For an electron in the l = 2 state:
(a) enumerate all the possible values of quantum numbers j and mj ;
(b) draw the corresponding vector diagrams;
(c) estimate the maximum value of the spin-orbit coupling energy E.
l=2∴n=3l=2\\\therefore n=3l=2∴n=3
(a) s=1/2s=1/2s=1/2
j=(n+s),(n−s)j=(n+s),(n-s)j=(n+s),(n−s)
j=(5/2),(3/2)j=(5/2),(3/2)j=(5/2),(3/2)
For j = 5/2
jm=j(j+1)ℏ2jm=j(j+1)\hbar^2jm=j(j+1)ℏ2
jm=354ℏ2jm=\dfrac{35}{4}\hbar^2jm=435ℏ2
For j = 3/2
jm=154ℏ2jm=\dfrac{15}{4}\hbar^2jm=415ℏ2
(c) ΔEso=αZℏ34m2c1r3[j(j+1)−l(l+1)−s(s+1)]\Delta E_{so}=\alpha\dfrac{Z\hbar^3}{4m^2c}\dfrac{1}{r^3}[j(j+1)-l(l+1)-s(s+1)]ΔEso=α4m2cZℏ3r31[j(j+1)−l(l+1)−s(s+1)]
s=1/2, l=2, j=5/2s=1/2,\space l=2,\space j=5/2s=1/2, l=2, j=5/2
ΔEso=αZℏ34m2c1r3[52(52+1)−2(2+1)−12(12+1)]\Delta E_{so}=\alpha\dfrac{Z\hbar^3}{4m^2c}\dfrac{1}{r^3}\bigg[\dfrac{5}{2}\bigg(\dfrac{5}{2}+1\bigg)-2(2+1)-\dfrac{1}{2}\bigg(\dfrac{1}{2}+1\bigg)\bigg]ΔEso=α4m2cZℏ3r31[25(25+1)−2(2+1)−21(21+1)]
ΔEso=2αZℏ34m2c1r3\Delta E_{so}=2\alpha\dfrac{Z\hbar^3}{4m^2c}\dfrac{1}{r^3}ΔEso=2α4m2cZℏ3r31
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