Question: In this problem you will supply the missing steps in the derivation of the formula E singlet = E 1s + E 2s + J
In this problem you will supply the missing steps in the derivation of the formula Esinglet= E1s+ E2s+ J + K for the singlet level of the 1s12s1configuration of He.
a. Expand Equation (22.17) to obtain
 [1s (1) 2s (2) + 2s (1) ls (2)]dT¡dr2 Esinglet [ls(1)2s (2) + 2s (1) Is (2)](Ĥ2)](https://dsd5zvtm8ll6.cloudfront.net/si.question.images/images/question_images/1525/4/1/6/6565aec02d001adb1525416644866.jpg)
b. Starting from the equations 1 2 HË i1s(i) = E s1s(i) and HË i 2s(i) = E s 2s(i), show that
![singlet = E + E. jus02C)- [Ls1)25(2) + 25(1)ls(2)]| 4πεο μ-2) [1s (1) 2s (2) + 2s (1)1s(2)] d, dr,](https://dsd5zvtm8ll6.cloudfront.net/si.question.images/images/question_images/1525/4/1/6/6565aec02d079d8e1525416645116.jpg)
c. Expand the previous equation using the definitions
 [1s (1) 2s (2)](https://dsd5zvtm8ll6.cloudfront.net/si.question.images/images/question_images/1525/4/1/6/6575aec02d1151451525416645369.jpg)
to obtain the desired result, Esinglet = E1s + E2s + J + K.
1 [[ls(1)2s (2) + 2s (1) Is (2)](H1) [1s (1) 2s (2) + 2s (1) ls (2)]dTdr2 Esinglet [ls(1)2s (2) + 2s (1) Is (2)](2) [1s (1) 2s (2) + 2s (1) 1s (2)]dTdr2 S/ [1s (1) 2s (2) + 2s (1) 1s (2)] 121(2) 4 -1 , [1s (1) 2s (2) + 2s (1) ls (2)]drdr2 singlet = E + E. jus02C)- [Ls1)25(2) + 25(1)ls(2)]| 4 -2) [1s (1) 2s (2) + 2s (1)1s(2)] d, dr,
Step by Step Solution
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a The integral can be expressed as the sum of three integrals each containing one of the three ... View full answer
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