Ncert solutions physics mtg class 12

JOIN ON TELEGRAM @iitjeeadv JOIN ON TELEGRAM @iitjeeadv NCERT SOLUTIONS PHYSICS CLASS 12 MTG Learning Media (P) Ltd New Delhi | Gurgaon www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Price ` 125 Published by : MTG Learning Media (P) Ltd., New Delhi Corporate Office : Plot 99, Sector 44 Institutional Area, Gurgaon, Haryana Phone : 0124 - 6601200 Web: mtg.in Email: info@mtg.in Registered Office : 406, Taj Apt., Ring Road, Near Safdarjung Hospital, New Delhi-110029 Information contained in this book has been obtained by mtg, from sources believed to be reliable However, neither mtg nor its authors guarantee the accuracy or completeness of any information published herein, and neither mtg nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information Copyright © MTG Learning Media (P) Ltd No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication This edition can be exported from India only by the publishers, MTG Learning Media (P) Ltd Visit www.mtg.in for buying books online www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Contents Chapters Page No Electric Charges and Fields Electrostatic Potential and Capacitance 15 Current Electricity 33 Moving Charges and Magnetism 46 Magnetism and Matter 61 Electromagnetic Induction 73 Alternating Current 86 Electromagnetic Waves 100 Ray Optics and Optical Instruments 108 10 Wave Optics 136 11 Dual Nature of Radiation and Matter 145 12 Atoms 164 13 Nuclei 171 14 Semiconductor Electronics: Materials, Devices 186 and Simple Circuits 15 Communication Systems www.pdfgrip.com 194 JOIN ON TELEGRAM @iitjeeadv www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Electric Charges and Fields Chapter 1 Electric Charges and Fields What is the force between two small charged spheres having charges of × 10–7C and × 10–7C placed 30 cm apart in air? × 10−7 × × 10−7 q1q2 = × 10 × or F = × 10–3 N 4πε r ( 30 × 10−2 )2 (repulsive in nature, as the two charges are like charges.) Soln F = The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in air is 0.2 N (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Soln (a) Given, F12 = 0.2 N, q1­= 0.4 × 10–6 C, q2 = –0.8 × 10–6 C q1q2 q1q2 0.4 × 10 −6 × 0.8 × 10 −6 or r2 = = × 109 × 4πε r 4πε F12 0.2 or r2 = 14.4 × 10–3 = 1.44 × 10–2 or r = 1.2 × 10–1 m = 12 cm (b) By Newton’s third law, F21 = F12 = 0.2 N attractive As, F12 = Ke is dimensionless Look up a table of Gme mp Physical Constants and determine the value of this ratio What does the ratio signify? Soln The ratio of electrostatic force to the gravitational force between an electron and a proton separated by a distance r from each other is Felec Ke e / r Ke = = Fgrav Gm m / r Gme mp e p Check that the ratio In terms of dimensions  Ke   Felec  MLT −2  = =1 So,   =   Gme mp   Fgrav  MLT −2 Thus, the ratio Ke is dimensionless Gme mp www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv 2 NCERT Solutions (Class XII) × 109 × (1.6 × 10 −19 )2 Ke = − 11 Gme mp 6.67 × 10 × 9.1 × 10 −31 kg × 1.67 × 10 −27 kg = 0.23 × 1040 = 2.3 × 1039 This ratio signifies that electrostatic forces are 1039 times stronger than gravitational forces (a) Explain the meaning of the statement ‘electric charge of a body is quantised’ (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges? Soln (a) The net charge possessed by a body is an integral multiple of charge of an electron i.e q = ± ne, where n = 0, 1, 2, 3, is the number of electrons lost or gained by the body and e = 1.6 × 10–19 C is charge of an electron This is called law of quantization of charge (b) At macroscopic level, charges are enormously large as compared to the charge of an electron, e = 1.6 × 10–19C Even a charge of nC contains nearly 1013 electronic charges So, at this large scale, charge can have a continuous value rather than discrete integral multiple of e, and hence, the quantization of electric charge can be ignored When a glass rod is rubbed with a silk cloth, charges appears on both A similar phenomenon is observed with many other pairs of bodies Explain how this observation is consistent with the law of conservation of charge Soln When a glass rod is rubbed with a silk cloth, electrons from the glass rod are transferred to the piece of silk cloth Due to this, the glass rod acquires positive (+) charge whereas the silk cloth acquires negative (–) charge Before rubbing, both the glass rod and silk cloth are neutral and after rubbing the net charge on both of them is also equal to zero Such similar phenomenon is observed with many other pairs of bodies Thus, in an isolated system of bodies, charge is neither created nor destroyed, it is simply transferred from one body to the other So, it is consistent with the law of conservation of charge Four point charges qA = + µC, qB = –5µC, qC = + µC and qD = –5 µC are located at the corners of a square ABCD of side 10 cm What is the force on a charge of µC placed at the centre of the square? Soln Forces of repulsion on mC charge at O due to mC charge, at A and C are equal and opposite Therefore, they cancel Similarly, forces of attraction on mC charge at O, due to –5 mC charges at B and at D are also equal and opposite Therefore, these also cancel Hence, the net force on the charge of l mC at O is zero www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Electric Charges and Fields (a) An electrostatic field line is a continuous curve That is, a field line cannot have sudden breaks Why not? (b) Explain why two field lines never cross each other at any point? q = –5 C Soln (a) q = +2 C B A A  O  FOD C  FOC FOB +1 C q = +2 C C B  FOA D q = –5 C D     As, F = − F  or  FOA + FOC = 0 OC OA So, the net force on charge of + µC at O is       F = FOA + FOC + FOB + FOD or F = .(i) A field line cannot have sudden breaks because the moving test charge never jumps from one position to the other (b) Two field lines never cross each other at any point, because if they so, we will obtain two tangents pointing in two different directions of electric field at a point, which is not possible Two point charges qA = µC and qB = –3 µC are located 20 cm apart in vacuum (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge? Soln (a) Electric fields at O due to the charges at A and B are EOA = EOB = × 10 −6 C q = × 10 × ( 20 / × 10 −2 m )2 4πε r or EOA = EOB = 27 × 105 N C–1 +3 C +1 EOA EOB A 10 cm O 10 cm B As, they are directed in same direction, so net electric field at midpoint O is E = EOA + EOB = 2EOB = × 27 × 105 or E = 5.4 × 106 N C–1 directed along OB (b) When test charge q0 = –1.5 × 10–9 C is placed at point O, it experiences a force F = q0E = 1.5 ×10–9 × 5.4 × 106 or F = 8.1 × 10–3N In direction opposite to that of E i.e along OA www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv 4 NCERT Solutions (Class XII) –7 A system has two charges qA = 2.5 × 10 C and qB = –2.5 × 10–7C located at point A (0, 0, –15  cm) and B (0, 0, +15 cm), respectively What are the total charge and electric dipole moment of the system? Soln The total charge of the system is, qnet = qA + qB or qnet = 2.5 × 10–7 – 2.5 × 10–7 or qnet = and the total electric dipole moment of the system is qB = – 2.5 × 10–7C qA = 2.5 × 10–7C A (0,0, – 15 cm) P B (0,0, + 15 cm) p = q.2a = 2.5 × 10–7C × × 15 × 10–2m or p = 75 × 10–9 C m or p = 7.5 × 10–8 C m directed along BA 10 An electric dipole with dipole moment × 10–9 C m is aligned at 30° with the direction of a uniform electric field of magnitude × 104 N C–1 Calculate the magnitude of the torque acting on the dipole Soln Torque on dipole is, τ = pE sin 30° or τ = × 10–9 × × 104 × or τ = 10 × 10–5 or τ = × 10–4 N m 11 A Polythene piece rubbed with wool is found to have a negative charge of × 10–7 C (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene? q × 10 −7 = 1.875 × 1012 = e 1.6 × 10 −19 so, 1.875 × 1012 electrons have transferred from wool to polythene, as polythene acquires negative charge (b) Yes, mass of 1.875 × 1012 electrons i.e., m = nme = 1.875 × 1012 × 9.1 × 10–31 kg = 1.71 × 10–18 kg has transferred from wool to polythene Soln (a) q = ne or n = 12 (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10–7C ? The radii of A and B are negligible compared to the distance of separation (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved? q1q2  Soln (a) F = 50 cm 4πε r A B −7 (6.5 × 10 ) or F = × 10 q1 = 6.5 × 10–7C q = 6.5 × 10–7C (0.5)2 or F = 1.520 × 10–2 N www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Electric Charges and Fields (b) q1′ = 2q1 , q2′ = 2q2 and r ′ = r 2q1 × 2q2 q1′ × q2′ = 2 πε 4πε (r ′ ) r   or F ′ = 16 × 1.520 × 10–2 = 24.32 × 10–2 N or F ′ = 0.24 N So, F ′ = 13 Suppose the spheres A and B in previous question have identical sizes A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both What is the new force of repulsion between A and B? Soln B A C q1 = 6.5 × 10–7 C, q2 = 6.5 × 10–7 C A q1 = 3.25 × 10–7 C B –7 q C, net = (3.25 + 6.5) × 10 C = 9.75 × 10–7 C A B q1 = 3.25 × 10–7 C, q2 = 4.875 × 10–7 C The new force of repulsion between A and B is q1′′× q2′′ 4πε (r ′′ )2 F ″ = or F ″ = × 109 × 3.25 × 10 −7 × 4.875 × 10 −7 (0.5) or F ″ = 5.7 × 10–3 N 14 Figure shows tracks of three charged particles in a uniform electrostatic field Give the signs of the three charges Which particle has the highest charge to mass ratio? Soln Particles and are negatively charged F = +q E as they experience forces in direction opposite  to that of electric field E , whereas particle is positively charged as it experience force in the E  direction of electric field E Particle-3 has the highest charge to mass ratio, as it shows maximum deflection in the electric field www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv 182 NCERT Solutions (Class XII) Hence initially 90% of the mixture is P-32 and 10% of the mixture is P-33 Let after time ‘t’ the mixture is left with 10% of P-32 and 90% of P-33 Half life of both P-32 and P-33 are given as 14.3 days and 25.3 days respectively Let ‘x’ be total mass undecayed initially and ‘y’ be total mass undecayed finally Let initial number of P-32 nuclides = 0.9 x Final number of P-32 nuclides = 0.1 y Similarly, initial number of P-33 nuclides = 0.1x Final number of P-33 nuclides = 0.9 y For isotope P-32 N = N0 − t /T1/ or 0.1 y = 0.9 x −t /14.3  .(i) For isotope P-33 N = N0 −t /T1/ or 0.9 y = 0.1 x −t / 25.3  .(ii) On dividing, we get t  t  +  − 1 −t /14.3 = −t / 25.3 or = 2 14.3 25.3  81  11   11  −t   t  14.3 × 25.3  =2  or 81 =  14.3 × 25.3  81 Taking log 11 t   11 × 0.3010 log e 81 = t  log e or 1.9082 =  25.3 × 14.3  14.3 × 25.3  t = 208.5 days  209 days 26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle Consider the following decay processes: 223 209 14 223 219 88 Ra → 82 Pb + C , 88 Ra → 86 Rn + 2He Calculate the Q-values for these decays and determine that both are energetically allowed Soln Let us calculate Q value for the given decay process For first decay process Q=m ( 223 88 Ra ) − m( 209 82 Pb ) − m( C) 14 Q = [223.01850 – 208.98107 – 14.00324] (c2)u = [0.034109] × 931.5 MeV = 31.85 MeV For the second decay process Q = m 223 Ra − m 219 Rn − m He ( 88 ) ( 86 ) ( ) www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Nuclei 183 Q = [223.01850 – 219.00948 – 4.00260] c u Q = 0.00642 × 931.5 MeV = 5.98 MeV Since, Q value is positive in both the cases, hence decay process in both ways are possible 27 Consider the fission of 238 92 U by fast neutrons In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments are 140 and 99 44 Ru Calculate Q for this fission process 58 Ce The relevant atomic and particle masses are: m ( 238 92 U ) = 238 05079 u, m( 140 58 Ce ) = 139 90543 u, m( 99 44 Ru ) = 98 90594 u Soln The fission of U-238 by fast neutrons into fragments Ce-140 and Ru99 with energy released Q, The Q value, Q = [m(U-238) + mn – m (Ce-140) – m(Ru-99)c2 Q = [238.05079 + 1.00867 – 139.90543 – 98.90594] amu × c2 Q = 0.24809 × 931.5 MeV = 231.09 MeV = 231.1 MeV 28 Consider the D-T reaction (deuterium - tritium fusion) + 13H → 24He + n (a) Calculate the energy released in MeV in this reaction from the data m(12H) = 2.014102 u, m(13H) = 3.016049 u (b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gases be heated to initiate the reaction? Soln (a) For the process 21H + 31H → 42He + n + Q Q = [m(21H) + m(31H) – m(42He) – mn] × 931 MeV = [2.014102 + 3.016049 – 4.002603 – 1.00867] × 931 MeV = 0.018878 × 931 = 17.58 MeV (b) Repulsive potential energy of two nuclei when they almost touch each other is q2 × 109 (1.6 × 10 −19 )2 = joule = 5.76 × 10 −14 joule = 4πε0 (2r) × × 10 −15 1H Classically, K.E atleast equal to this amount is required to overcome Coulomb repulsion Using the relation 5.76 × 10 −14 (K.E.) = = 1.39 × 109 K K.E = × kT ⇒ T = −23 3k × 1.38 × 10 29 Obtain the maximum kinetic energy of β-particles and the radiation frequencies of γ-decays in the decay scheme shown in figure You are given 198 that m(198 79Au) = 197.968233 u, m( 80Hg) = 197.966760 u www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv 184 NCERT Solutions (Class XII) Soln Energy corresponding to γ1 E1 = 1.088 – = 1.088 MeV = 1.088 × 1.6 × 10–13 joule \ Frequency, υ(γ1) = Similarly, u(γ ) = and u(γ ) = E1 1.088 × 1.6 × 10 −13 = = 2.626 × 10 20 Hz h 6.63 × 10 −34 E2 0.412 × 1.6 × 10 −13 = = 9.95 × 1019 Hz h 6.63 × 10 −34 E3 (1.088 − 0.412) × 1.6 × 10 −13 = = 1.631 × 1020 Hz h 6.63 × 10 −34 Maximum K.E of β1 particle 198 Kmax(β1) = [m(198 79Au) – mass of second excited state of 80Hg] × 931 MeV 1.088   × 931 MeV =  m(198 Au − m(198 80 Hg) −  79 931  = 931 [197.968233 – 197.966760] – 1.088 MeV = 1.371 – 1.088 = 0.283 MeV Similarly, Kmax(β2) = 0.957 MeV 30 Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within the Sun and (b) the fission of 1.0 kg of 235U in a fission reactor Soln (a) In the fusion reactions taking place within core of sun, hydrogen nuclei combines to form a helium nucleus with the release of 26 MeV of energy 11H → 42He + 2e + + 26 MeV Number of atoms in kg of 11H , n= 1000 g × × 10 23 1000 g = × × 10 23 = × 1026 atoms Atomic mass 1g www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Nuclei 185 Energy released in the fusion of kg of 11H , × 10 26 × 26 MeV = 39 × 1026 MeV (b) Energy released per fission of U-235 is 200 MeV Number of atoms in kg of U-235, 1000 g × × 10 23 n= = 25.53 × 10 23 atoms 235 g Total energy released for fission of kg of uranium, E2 = 25.53 × 1023 × 200 MeV = 5.1 × 1026 MeV E1 = E1 39 × 10 26 = = 7.65 ≈ E2 5.1 × 10 26 So the energy released in fusion of kg of Hydrogen is nearly times the energy released in fission of kg of uranium-235 31 Suppose India has a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which is to be obtained from nuclear power plants Suppose we are given that, on an average, the efficiency of utilization (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25% How much amount of fissionable uranium would our country need per year? Take the heat energy per fission of 235U to be about 200 MeV Soln 10% of total power 200,000 MW to be obtained from nuclear power plant by 2020 AD So, power from nuclear plants = × 105 × 0.1 MW = × 104 MW = × 1010 W With efficiency of power plants 25% only, the energy converted into electrical energy per fission 25 = × 200 = 50 Mev = 50 × 1.6 × 10–13 Joule = × 10–3 100 Total energy to be produced = × 104 MW = × 1010 joule/sec × 10 24 × 36 × 24 × 365 = × 1010 × 60 × 60 × 24 × 365 joule/year = Mass of 6.023 × 1023 atoms of 235 U = 235 g = 235 × 10–3 kg × 36 × 24 × 365 × 10 24 atoms Mass of 235 × 10 −3 × 36 × 24 × 365 × 10 24 = × = 3.08 × 104 kg 23 6.023 × 10 Hence mass of uranium needed per year= 3.08 × 104 kg a  b www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv 186 Chapter 14 NCERT Solutions (Class XII) Semiconductor Electronics: Materials, Devices and Simple Circuits In an n-type silicon, which of the following statement is true? (a) Electrons are majority carriers and trivalent atoms are the dopants (b) Electrons are minority carriers and pentavalent atoms are the dopants (c) Holes are minority carriers and pentavalent atoms are the dopants (d) Holes are majority carriers and trivalent atoms are the dopants Soln (c) : For n-type silicon statement (c) is true Which of the statements given in previous question is true for p-type semiconductors? Soln (d) : For p-type semiconductors statement (d) is true Carbon, silicon and germanium have four valence electrons each These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge Soln (c) : The energy band gap is largest for carbon, less for silicon and least for germanium So, the correct statement is (c) (Eg)C > (Eg)Si > (Eg)Ge In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them (b) they move across the junction by the potential difference (c) hole concentration in p-region is more as compared to n-region (d) all of the above Soln (c) : In the unbiased p-n junction, holes diffuse from the p-region to n-region because holes concentration in the p-region is high as compared to n-region www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Semiconductor Electronics: Materials, Devices and Simple Circuits 187 When a forward bias is applied to a p-n junction, it (a) raises the potential barrier (b) reduces the majority carrier current to zero (c) lowers the potential barrier (d) none of the above Soln (c) : Under forward biasing the movement of majority charge carriers across the junction reduces the width of depletion layer or lowers the potential barrier For transistor action, which of the following statements is/are correct: (a) Base, emitter and collector regions should have similar size and doping concentrations (b) The base region must be very thin and lightly doped (c) The emitter junction is forward biased and collector junction is reverse biased (d) Both the emitter junction as well as the collector junction are forward biased Soln (b, c) For a transistor circuit in action, statements (b) and (c) are correct For a transistor amplifier, the voltage gain (a) remains constant for all frequencies (b) is high at high and low frequencies and constant in the middle frequency range (c) is low at high and low frequencies and constant at mid frequencies (d) none of the above Soln (c) : The voltage gain in a transistor as amplifier is low at high and low frequencies and constant at mid frequencies In half-wave rectification, what is the output frequency if the input frequency is 50 Hz What is the output frequency of a full-wave-rectifier for the same input frequency? Soln In half wave rectification, only one ripple is obtained per cycle in the output www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv 188 NCERT Solutions (Class XII) Output frequency of a half wave rectifier = input frequency = 50 Hz In full wave rectification, two ripples are obtained per cycle in the output Output frequency = × input frequency = × 50 = 100 Hz For a CE-transistor amplifier, the audio signal voltage across the collector resistance of kΩ is V Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is kΩ Soln Here RC = kΩ = 2000 Ω, RB = kΩ = 1000 Ω, β = 100, V0 = V V βR 2000 Voltage gain, = C or = 100 × Vi RB Vi 1000 ∴ Input signal voltage, Vi = 0.01 V (i) V 0.01 V 0.01 V (ii) I B = i = = × 10–5 A = 10 m A = RB 1000 Ω × 10 Ω 10 Two amplifiers are connected one after the other in series (cascaded) The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20 If the input signal is 0.01 volt, calculate the output ac signal Soln Total voltage gain can be calculated as Av = Av1 × Av2 = 10 × 20 = 200 So, output voltage, Av = Voutput Vinput V or 200 = out or Vout = V 0.01 11 A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV Can it detect a wavelength of 6000 nm? Soln Energy of the incident photon with a band gap of 6000 nm E= 6.63 × 10 −34 × × 10 = 0.207 eV hc hc J = eV = λ eλ 1.6 × 10 −19 × × 10 −6 www.pdfgrip.com JOIN ON TELEGRAM @iitjeeadv Semiconductor Electronics: Materials, Devices and Simple Circuits 189 The photodiode need an energy of 2.8 eV to give response to incident light As E < Eg, the given photodiode cannot detect the radiation of wavelength 6000 nm 12 The number of silicon atoms per m3 is × 1028 This is doped simultaneously with × 1022 atoms per m3 of Arsenic and × 1020 per m3 atoms of Indium Calculate the number of electrons and holes Given that ni = 1.5 × 1016 m–3 Is the material n-type or p-type? Soln We know that for each atom doped of Arsenic, one free electron is received Similarly, for each atom doped of indium, a vacancy is created So, the number of free electrons introduced by pentavalent impurity added ne = NAs = × 1022 m–3 The number of holes introduced by trivalent impurity added nh = NIn = × 1020 m–3 We know the relation, nenh = ni2 (i) Now net electrons, n′e = ne – nh = × 1022 – × 1020 = 4.95 × 1022 m–3 (ii) Now using equation (i), net holes n (1.5 × 1016 ) = 4.5 × 10 m −3 n′ h = i = n′ e 4.95 × 10 22 So, n′e >> n′h, the material is of n-type 13 In an intrinsic semiconductor the energy gap Eg is 1.2 eV Its hole mobility is much smaller than electron mobility and independent of temperature What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration ni is  Eg  given by ni = n0 exp  − 2k T  , where n0 is a constant B kB = 8.62 × 10–5 eV K–1 Soln Conductivity is given by σ = e (ne µe + nh µh) For intrinsic semiconductor ne = nh = ni Also mobility of holes (µh)