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7.%20Atomic and Molecular Physics NET-JRF%20June%202011-Dec%202014

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Brijmohan
Banaras Hindu University (BHU), Varanasi
PhD Condensed matter Physics
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fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES ATOMIC AND MOLECULAR PHYSICS NET/JRF (JUNE-2011) Q1. Consider the energy level diagram (as shown in the figure below) of a typical three level ruby laser system with 1.6 1019 Chromium ions per cubic centimeter. All the atoms excited by the 0.4 m radiation decay rapidly to level E2 which has a lifetime = 3 ms. E3 0.4 m 0.7 m E2 E1 A. Assuming that there is no radiation of wavelength 0.7 m present in the pumping cycle and that the pumping rate is R atoms per cm3, the population density in the level N2 builds up as: (a) N2 (t) = R (et/ 1) (c) N 2 t Ans: Rt 2 (b) N2 (t) = R (1 e-t/ ) 1 e t / (d) N2 (t) = R t (b) N 2 t R 1 e t / B. The minimum pump power required (per cubic centimeter) to bring the system to transparency, i.e. zero gain, is (a) 1.52 kW (b) 2.64 kW (c) 0.76 kW (d) 1.32 kW Ans: (c) Solution: The Minimum Power required to achieve zero gain is P N hv N hc 1.6 10 19 6.6 10 34 3 10 8 754 W cm 3 6 3 2 0.7 10 3 10 2 2 P 0.76kW per cubic centimeter Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 1 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NET/JRF (DEC-2011) Q2. Given that the ground state energy of the hydrogen atom is 13.6 eV, the ground state energy of positronium (which is a bound state of an electron and a positron) is (b) 6.8 eV (a) + 6.8 eV Ans: (c) 13.6 eV (b) The energy expression for Positronium atom is E n For n = 1, E1 Q3. (d) 27.2 eV 13.6 eV 6.8eV , 2 13.6 eV 2n 2 E1 6.8 eV A laser operating at 500 nm is used to excite a molecule. If the Stokes line is observed at 770 cm-1, the approximate positions of the Stokes and the anti-Stokes lines are (a) 481.5 nm and 520 nm (b) 481.5 nm and 500 nm (c) 500 nm and 520 nm (d) 500 nm and 600 nm Ans: Solution: Given 0 500 nm 5 10 5 , v stoke 770 cm 1 v 0 20 ,000 cm 1 Raman shift v v0 v stoke 19230 cm 1 Wave number of anti-stokes line is v ami stoke v v0 = 39,230 cm-1 In wavelength term anti stoke 2.549 10 7 254 .9 nm and stoke 12987 nm Q4. If the hyperfine interaction in an atom is given by H aS e S p where S e and S p denote the electron and proton spins, respectively, the splitting between the 3 S1 and 1 S 0 state is (a) a 2 / 2 Ans: (b) a 2 (c) a 2 / 2 (d) 2a 2 (b) 1 2 2 Solution: Total spin is S S e S p S 2 S e2 S p 2S e S p S e S p S 2 S e2 S p 2 a 3 2 2 where S e2 S p S S 1 2 2 H aS e S p S 2 S e2 S p 2 4 H a 2 3 2 3 2 a 2 3 2 S S 2 4 4 2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 2 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES For 3 S1 : S 1 S 2 S S 1 2 2 2 , For 1 S 0 : S 0 S 2 S S 1 2 0 2 H1 a 3 a 2 2 2 2 2 4 and H 2 a 3 3 2 0 2 a 2 2 4 The splitting between Q5. 3 for 1 S 0 1 3 S1 and 1 S 0 is H H 1 H 2 a 2 a 2 4 4 The ratio of intensities of the D1 and D2 lines of sodium at high temperature is (a) 1:1 Ans: for 3 S1 (b) 2:3 (c) 1:3 (d) 1:2 (d) Solution: The electronic transition for D2 and D1 line is D2 : 2 p3 / 2 2 S1 / 2 , D21 : 2 p 3 / 2 2 S1 / 2 Q6. 3 1 I D2 2 J 2 1 4 2 2 1 I D1 2 J 1 1 2 1 2 1 2 2 An atom of mass M can be excited to a state of mass M by photon capture. The frequency of a photon which can cause this transition is (a) Ans: c 2 2h (b) c 2 h (c) 2 c 2 2 Mh (d) c 2 2M 2 Mh (d) Solution: The conversation law of energy and Momentum give Mc2 h M c 4 p 2 c 2 2 1/ 2 and h p c M 2 c 4 h 2 2 2Mc2 h M 2 c 4 2 c 4 2M c 4 h 2 2 2Mc2 h 2 c 4 2M c 4 c 2 2 Mc 2 h 2 M c 4 1 h 2M c 2 1 2M . 2 Mh 2M Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 3 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NET/JRF (JUNE-2012) Q7. 12 16 C O is at 3.842 cm-1 while that of The first absorption spectrum of 13 16 C O is at 3.673 cm-1. The ratio of their moments of inertia is (a) 1.851 Ans: (b) 1.286 (c) 1.046 (d) 1.038 (c) Solution: For 12 C 16 O : 28 1 3.842 cm 1 B1 1.921 cm 1 2 B 2 3.673 cm 1 B 2 1.8365 cm 1 h I 2 B1 1.921 1.046 Where, B 2 I1 B2 1.8365 8 IC For Q8. 13 C 16 O : The spin-orbit interaction in an atom is given by H = a L.S, where L and S denote the orbital and spin angular momenta, respectively, of the electron. The splitting between the levels 2P3/2 and 2P1/2 is (a) 3 2 a 2 (b) 1 2 a 2 (c) 3a 2 2 (d) 5 2 a 2 Ans: Solution: Given H aL S where J L S . 1 a J 2 L2 S 2 2 L S L S J 2 L2 S 2 H J 2 L2 S 2 2 2 For 3 P3 / 2 : S 1 which gives S 2 S S 1 2 2 2 L = 1 which gives L2 L L 1 2 2 2 J a 15 a 3 15 which gives J 2 J J 1 2 2 H 1 2 2 2 2 2 4 8 2 4 For 2 P3 / 2 : S 1 3 which gives S 2 S S 1 2 2 2 4 L = 1 which gives L2 L L 1 2 2 2 J a 3 3 1 3 which gives J 2 J J 1 2 2 H 2 2 2 a 2 2 4 4 2 4 a 7 2 1 8 2 H H 1 H 2 2 a 2 a H a 8 8 8 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 4 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q9. The spectral line corresponding to an atomic transition from J = 1 to J = 0 states splits in a magnetic field of 1 kG into three components separated by 1.6 10-3 . If the zero field spectral line corresponds to 1849 , what is the g-factor corresponding to the J = 1 state? (You may use hc 0 (a) 2 Ans: 2 10 4 cm) (b) 3/2 (c) 1 (d) 1/2 (c) Solution: The Zeeman splitting is E gM J B B g B B for MJ = +1 Given, Zeeman splitting separations g c 2 3 108 2 c 1.6 10 3 1.6 10 1 0.1404 1010 1849 10 10 2 E 6.625 10 34 0.1404 1010 1.00 B B 9.27 10 24 0.1 g 1.0 NET/JRF (DEC-2012) Q10. Consider the energy level diagram shown below, which corresponds to the molecular 2 nitrogen laser. 21 R 1 1 If the pump rate R is 10 20 atoms cm -3 s -1 0 and the decay routes are as shown with 21 20 ns and 1 1 s , the equilibrium populations of states 2 and 1 are, respectively, (a) 1014 cm-3 and 2 1012 cm-3 (c) 2 1012 cm-3 and 2 106 cm-3 Ans: (b) 2 1012 cm-3 and 1014 cm-3. (d) zero and 1020 cm-3 (b) Solution: dN 2 N dN1 N 2 N 1 R 2 and . dt 21 dt 21 1 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 5 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Under equilibrium condition dN 2 dN1 0 dt dt N 2 21 R 20 20 10 9 2 10 12 cm 3 N1 Q11. 1 N 2 10 6 2 1012 cm 3 1014 cm 3 9 21 20 10 Consider a hydrogen atom undergoing a 2P 1S transition. The lifetime tsp of the 2P state for spontaneous emission is 1.6 ns and the energy difference between the levels is 10.2 eV. Assuming that the refractive index of the medium n0 = 1, the ratio of Einstein coefficients for stimulated and spontaneous emission B21 / A21 is given by (a) 0.683 1012 m3J-1s-1 (c) 6.83 1012 m3J-1s-1 Ans: (b) 0.146 10-12 Jsm-3 . (d) 1.463 10-12 Jsm-3 . (a) Solution: n 1, E 10.2 eV and Q12. B21 2c3 2 2 c 3 0.67 1012 . 3 3 3 3 A21 n E n0 Consider a He-Ne laser cavity consisting of two mirrors of reflectivities R1 = 1 and R2 = 0.98. The mirrors are separated by a distance d = 20 cm and the medium in between has a refractive index n0 = 1 and absorption coefficient = 0. The values of the separation between the modes and the width p of each mode of the laser cavity are: (a) 75 kHz, p 24 kHz (c) 750 MHz , p 2.4MHz Ans: (b) 100 kHz, p 100 kHz (d) 2.4MHz , p 750 MHz (a) Solution: Mode separation c 750 MHz 2dn0 where c 3 108 m / sec , c 20 10 2 m and n0 1 . Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 6 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Width of each mode p 1 where t c 2 t c 2n0 d 1 c ln R R e 2 d 1 2 . Note: In this question there is no need to calculate p since in the given options there is only one option with 750 MHz and i.e option (a). You can calculate p without calculator so use your common sense. Q13. NET/JRF (JUNE-2013) A muon from cosmic rays is trapped by a proton to form a hydrogen-like atom. Given that a muon is approximately 200 times heavier than an electron, the longest wavelength of the spectral line (in the analogue of the Lyman series) of such an atom will be o o (a) 5.62 A Ans: (b) 6.67 A o o (c) 3.75 A (d) 13.3 A (b) Solution: In case of muonic atom, the reduce mass is m ' m ' E1 E1 ' En m n 2 180 n 2 e m m p m m p 180me where, E1 13 .6eV ' For ground state of muonic atom n=1, E1 180 E1 ' For first excited state of muonic atom n=2, E 2 45E1 The longest wavelength of the photon corresponds to the transition between first and ground state of muonic atom. The energy difference between first excited and ground state is ' E E 2 E1' 135 E1 1836 eV 2938 10 19 J In term of wavelength E hv hc o hc 6.63 10 34 3 10 8 6.67 10 10 m 6.67 A 19 E 2938 10 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 7 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q14. Consider the hydrogen-deuterium molecule HD. If the mean distance between the two atoms is 0.08 nm and the mass of the hydrogen atom is 938 MeV / c 2 , then the energy difference E between the two lowest rotational states is approximately Ans: (c) 2 10 2 eV (b) 10 2 eV (a) 10 1 eV (d) 10 3 eV (b) Solution: Rotational energy expression E h2 8 2 I J J 1 BJ J 1 Difference between two lowest energy levels is E 2B where Here, B h2 8 I 2 2 and 2I I r 2 MHMD M 2M H 2 H MH , M H M D M H 2M H 3 1.01 10 34 J sec 1.01 10 34 2 3 938 MeV/c2 1 1019 eV sec 6.3 10 16 eV sec 1.6 6.3 10 16 eV sec 2 E , E 9.2 10 3 eV 10 2 eV 2 2 I 938 106 eV / c 2 0.08 10 9 m 3 2 Q15. The electronic energy levels in a hydrogen atom are given by E n 13 .6 / n 2 eV. If a selective excitation to the n 100 level is to be made using a laser, the maximum allowed frequency line-width of the laser is (a) 6.5 MHz Ans: (b) 6.5 GHz (c) 6.5 Hz (d) 6.5 kHz (b) Solution: En 13.6 / n 2 En 2 13.6 2 13.6 eV h 1.6 10 19 6.5 GHz . 3 3 n n Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 8 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q16. Consider the laser resonator cavity shown in the figure. If I1 is the intensity of the radiation at mirror M 1 and R1 1 R2 R l is the gain coefficient of the medium between the mirrors, then the energy density of photons in the plane P at P M1 distance x from M 1 is (a) I 1 / c e x M2 (b) I 1 / c e x (c) I 1 / c e x e x Ans: x a (d) I 1 / c e 2 x (c) NET/JRF (DEC-2013) Q17. A perturbation V pert aL2 is added to the Hydrogen atom potential. The shift in the energy level of the 2P state, when the effects of spin are neglected up to second order in a , is (a) 0 Ans: (b) 2a 2 a 2 4 3 (d) a 2 a 2 4 2 (c) 2a 2 (c) Solution: For 2P state, L=1 The shift in the energy due to perturbation V pert aL2 is E aL( L 1) 2 a1(1 1) 2 2a 2 Q18. A gas laser cavity has been designed to operate at 0.5 m with a cavity length of 1m . With this set-up, the frequency is found to be larger than the desired frequency by 100 Hz. The change in the effective length of the cavity required to retune the laser is (a) 0.334 10 12 m (b) 0.334 10 12 m Ans: (c) 0.167 10 12 m (d) 0.167 10 12 m (d) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 9 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q19. The spectroscopic symbol for the ground state of 13 Al is 2 P1/ 2 . Under the action of a strong magnetic field (when L S coupling can be neglected) the ground state energy level will split into (a) 3 levels Ans: (b) 4 levels (c) 5 levels (d) 6 levels (d) Solution: In extremely strong magnetic field coupling between L-S breaks down. J is no longer a valid quantum number. The ground state energy level will split into (2S+1)(2L+1) = 6 NET/JRF (JUNE-2014) Q20. A spectral line due to a transition from an electronic state p to an s state splits into three Zeeman lines in the presence of a strong magnetic field. At intermediate field strengths the number of spectral lines is (a) 10 Ans: (b) 3 (c) 6 (d) 9 (a) Solution: For p state: l 1, s 1/ 2 : j 1/ 2 & 3/ 2 . This gives two spectral terms 2P3/2 & 2P1/2 For s state: l= 0, s = 1/2: j = 1/2 : This gives spectral terms 2S1/2 The terms 2P3/2 and 2S1/2 corresponding to J = 3/2 & J = 1/2 will break into 2J+1 Zeeman levels, which is 4 and 2 respectively. For 2P3/2 Mj = - 3/2 For 2S1/2 Mj = -1/2 -1/2 +1/2 +3/2 +1/2 The selection rule is MJ = 0, 1 (MJ = 0 MJ = 0 If J = 0) MJ = 0 gives component, MJ = 1 gives component Number of component = 2, Numbers of + components = 2 Number of - components = 2 The terms 2P1/2 and 2S1/2 corresponding to J = 1/2 & J = 1/2 will break into 2J+1 Zeeman levels, which is 2 & 2 respectively. Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 10 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES For 2P1/2 Mj = -1/2 +1/2, For 2S1/2 Mj = -1/2 The selection rule is MJ = 0, 1 +1/2 (MJ = 0 MJ = 0 If J = 0) MJ = 0 gives component, MJ = 1 gives component Number of component = 2, Numbers of + components = 1 Number of - components = 1 Thus, total number of Zeeman component = 10 Q21. A double slit interference experiment uses a laser emitting light of two adjacent frequencies v1 and v2 v1 v2 . The minimum path difference between the interfering beams for which the interference pattern disappears is (a) Ans: c v 2 v1 (b) c v 2 v1 (c) c 2 v 2 v1 (d) c 2 v 2 v1 (c) Solution: The condition of maximum intensity for interfering laser beam is: d sin n 1 The condition of dark intensity for interfering laser beam is: d sin n 2 For interference pattern to vanish the minimum path difference should be /2 The spectral bandwidth of laser is defined as v c 2 whereas v c 2 For two closely spaced line of wavelength 1 and 2 c 2 1 c c 1 2 1 1 c c 2 1 v 1 2 21 1 21 1 Since, for interference pattern to vanish for two closely spaced line of wavelength 1 and 2, the minimum path difference should be = Q22. 2 c c 2 v 2 2 1 How much does the total angular momentum quantum number J change in the transition of Cr 3d 6 atom as it ionize to Cr 2 3d 4 ? (a) Increases by 2 (b) Decreases by 2 (c) Decreases by 4 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com (d) Does not change Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 11 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (c) Solution: In Cr 3d 6 state M L 2 1 0 1 2 In this configuration, S 4 1 2 and L 2 2 This is the case of more than half filled subshell, thus state with highest J value will have the lowest energy. The hight J - value is J L S y Thus, the ground state spectral term for this configuration is 5 D4 . Now in Cr 2 3d 4 M L 2 1 0 1 2 In this configuration, S 4 1 2 and L 2 2 Since this is the case of less than half filled subshell, thus, state with lowest J value will have the lowest energy. The lowest J - value is J L 5 2 2 0 . Thus the ground state spectral term for this configuration is the J - value decreases from J 4 to J 0 . Thus correct answer is option (c). NET/JRF (DEC-2014) Q23. An atomic transition 1 P 1S in a magnetic field 1 Tesla shows Zeeman splitting. Given that the Bohr magneton B 9.27 10 24 J / T , and the wavelength corresponding to the transition is 250 nm, the separation in the Zeeman spectral lines is approximately (a) 0.01 nm (b) 0.1 nm (c) 1.0 nm Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com (d) 10 nm Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 12 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (a) Solution: This is the case of Normal Zeeman effect. The Zeeman separation in terms of frequency is B B h In terms of wavelength it is 2 c where B is Bohr magneton 2 c B B h 250 10 m 9.27 10 J / T 1T 3 10 3 10 m / s 6.625 10 Js 2 9 8 24 34 12 m 0.003 nm None of the answer is matching correctly. But best suitable answer is option (a) Q24. If the leading anharmonic correction to the energy of n -th vibrational level of a diatomic 2 1 molecule is xe n with xe 0.001 , the total number of energy levels possible 2 is approximately (a) 500 Ans: (b) 1000 (c) 250 (d) 750 (a) 2 1 1 Solution: The energy of anharmonic oscillator is Ev v xe v 2 2 where v 0, 1, 2, .....vmax is vibrational quantum number E Now, dEv dv v vmax 1 0 2 xe vmax 0 2 1 1 2 xe vmax 1 2 xe vmax 2 2 vmax v vmax r 1 1 1 1 500 2 xe 2 2 xe 2 0.001 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 13 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q25. The effective spin-spin interaction between the electron spin S e and the proton spin S p in the ground state of the Hydrogen atom is given by H aS e S p . As a result of this interaction, the energy levels split by an amount (a) Ans: 1 2 a 2 (b) 2a 2 (c) a 2 (d) 3 2 a 2 (c) Solution: The Hamiltonian is given as H aSe S p where S e and S p are electron and proton spin. The coupling between S e and S p gives net resultant spin angular Se momentum S Se S p 1 2 2 S 2 Se2 S p 2Se S p Se S p S 2 Se2 S p 2 H Sp S a 2 2 S Se2 S p 2 2 where S 2 S S 1 2 , Se2 Se Se 1 2 , S p S p S p 1 2 Since Se 1 1 and S p S 0, 1 2 2 For S 0 (singlet state) H1 a 3 3 2 3 2 0 a 2 4 4 4 For S 1 (Triplet state) 1 2 a 4 12 s1/2 a 3 3 1 H 2 2 2 2 a 2 2 4 4 4 H H 2 H1 3 a 2 4 F 1 F 0 1 2 3 2 a a H a 2 4 4 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 14

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