 eV.The atomic number of atom is max min (A) 2 (B) 5 (C) 4 (D) None of these Solution Ans. (A) Maximum energy is liberated for transition En E1 and minimum energy for En En–1 Hence E1 E1 52.224eV and E1 E1 1.224eV E = –54.4eV and n=5 n2 n2 (n 1)2 1 Now E =– 13.6Z2 = –54.4 eV. Hence Z = 2 1 12 Example#12 The positions of 2 D, 4 He and 7 L i are shown on the binding energy curve as shown in figure. 1 2 3 Binding energy per nucleon (MeV)8 42He 7 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 6 37Li 5 4 3 2 1 12D 2 4 6 8 10 Mass Number (A) The energy released in the fusion reaction. 2 D 7 Li 2 4 H e 1 n 1 3 2 0 (A) 20 MeV (B) 16 MeV (C) 8 MeV (D) 1.6 MeV 40 E
Solution JEE-Physics Released energy = 2 × 4 × 7 – 2 × 1 – 7 × 5.4 = 16 MeV Ans. (B) Example#13 How many head-on elastic collisions must a neutron have with deuterium nuclei to reduce it energy from 6.561 MeV to 1 keV ? (A) 4 (B) 8 (C) 3 (D) 5 Solution Ans. (A) Energy loss 4m1m2 4 (1 )(2 ) 8 Initial KE (m1 m2 )2 (1 2)2 9 8 88 After 1st collision E1 9 E 0 , After 2nd collision E2 9 E1 , After nth collision E n 9 E n1 Adding all the losses 8 81 E =E1 + E2 + ....... + En = (E + E + ...... E) ; here E = E – E1 = E – E= E 9 0 1 n–1 1 0 0 90 90 8 1 1 2 E = E – E2 = E – E= E= E and so on 2 1 1 91 91 0 9 E = 8 1 E0 1 2 E0 .... 1 n 1 E0 = 8 11911n 1 1 E 0 9 E 0 9 9 9 9 9n E0 9 E = 6.561 MeV, E = (6.561 – 0.001) MeV 6.561 0.001 1 1 1 1 n=4 0 6.561 9n 6561 = 9n Example#14 The figure shows the variation of photo current with anode potential for a photosensitive surface for three different radiations. Let I , I and I be the intensities and f , f and f be the frequencies for the ab c ab c curves a, b and c respectively. Choose the incorrect relation. Photo current c a b O Anode potential NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 (A) f = f (B) I < I (C) f < f (D) I > I ab ab cb cb Solution Ans. (C) Here I = I > I and f > f = f BC a cab Example#15 Statement-1: Radioactive nuclei emit -particles (fast moving electrons) and Statement-2: Electrons exist inside the nucleus. (A) Statement–1 is True, Statement–2 is True ; Statement–2 is a correct explanation for Statement–1 (B) Statement–1 is True, Statement–2 is True ; Statement–2 is NOT a correct explanation for Statement–1 (C) Statement–1 is True, Statement–2 is False. (D) Statement–1 is False, Statement–2 is True. Solution Ans. (C) E 41
JEE-Physics Example#16 A vessel of 831cc contains 3 H at 0.6 atm and 27°C. If half life of 3 H is 12.3 years then the 1 1 activity of the gas is- (A) 3.04 × 1013 dps (B) 582 Ci (C) 2.15 × 1013 dps (D) 823 Ci Solution Ans. (B,C) Number of moles of gas n PV (0.6 105 )(831 106 ) =0.02 RT (8.31)(300) (0.693)nN A (0.693)(0.02)(6.02 1023 ) 2.15 1013 Activity = N = T1 / 2 12.3 3.15 107 = 2.15 × 1013 dps 3.7 1010 =582 Ci Example#17 Choose the CORRECT statement(s) (A) Mass of products formed is less than the original mass in nuclear fission and nuclear fusion reactions. (B) Binding energy per nucleon increases in -decay and -decay. (C) Mass number is conserved in all nuclear reactions. (D) Atomic number is conserved in all nuclear reactions. Solution Ans. (ABC) Fusion and fission are always exothermic and & decay will result in more stable product. Mass number is conserved but atomic number is not conserved. N X M Y 0 Z, N = M + 0 and A may not be equal to B + C A B C Example#18 to 20 Einstein in 1905 propounded the special theory of relativity and in 1915 proposed the general theory of relativity. The special theory deals with inertial frames of reference. The general theory of relativity deals with problems in which one frame of reference. He assumed that a fixed frame is accelerated w.r.t. another frame of reference of reference cannot be located. Postulates of special theory of relativity • The laws of physics have the same form in all inertial systems. • The velocity of light in empty space is a universal constant the same for all observers. Einstein proved the following facts based on his theory of special relativity. Let v be the velocity of the speceship w.r.t. a given frame of reference. The observations are made by an observer in that reference frame. • All clocks on the spaceship will go slow by a factor 1 v2 / c2 • All objects on the spaceship will have contracted in length by a factor 1 v2 / c2 • The mass of the spaceship increases by a factor 1 v2 / c2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 • Mass and energy are interconvertable E = mc2 • The speed of a material object can never exceed the velocity of light. • If two objects A and B are moving with velocity u and v w.r.t. each other along the x-axis, the relative uv velocity of A w.r.t. B = 1 uv / c2 1 8 . One cosmic ray particle approaches the earth along its axis with a velocity of 0.9c towards the north pole and another one with a velocity of 0.5c towards the south pole. The relative speed of approach of one particle w.r.t. another is- (A) 1.4 c (B) 0.9655 c (C) 0.8888c (D) c 1 9 . The momentum of an electron moving with a speed 0.6 c is (Rest mass of electron is 9.1 × 10–31kg) (A) 1.6 × 10–22 kgms–1 (B) 2 × 10–22 kgms–1 (C) 5.46 × 10–31 kgms–1 (D) 5.46 × 10–22 kgms–1 42 E
JEE-Physics 2 0 . A stationary body explodes into two fragments each of rest mass 1kg that move apart at speeds of 0.6c relative to the original body. The rest mass of the original body is- (A) 2 kg (B) 2.5 kg (C) 1.6 kg (D) 2.25 kg Solution 1 8 . Ans. (B) u v 0.9c 0.5c 1.4c 0.9655c uv (0.9c)(0.5c) 1.45 Relative speed = 1 c2 1 c2 1 9 . Ans. (B) (9.1 1 0 31 ) 3 3 10 8 5 p = mv = m0v = 2 × 10–22 kg ms–1 1 v2 / c2 3 2 5 1 2 0 . Ans. (B) m0c2 = m01c2 m 02 c2 11 m = 2.5kg 2 2 0 0.8 0.8 v v 1 c 1 c Example#21 to 23 A mercury arc lamp provides 0.1 watt of ultra-violet radiation at a wavelength of = 2537 Å only. The photo tube (cathode of photo electric device) consists of potassium and has an effective area of 4 cm2. The cathode is located at a distance of 1m from the radiation source. The work function for potassium is 0 = 2.22 eV. 2 1 . According to classical theory, the radiation from arc lamp spreads out uniformly in space as spherical wave. What time of exposure to the radiation should be required for a potassium atom (radius 2Å) in the cathode to accumulate sufficient energy to eject a photo-electron ? (A) 352 second (B) 176 second (C) 704 seconds (D) No time lag 2 2 . To what saturation current does the flux of photons at the cathode corresponds if the photo conversion efficiency is 5%. (A) 32.5 nA (B) 10.15 nA (C) 65 nA (D) 3.25 nA 2 3 . What is the cut off potential V ? 0 (A) 26.9 V (B) 2.69 V (C) 1.35 V (D) 5.33 V Solution NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 21. Ans. (A) 0.1 UV energy flux at a distance of 1m 4 12 cross section (effective area) of atom = (2 10-10)2 = 4 10-20 m2 Energy required to eject a photoelectron from potassium = 2.2 eV 2.2 1.6 10-19 J. Exposure time = 2.2 1.6 1019 = 352 seconds. 0.1 4 12 4 1020 22. Ans. (A) Flux of photon at the cathode = 0.1 1 = 1.015 1016 photons/ sec m2 4 12 photon energy Saturation current = (photon flux effective area of cathode) 5/100 1.6 10-19 = 3.25 10-8 A. E 43
JEE-Physics 23. Ans. (B) (4.897 2.22)eV Cut off potential = e = 2.69 volts. Example#24 With respect to photoelectric experiment, match the entries of Column I with the entires of Column II. Column I Column II (A) If (frequency) is increased keeping (P) Stopping potential increases I (intensity) and (work function) constant (B) If I is increased keeping and constant (Q) Saturation photocurrent increases (C) If the distance between anode and (R) Maximum KE of the photoelectrons increase cathode increases. (D) If is decreased keeping and (S) Stopping potential remains the same I constant Solution Ans. (A)(P,R); (B)(Q,S); (C)(S); (D)(P,R) From eV = h – and K= h – . If increases keeping constant, then V and K both increase. 0 max 0 max If decreases keeping constant, then V0 and Kmax increase. If I increases more photoelectrons would be liberated, hence saturation photocurrent increases. If separation between cathode and anode is increased, then there is no effect on , K or current. max 0 Example#25 A sample of hydrogen gas is excited by means of a monochromatic radiation. In the subsequent emission spectrum, 10 different wavelengths are obtained, all of which have energies greater than or equal to the energy of the absorbed radiation. Find the initial quantum number of the state (before absorbing radiation). Solution Ans. 4 10 emission lines final state n = 5 If the initial state were not n = 4, in the emission spectrum, some lines with energies less than that of absorbed radiation would have been observed. initial state n = 4. n=5 n=4 n=3 n=2 n=1 Example#26 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding 24.2 to the largest wavelength of the Balmer series. The K.E. of colliding electron will be N eV. Find the value of N. Solution Ans. 2 Kinetic energy of electron = 13.6 1 1 eV = 12.1 eV. 9 44 E
JEE-Physics Example#27 . Neutrons in thermal equilibrium with matter at 27°C can be thought to behave like ideal gas. Assuming them to have a speed of v, what is their De broglie wavelength (in nm). Fill 156 in the OMR sheet. [Take m rms 11 n = 1.69 × 10–27 kg. k = 1.44 × 10–23 J/K, h = 6.60 × 10–34 Jsec] Solution Ans. 2 v rms 3kT h h h 6.6 1034 2.2 1010 ; mp m n v rms 3 1.44 1023 1.69 1027 300 1.2 1.3 3kTm n 156 156 2.2 10 10 220 1010 22 nm 2 11 11 1.2 1.3 11 11 Example#28 Electromagnetic waves of wavelength 1242 Å are incident on a metal of work function 2eV. The target metal is connected to a 5 volt cell, as shown. The electrons pass through hole A into a gas of hydrogen atoms in their ground state. Find the number of spectral lines emitted when hydrogen atoms come back to their ground states after having been excited by the electrons. Assume all excitations in H-atoms from ground state only. (hc = 12420 eVÅ) H A atoms metal target 5 volt Solution Ans. 6 kE = 12420 Å eV – 2 eV = 8 eV max 1242 Å NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-12\\Modern Physics\\Eng\\01_Modern Physics.p65 12.75 n=4 – 0.85 eV 12.1 n=3 – 1.51 eV 10.2 n=2 – 3.4 eV n=1 – 13.6 eV The electrons are emitted with kinetic energy varying from zero to 8 eV. When accelerated with 5 volt potential difference their energies increases by 5eV. Hence hydrogen will get photons of energies in the range from 5 eV to 13 eV. So maximum possible transactions are upto n = 4. Hence number of spectra lines is 4C =6 2 E 45
JEE-Physics CAPACITANCE CONCEPT OF CAPACITANCE Capacitance of a conductor is a measure of ability of the conductor to store charge on it. When a conductor is charged then its potential rises. The increase in potential is directly proportional to the charge given to the conductor. QV Q = CV The constant C is known as the capacity of the conductor. Capacitance is a scalar quantity with dimension CQ Q2 A2T2 = M–1 L–2 T4 A2 V M1L2 T 2 W Unit :- farad, coulomb/volt The capacity of a conductor is independent of the charge given or its potential raised. It is also independent of nature of material and thickness of the conductor. Theoretically infinite amount of charge can be given to a conductor. But practically the electric field becomes so large that it causes ionisation of medium surrounding it. The charge on conductor leaks reducing its potential. THE CAPACITANCE OF A SPHERICAL CONDUCTOR ++ When a charge Q is given to a isolated spherical conductor then its potential rises. + NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 1Q Q O V ++ C V 40R ++ 4 0 R + + R + + If conductor is placed in a medium then C = 4R = 4 R medium r Capacitance depends upon : • Size and Shape of Conductor • Surrounding medium • Presence of other conductors nearby CONDENSER/CAPACITOR The pair of conductor of opposite charges on which sufficient quantity of charge may be accommodated is defined as condenser. • Principle of a Condenser It is based on the fact that capacitance can be increased by reducing potential keeping the charge M constant. + +Q Consider a conducting plate M which is given a charge Q such that its potential rises to V then ++ ++ Q ++ C ++ ++ V ++ Let us place another identical conducting plate N parallel to it such that charge is induced M N on plate N (as shown in figure). If V is the potential at M due to induced negative charge ++ –+ ++ –+ – ++ –+ ++ –+ on N and V is the potential at M due to induced positive charge on N, then ++ –+ + ++ –+ ++ –+ QQ C' V ' V V V Since V' < V (as the induced negative charge lies closer to the plate M in comparison to M N induced positive charge). C' > C Fur ther, if N is earthed from the outer side ++ – ++ – (see figure) then V\" = V – V ( the entire positive charge flows to the earth) ++ – +– ++ – ++ – QQ C\" >> C ++ – C\" ++ – V \" V V If an identical earthed conductor is placed in the vicinity of a charged conductor then the capacitance of the charged conductor increases appreciable. This is the principle of a parallel plate capacitor. E1
JEE-Physics ENERGY STORED IN A CHARGED CONDUCTOR/CAPACITOR Let C is capacitance of a conductor. On being connected to a battery. It charges to a potential V from zero potential. If q is charge on the conductor at that time then q = CV. Let battery supplies small amount of charge dq to the conductor at constant potential V. Then small amount of work done by the battery against the force exerted by exsiting charge is q Qq 1 q2 Q Q2 dW Vdq dq W dq W CC 2C 0 C 2 0 where Q is the final charge acquired by the conductor. This work done is stored as potential energy, so Q2 = 1 (CV )2 1 CV2 = 1 Q V 2 1 U Q2 1 CV2 1 U 2C 2 2 V = QV QV 2C 2 2 2C 2 GOLDEN KEY POINTS • As the potential of the Earth is assumed to be zero, capacity of earth or a conductor C= connceted to earth will be infinite qq C V0 • Actual capacity of the Earth C = 4R = 1 64 105 711 F 9 109 1 • Work done by battery W = (charge given by battery) × (emf) = QV but Energy stored in conductor QV b2 so 50% energy supplied by the battery is lost in form of heat. REDISTRIBUTION OF CHARGES AND LOSS OF ENERGY When two charged conductors are connected by a conducting wire then charge flows from a conductor at higher potential to that at lower potential. This flow of charge stops when the potential of two conductors became equal. Let the amounts of charges after the conductors are connected are Q ' and Q ' respectively and potential is V 12 then +++ + +++ + + + +++ + + V1 +V2 + V V+ ++ + C1 + Q1 + C1 + Q'1 ++C2 + Q2 C2 + Q'2 ++ + + + NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 + + + + + + (Before connection) (After connection) • Common potential According to law of Conservation of charge Qbefore connection = Q after connection C1V1 + C2V2 = C1V + C2V Common potential after connection V C1 V1 C2 V2 C1 C2 E 2
JEE-Physics • Charges after connection Q' = CV = C Q 1 Q2 C1 C1 Q (Q : Total charge on system) 1 1 1 C 1 C2 C 2 Q' = CV = C Q 1 Q2 C1 C2 2 2 2 C 1 C2 C Q 2 Ratio of the charges after redistribution Q1 ' C1V R1 (in case of spherical conductors) Q2 ' C2V R2 • Loss of energy in redistribution When charge flows through the conducting wire then energy is lost mainly on account of Joule effect, electrical energy is converted into heat energy, so change in energy of this system, U = U – 1 C1 V 2 1 C2V2 1 C1 V12 1 C 2 V22 U 1 C1C2 ( V1 V2 )2 f Ui 2 2 2 2 C1 C 2 2 Here negative sign indicates that energy of the system decreases in the process.SOLVED EXAMPLES Example A conductor gets a charge of 50 C when it is connected to a battery of e.m.f. 5 V. Calculate capacity of the conductor. Solution Q 50 10 6 Capacity of the conductor C 10F V5 Example The capacity of a spherical capacitor in air is 50 F and on immersing it into oil it becomes 110 F. Calculate the dielectric constant of oil. Solution Dielectric constant of oil r C medium 110 2.2 C air 50 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 Example A radio active source in the form of a metal sphere of diameter 10–3m emits particles at a constant rate of 6.25 × 1010 particles per second. If the source is electrically insulated, how long will it take for its potential to rise by 1.0 volt, assuming that 80% of emitted particles escape from the surface. Solution Capacitance of sphere C = 0.5 103 = 1 1012 F 4 0R 9 × 109 18 Rate to escape of charge from surface 80 6.25 1010 1.6 1019 = 8 × 10–9 C/s 100 10–9) t and q = CV 8 × 10–9 × t= 1 1012 1 t = 10 12 10 3 therefore q = (8 × 18 8 109 18 = 144 = 6.95 s E3
JEE-Physics Example The plates of a capacitor are charged to a potential difference of 100 V and then connected across a resister. The potential difference across the capacitor decays exponentially with respect to time. After one second the potential difference between the plates of the capacitor is 80 V. What is the fraction of the stored energy which has been dissipated ? Solution 11 CV2 Energy losses U = CV 2 – 2 02 U 1 C V02 1 CV2 V02 V 2 (100)2 (80)2 20 180 9 2 2 V02 (1 0 0 )2 Fractional energy loss = = = = U0 1 (100 )2 25 2 C V02 Example Two uniformly charged spherical drops at potential V coalesce to form a larger drop. If capacity of each smaller drop is C then find capacity and potential of larger drop. Solution When drops coalesce to form a larger drop then total charge and volume remains conserved. If r is radius and q is charge on smaller drop then C = 4 0 r and q = CV Equating volume we get 4 R3 = 2 × 4 r3 R = 21/3 r 3 3 Capacitance of larger drop Charge on larger drop C' = 4 0R = 21/3 C Q = 2q = 2CV Potential of larger drop V' = Q 2CV = 22/3 V C ' 21 / 3 C PAR ALLEL PL ATE CAPACITOR + – M Edge effect area=A + –N (i) Capacitance + – M P E+ N + – It consists of two metallic plates M and N each of area A at E– + – + – separation d. Plate M is positively charged and plate N is d + – earthed. If r is the dielectric constant of the material medium + – and E is the field at a point P that exists between the two + – plates, then + – + – = E I step : Finding electric field E=E +E 2 + 2 = = [= ] +– 0r 0r II step : Finding potential difference V = Ed = qd ( E V and q NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 d= ) d A 0r A0r III step : Finding capacitance C q r 0 A Vd If medium between the plates is air or vacuum, then r= 1 C0 0 A d so C rC0 KC0 (where r = K = dielectric constant) 4
JEE-Physics (ii) Force between the plates The two plates of capacitor attract each other because they are oppositely charged. Electric field due to positive plate E = = Q 20 20 A Force on negative charge –Q is F = –Q E = – Q2 20 A Magnitude of force Q2 1 0 A E2 F = 20 A = 2 Force per unit area or energy density or electrostatic pressure F 1 E2 A u p 2 0 SPHERICAL CAPACITOR + (i) Outer sphere is earthed + R1 + + + When a charge Q is given to inner sphere it is uniformly distributed on its surface A charge –Q is induced on inner surface of outer sphere. The charge + + +Q induced on outer surface of outer sphere flows to earth as it is grounded. ++ + E = 0 for r < R1 and E = 0 for r > R2 R2 Potential of inner sphere Q Q Q R2 R1 V1 = 40R1 + R1R2 40R 2 4 0 As outer surface is earthed so potential V2 = 0 Potential difference between plates Q (R2 R1 ) V = V1 – V2 = 40 R1R2 So C = Q = 4 0 R1R2 (in air or vacuum) V R2 R1 R1R2 In presence of medium between plate C = 4 r 0 R2 R1 (ii) Inner sphere is earthed qout + (Q–Q') on outer + Surface Here the system is equivalent to a spherical capacitor of inner and + outer radii R1 and R2 respectively and a spherical conductor of radius (+Q') + + R2 in parallel. This is because charge Q given to outer sphere + distributes in such a way that for the outer sphere. (–Q') ++ qin ++ NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 Q'= R1 Q and Insi–dveea+crehveae+rqaguneadsl++++ R2 + + + + Charge on the inner side is + Charge on the outer side is Q – R1 (R2 R1 ) Q' Q – Q' R2 Q = R2 Q –Q' R2 + So total capacity of the system. R1R2 + 4 0 R2 = 4 0R 2 C = 4 0 R2 R1 2 R2 R1 E5
JEE-Physics CYLINDRICAL CAPACITOR + + When a charge Q is given to inner cylinder it is uniformly distributed on its surface. A charge –Q is induced on inner surface of outer cylinder. The charge +Q induced + + on outer surface of outer cylinder flows to earth as it is grounded + +R2 + Electrical field between cylinders E = = Q / + R1 + 20r 20r + + + R2 Q Q n R2 + V = R1 20r dr 2 0 R1 + Potential difference between plates = Capacitance C = Q = 2 0 V loge (R2 / R1 ) In presence of medium C 20r loge (R2 / R1 ) Example The stratosphere acts as a conducting layer for the earth. If the stratosphere extends beyond 50 km from the surface of earth, then calculate the capacitance of the spherical capacitor formed between stratosphere and earth's surface. Take radius of earth as 6400 km. Solution The capacitance of a spherical capacitor is C = 40 ab ba b = radius of the top of stratosphere layer = 6400 km + 50 km = 6450 km = 6.45 × 106 m a = radius of earth = 6400 km = 6.4 × 106 m C 1 6.45 106 6.4 106 = 0.092 F 9 109 6.45 106 6.4 106 Example NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Solution = 15 cm = 15 × 10–2 m; a = 1.4cm = 1.4 × 10–2 m ; b=1.5 cm = 1.5 × 10–2m; q = 3.5 µC = 3.5 × 10–6C Capacitance C 2 0 = 2 8.854 10 12 15 10 2 = 1.21 × 10–8 F 2.30 3 log10 b 2.303 log10 1.5 10 2 a 1.4 10 2 Since the outer cylinder is earthed, the potential of the inner cylinder will be equal to the potential difference between them. Potential of inner cylinder, is V q 3.5 106 2.89 104 V C 1.2 1010 6E
JEE-Physics GOLDEN KEY POINTS • If one of the plates of parallel plate capacitor slides AA relatively than C decrease (As overlapping area decreases). d • If both the plates of parallel plate capacitor are touched Edge effect each other resultant charge and potential became zero. M –N • Electric field between the plates of a capacitor is shown + – in figure. Non-uniformity of electric field at the boundaries + – E = uniform in the centre of the plates is negligible if the distance between the + plates is very small as compared to the length of the + – plates. + – + – E = non-uniform at the edges + + – + – + – – COMBINATION OF CAPACITOR • Capacitor in series: A +Q –Q +Q –Q +Q –Q B In this arrangement of capacitors the charge has no alternative path(s) C1 C2 C3 to flow. V1 V2 V3 +– (i) The charges on each capacitor are equal V i.e. Q = C1V1 = C2V2 = C3V3 (ii) The total potential difference across AB is shared by the capacitors in the inverse ratio of the capacitances V = V + V + V 123 If C is the net capacitance of the series combination, then S Q QQQ 1 1 1 1 CS C1 C2 C3 CS C1 C2 C3 • Capacitors in parallel NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 In such in arrangement of capacitors the charge has an alternative path(s) to flow. +Q1 –Q1 (i) The potential difference across each capacitor is same and equal the C1 total potential applied. i.e. V = V = V = V V Q1 Q2 Q3 A +Q2 –Q2 B 123 C1 C2 C3 C2 +Q3 –Q3 C3 V (ii) The total charge Q is shared by each capacitor in the direct ratio of the capacitances. Q=Q +Q +Q 123 If CP is the net capacitance for the parallel combination of capacitors : CV = CV + CV + CV C = C + C + C P 1 2 3 P 1 2 3 E7
JEE-Physics GOLDEN KEY POINTS • For a given voltage to store maximum energy capacitors should be connected in parallel. • If N identical capacitors each having breakdown voltage V are joined in (i) series then the break down voltage of the combination is equal to NV (ii) parallel then the breakdown voltage of the combination is equal to V. • Two capacitors are connected in series with a battery. Now battery is removed and loose wires connected together then final charge on each capacitor is zero. • If N identical capacitors are connected then C series C , Cparallel NC N • In DC capacitor's offers infinite resistance in steady state, so there will be no current flows through capacitor branch. Example Capacitor C, 2C, 4C, ... are connected in parallel, then what will be their effective capacitance ? Solution Let the resultant capacitance be C = C + 2C + 4C +... = C[1 +2 + 4 +... ] = C × resultant Example An infinite number of capacitors of capacitance C, 4C, 16C ... are connected in series then what will be their resultant capacitance ? Solution Let the equivalent capacitance of the combination = C eq 1 1 1 1 ... 1 1 1 ... 1 (this is G. P.series) C eq 4 C C 4C 16C 16 a 1 1 11 C eq 3 S 1 r first term a = 1, common ratio r = 4 C 1 4 C eq 1 C 4 EFFECT OF DIELECTRIC NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 • The insulators in which microscopic local displacement of charges takes place in presence of electric field are known as dielectrics. • Dielectrics are non conductors upto certain value of field depending on its nature. If the field exceeds this limiting value called dielectric strength they lose their insulating property and begin to conduct. • Dielectric strength is defined as the maximum value of electric field that a dielectric can tolerate without breakdown. Unit is volt/metre. Dimensions M1L1T–3A–1 Polar dielectrics • In absence of external field the centres of positive and negative charge do not coincide-due to asymmetric shape of molecules. • Each molecule has permanent dipole moment. • The dipole are randomly oriented so average dipole moment per unit volume of polar dielectric in absence of external field is nearly zero. • In presence of external field dipoles tends to align in direction of field. E Ex. Water, Alcohol, CO2, HC, NH3 8
JEE-Physics Non polar dielectrics • In absence of external field the centre of positive and negative charge coincides in these atoms or molecules because they are symmetric. • The dipole moment is zero in normal state. • In presence of external field they acquire induced dipole moment. Ex. Nitrogen, Oxygen, Benzene, Methane Polarisation : The alignment of dipole moments of permanent or induced dipoles in the direction applied electric field is called polarisation. • Polarisation vector P This is a vector quantity which describes the extent to which molecules of dielectric become polarized by an electric field or oriented in direction of field. = the dipole moment per unit volume of dielectric = P np where n is number of atoms per unit volume of dielectric and p is dipole moment of an atom or molecule. qid = qi = i = induced surface charge density. P= np = Ad A Unit of is C/m2 Dimension is L–2T1A1 P E0 E Ei i i - Let E0, V0, C0 be electric field, potential difference and capacitance in absence of dielectric. Let E, V, C are electric field, potential difference and capacitance in presence of dielectric respectively. Electric field in absence of dielectric E0 = V0 Q d 0 0 A Electric field in presence of dielectric i Q Qi V E = E0 – Ei = 0 = 0 = d NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 Capacitance in absence of dielectric C0 = Q V0 Capacitance in presence of dielectric C= Q Qi V The dielectric constant or relative permittivity K or r = E0 = V0 C Q = = E V = = i 0 C0 Q Qi Q 1 1 From K= Q Qi Qi = Q (1 – ) and K= i = (1 – ) K i K E9
JEE-Physics C CAPACITY OF DIFFERENT CONFIGUR ATION In case of parallel plate capacitor C = 0 A d If capacitor is partially filled with dielectric When the dielectric is filed partially between plates, the thickness of dielatric slab is t(t < d). If no slab is introduced between the plates of the capacitor, then a field E given by E0 , exists in a space d. 0 0 On inserting the slab of thickness t, a field E E0 exists inside the slab of thickness t and E0 r a field E exists in remaining space (d – t). If V is total potential thenV = E (d – t) + E t 00 E E0 r t r E= E0 d V = E0 d t t E = Dielectric constant r E0 E0 V t t q t q 0 A 0A ...(i) 0 d r A0 d t r C V d 1 d t 1 1 t 1 r r If medium is fully present between the space. Cmed t = d r Now from equation (i) C 0rA medium d If capacitor is partialy filled by a conducting slab of thickness (t< d). E0 E=0 d 0A t r= r = for conductor C 0 A E0 1 1 dt d t DISTANCE AND AREA DIVISION BY DIELECTRIC r1 r2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 • Distance Division A B (i) Distance is divided and area remains same. (ii) Capacitors are in series. (iii) Individual capacitances are C1 0 r1 A , C2 0r2 A d1 d2 d1 d2 d These two in series 1 1 1 1 d1 d2 1 1 d 1 r2 d2 r1 C 0A r1 r2 C C1 C2 C 0 r1 A 0 r2 A C 0 A d1r2 d2 r1 r1 r2 Special case : If d =d = d C 0 A 2r1 r2 12 2 d r1 r2 10 E
JEE-Physics • Area Division (i) Area is divided and distance remains same. (ii) Capacitors are in parallel. (iii) Individual capacitances are C1 0r 1 A 1 C2 0r 2 A 2 r1 A1 d d r2 These two in parallel so C =C + C2 0 r1 A 1 0 r 2 A 2 0 (r1 A 1 r2A2 ) A2 1 dd d Special case : If A = A = A Then C= 0 A r1 r2 d 2 1 22 • Variable Dielectric Constant : If the dielectric constant is variable, then equivalent capacitance can be obtained by selecting an element as per the given condition and then integrating. (i) If different elements are in parallel, then C = dC , where dC = capacitance of selected differential element. (ii) If different element are in series, then 1 d 1 is solved to get equivalent capacitance C. C C FORCE ON A DIELECTRIC IN A CAPACITOR Consider a differential displacement dx of the dielectric as shown in figure always keeping the net force on it zero so that the dielectric moves slowly without acceleration. Then, W + W =0, where W denotes the Electrostatic F F work done by external agent in displacement dx dx x F Force exerted by an external agent Q2 1 Q2 Q2 Q2 dC W = –W W = U F.dx 2 d C W –F.dx = 2C2 dC F 2C2 dx F Electrostatic F 2C This is also true for the force between the plates of the capacitor. If the capacitor has battery connected to it, then as the p.d. across the plates is maintained constant. V = Q F 1 V 2 dC . C 2 dx NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 Example A parallel plate capacitor is half filled with a dielectric (K) of mass M. Capacitor is attached with a cell of emf E. Plates are held fixed on smooth insulating horizontal surface. A bullet of mass M hits the dielectric elastically and its found that dielectric just leaves out the capacitor. Find speed of bullet. b 2a v0 E M E 11
JEE-Physics Solution Since collision is elastic Velocity of dielectric after collision is v. 0 Dielectric will move and when it is coming out of capacitor a force is applied on it by the capacitor F dU E20b(K 1) dx 2d Which decreases its speed to zero, till it comes out it travels a distance a. 1 E20b(K 1)a 0ab(K 1) 1 / 2 2 2d M v 2 v0 E 0 Md GOLDEN KEY POINTS Example NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 A capacitor has two circular plates whose radius are 8cm and distance between them is 1mm. When mica (dielectric constant = 6) is placed between the plates, calculate the capacitance of this capacitor and the energy stored when it is given potential of 150 volt. Solution Area of plate r2 = × (8 × 10–2)2 = 0.0201 m2 and d = 1mm = 1 × 10–3 m Capacity of capacitor C 0r A 8.85 1012 6 0.0201 = 1.068 × 10–9 F Potential difference d 1 103 V = 150 volt Energy stored U 1 CV 2 1 (1.068 109 ) (150)2 = 1.2 × 10–5 J 22 12 E
JEE-Physics Example A parallel-plate capacitor is formed by two plates, each of area 100 cm2, separated by a distance of 1mm. A dielectric of dielectric constant 5.0 and dielectric strength 1.9 × 107 V/m is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing any dielectric breakdown. Solution If the charge on the capacitor = Q QQ the surface charge density = and the electric field = . A KA0 This electric field should not exceed the dielectric strength 1.9 × 107 V/m. if the maximum charge which can be given is Q then Q 1.9 107 V/m KA0 A = 100 cm2 = 10–2 m2 Q = (5.0) × (10–2) × (8.85 × 10–12) × (1.9 × 107) = 8.4 × 10–6 C. Example The distance between the plates of a parallel-plate capacitor is 0.05 m. A field of 3 × 104 V/m is established between the plates. It is disconnected from the battery and an uncharged metal plate of thickness 0.01 m is inserted into the (i) before the introduction of the metal plate and (ii) after its introduction. What would be the potential difference if a plate of dielectric constant K = 2 is introduced in place of metal plate ? Solution (i) In case of a capacitor as E = (V/d), the potential difference between the plates before the introduction of metal plate V = E × d = 3 × 104 × 0.05 = 1.5 kV (ii) Now as after charging battery is removed , capacitor is isolated so q = constant. If C' and V' are the capacity and potential after the introduction of plate q = CV = C'V' i.e., V' C V C' NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 And as C 0 A and C ' (d 0 A / K) , V ' (d t) (t / K) V d t) (t d VM d t V 0 .05 0 .0 1 1.5 1.2 kV d 0 .0 5 So in case of metal plate as K = , introduced VD (0.05 0.01) (0001 / 2) 1.5 1.35 kV 0.05 And if instead of metal plate, dielectric with K = 2 is E 13
JEE-Physics CHARGING & DISCHARGING OF A CAPACITOR Charging Discharging • When a capacitor, resistance, battery, and • When a charged capacitor, resistance and keys is key is conected in series and key is closed, conected in series and key is closed. Then energy then stored in capacitor is used to circulate current in the circuit. VR VC VC VR R CS CR V S –+ • Charge at any instant • Charge at any instant Q Q dQ R V +V =0 V = V + V IR C dt CR C RC Q CV 1 et RC = Q0 1 et RC Q=Q e–t/RC 0 At t = = RC = time constant At t = = RC = time constant Q = Q [1 – e–1] = 0.632 Q Q = Q e–1 = 0.368 Q 00 00 So, in charging, charge increases to So, in discharging, charge decreases to 36.8% of the initial charge in the time equal to . 63.2% of charge in the time equal to . • Current at any instant • Current at any instant i = dQ/dt = i0et/RC {i = Q /RC} i = dQ/dt = i0et / RC {i = Q /RC} 00 00 • Potential at any instant • Potential at any instant V V0 (1 e t RC ) V V0 e t RC Example Find the time constant for given circuit if R1 = 4 , R2 = 12, C1 = 3F and C2 = 6F. C1 C2 R2 NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 V0 R1 Solution Given circuit can be reduced to : C = C1C2 36 2F , R R1R 2 4 12 3 C1 C2 36 = R1 R2 4 12 C Time constant = RC = (3) (2 × 10–6) = 6s V0 R 14 E
JEE-Physics Example A capacitor of 2.5 F is charged through a series resistor of 4M. In what time the potential drop across the the capacitor will become 3 times that of the resistor. (Given : n2 = 0.693) Solution 4M 2.5 F VR VC V0 VC = V0(1 – e–t/RC) VC = 3VR V0 = VC + VC 3 3 VC = 4 V0 33 1 4=et/RC 4 V0 = V0(1 – e–t/RC) = 1–e–t/RC =e–t/RC 4 4 t n4 t = RC n4 = 2RC n2 = 2 × 4 × 106 × 2.5 × 10–6 × 0.693 = 13.86 s RC NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 E 15
JEE-Physics SOME WORKED OUT EXAMPLES Example#1 The switch in circuit shifts from 1 to 2 when V > 2V/3 and goes back to 1 from 2 when V < V/3. The CC voltmeter reads voltage as plotted. What is the period T of the wave form in terms of R and C? 1 Voltage 2 2V/3 t2 t1 C VC T V/3 V time (A) RC n3 (B) 2RC n 2 (C) RC n3 (D) RC n3 2 3 Solution Ans. (B) During time 't ' capacitor is discharging with the help of resistor 'R' q = q e–t/RC 2 0 V = V e –t/RC [ Q=CV] 0 As V = 2V V = RC n2 ;V= ;t 03 32 During time 't ' capacitor is charging with the help of battery. 1 q = q (1-e–t/RC) or V = V (1-e )–t/RC 0 0 as V = 2V V = RC n2 ;V= ;t 03 31 T = t1 + t2 = 2RC n2 Example#2 Seven capacitors, each of capacitance 2F are to be connected to obtain a capacitance of 10/11 F. Which of the following combinations is possible ? (A) 5 in parallel 2 in series (B) 4 in parallel 3 in series (C) 3 in parallel 4 in series (D) 2 in parallel 5 in series NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit-07\\Capacitance\\English\\theory.p65 Solution Ans. (A) 5(2F) in series with 2F series with 1F, Ceq= 10 1 10 F 2 , 10F in 10 1 11 Example#3 In the circuit shown, if potential of A is 10V, then potential of B is - 1F B 1F (A) 25/3 V (B) 50/3 V 1F (C) 100/3 V (D) 50 V A 10V 16 E
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