1000 Solved Problems in Modern Physics

104 2 Quantum Mechanics – I 2.9 In atomic physics the atomic units are as follows: (i) (a) The Bohr radius^2 /mee^2 is used as ...

2.3 Solutions 105 (b) Equating the coulomb force to the centripetal force Ze^2 / 4 πεor^2 =mv^2 /r (2) Solving (1) and (2) v=Ze^ ...

106 2 Quantum Mechanics – I ΔEn, 1 = α^2 mec^2 4 ( 1 12 − 1 n^2 ) ,n= 2 , 3 , 4 , 5 Thus, ΔE 21 = ( 1 137 ) 2 ( 0. 511 × 106 4 ) ...

2.3 Solutions 107 mv^2 /r=kr (2) solving (1) and (2) v^2 = ( n m )( k m )^12 (3) r^2 = n √ km (4) (c) E=U+T= 1 2 kr (^2) + 1 2 ...

108 2 Quantum Mechanics – I f=me^4 / 4 n^3 h^3 ε 02 ν= me^4 8 ε 02 h^3 ( 1 n^2 f − 1 n^2 i ) = me^4 8 ε 02 h^3 { (ni−nf)(ni+nf) ...

2.3 Solutions 109 Iron:EK= 13 .6(26−1)^2 = 8 ,500 eV λK= 1 , 241 8 , 500 = 0 .146 nm= 1. 46 A ̊ 2.25 The minimum wavelength of t ...

110 2 Quantum Mechanics – I (^3) d→ (^2) P 1 / 2 , (^3) d 3 / 2 → (^2) P 3 / 2 (^3) d 5 / 2 → (^2) P 3 / 2 In all there are seve ...

2.3 Solutions 111 2.34 By Moseley’s law 1 λ =A(Z 1 −1)^2 =A(19−1)^2 (1) 1 λ/ 4 =A(Z−1)^2 (2) Dividing (2) by (1) and solving for ...

112 2 Quantum Mechanics – I 2.37 (a) By definition the magnetic moment of electron is given by the product of the charge and the ...

2.3 Solutions 113 From the geometry of the figure, EF CF = DA CA or s L+ 2 l = h l/ 2 →h= s.l 2 L+l (8) Eliminatinghbetween (7) ...

114 2 Quantum Mechanics – I Magnetic moment for both hydrogen and sodium is 1 Bohr magneton,μB= e 2 me = 9. 27 × 10 −^24 JT−^1 ...

2.3 Solutions 115 The magnetic momentμ 0 produced is equal to this current multiplied by the area enclosed. μB=iA=ωe. πr^2 2 π ( ...

116 2 Quantum Mechanics – I Term SPDF l 0123 Parity=(−1)l + 1 − 1 + 1 − 1 2.47 J=l+s= 0 + 1 / 2 = 1 / 2 F=I+J,I+J− 1 ,...I−J = 2 ...

2.3 Solutions 117 Further from (2) d(Δω)=dω. The relative distribution of Doppler shift is dN N = exp [ − ( Δω ΔωD ) 2 ] √ π dω ...

118 2 Quantum Mechanics – I The splitting of levels as in sodium is shown in Fig. 2.5. Transitions take place with the selection ...

2.3 Solutions 119 2.52ΔE=μBBΔm=μBB(becauseΔm=±1) =(9. 27 × 10 −^24 )(1.0)= 9. 27 × 10 −^24 J= 5. 79 × 10 −^5 eV. The splitting o ...

120 2 Quantum Mechanics – I 2.56 n(E 2 ) n(E 1 ) =e−(E^2 −E^1 )/kT= 1 10 T= E 2 −E 1 kln 10 = 2. 26 8. 625 × 10 −^5 × 2. 3 = 1. ...

2.3 Solutions 121 deciding the model of the nucleus, that is discarding the electron–proton hypothesis. Consider the nitrogen nu ...

122 2 Quantum Mechanics – I μ=m 1 / 2 = 0. 5 mp N 0 /N 1 =(1/3) exp ( ^2 /μr^2 kT ) =(1/3) exp ( ^2 c^2 /μc^2 r^2 kT ) = 1 3 e ...

2.3 Solutions 123 EJ=[J(J+1)^2 c^2 ]/(2)(0.5)(Mpc^2 )r^2 E 2 =(2)(3)(197.3)^2 (10−^15 )^2 /(940)×(10−^10 )^2 = 0. 264 × 10 −^10 ...