1000 Solved Problems in Modern Physics

224 3 Quantum Mechanics – II whereF(r)=rf(r) Lzψ 1 =−i ∂ ∂φ (cosφ+isinφ)sinθF(r) =−i(−sinφ+icosφ)sinθF(r) =(cosφ+isinφ)sinθF( ...

3.3 Solutions 225 3.90 With reference to the Table 3.2, the functionψ(r,θ,φ)isthe2pfunction for the hydrogen atom. (a)Lz=−i∂φ∂ ...

226 3 Quantum Mechanics – II Exchange ofx →−x, y →−y, z →−zimpliesθ →π−θand φ→π+φ, so that P 1 m(cosθ)→(−1)l+m andeimφ→(−1)meimφ ...

3.3 Solutions 227 3.93 u(θ,φ)=^1 / 4 √ 15 π sin 2θcos 2φ =^1 / 4 √ 15 π sin^2 θ [ (e^2 iφ+e−^2 iφ) 2 ] (1) ButY 2 + 2 (θ,φ)= √ 1 ...

228 3 Quantum Mechanics – II (b) First we show that the wavefunction is normalized. ∫ |ψ|^2 dr= 1 4 π ∫∞ 0 |g(r)|^2 r^2 dr ∫π 0 ...

3.3 Solutions 229 3.3.7 Approximate Methods ......................... 3.97ΔE=<ψ|δU|ψ> δU=U(interaction energy of electron ...

230 3 Quantum Mechanics – II =^1 / 2 mω^2 [(x+qE/mω^2 )^2 −q^2 E^2 /m^2 ω^4 ] =^1 / 2 mω^2 ( x+ qE mω^2 ) 2 − q^2 E^2 2 mω^2 (2) ...

3.3 Solutions 231 ∂E ∂Z =0; this yieldsZ= 27 16 E ( 27 16 ) =− ( 2 e^2 2 a 0 )( 27 16 ) 2 =−(2)(13.6) ( 27 16 ) 2 =− 77 .45 eV T ...

232 3 Quantum Mechanics – II 2 S(m=0);ψ 2 s(0)=(4π)− (^12) ( 1 2 a )^32 ( 2 − r a ) exp ( − r a ) 2 P(m=0);ψ 2 p(0)=(4π)− (^12) ...

3.3 Solutions 233 First order correction is ΔE= ∫ ψ 0 ∗H′ψdτ ∫ ψ 0 ∗ψdτ ∫ ψ 0 ∗H′ψdτ=K^2 W ∫ a 2 0 sin^2 (n 1 πx a ) dx ∫a/ 2 0 ...

234 3 Quantum Mechanics – II Nowψican be expanded as a sum of partial waves ψi=eikrcosθ= ∑∞ l= 0 Aljl(kr)pl(cosθ)(8) wherejl(kr) ...

3.3 Solutions 235 oreikrf(θ)= ∑ Blpl(cosθ) [ ei(kr− π 2 l+δl) −e−i(kr− π 2 l+δl)] 2 ik − ∑ il(2l+1)pl(cosθ) [ ei(kr− π 2 l) −e−i ...

236 3 Quantum Mechanics – II σL(45◦)= 4 |f(90◦)|CM. Thus quantum mechanics explains the experimental result 3.106σ= ( 4 π k^2 )∑ ...

3.3 Solutions 237 3.109 By Problem 3.104 σ(θ)= 1 k^2 [ sin^2 δ 0 +6sinδ 0 sinδ 1 cos(δ 1 −δ 0 ) cosθ+9sin^2 δ 1 cos^2 θ ] (1) We ...

238 3 Quantum Mechanics – II ∑R/λ l= 0 (2l+1)=(R/λ-)^2 ∴σr=σs=πλ-^2 ( R λ-^2 ) =πR^2 The total cross-section σt=σr+σs= 2 πR^2 wh ...

3.3 Solutions 239 tankR≈kR+ (kR)^3 3 +··· We therefore have tanδ 0 ≈ k [ D−R− ( (kR)^3 3 k )] 1 +k^2 DR If the inequalitieskR< ...

240 3 Quantum Mechanics – II binding energyWof the deuteron, which is much smaller than the well depth (∼25 MeV). In the deutero ...

3.3 Solutions 241 Putα= 1 /aandβ=q/ F(q)≈ 4 πA ∫∞ 0 re−αr sinβr β dr I=− ∂ ∂α ∫∞ 0 e−αrsinβrdr = − 1 β ∂ ∂α β α^2 +β^2 = 1 β 2 ...

242 3 Quantum Mechanics – II But the total cross-section is given by σt= 4 π k^2 (2l+1) sin^2 δl. It follows thatIm f(0)=kσt/ 4 ...

3.3 Solutions 243 and minima are smeared out, just as in the case of optical diffraction from a diffuse boundary of objects char ...