Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
Question
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Chapter 12, Problem 12.45E
Interpretation Introduction

Interpretation:

The variation theory treatment of H using ϕ=ekr2 as an unnormalized trial function is to be performed and specific expression for k is to be stated.

Concept introduction:

According to the variation theory, the lower the energy of a system the better is the approximation. Variation theory is based on the fact that any test system has average energy equal to or greater than the ground state energy of that system. The advantage of the variation theory is that any wavefunction can be taken for any test system.

Expert Solution & Answer
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Answer to Problem 12.45E

The variation theory treatment of H using ϕ=ekr2 as an unnormalized trial function has been performed and specific expression for k is e4μ218π3ε024.

Explanation of Solution

The expression for trial function can be written as given below.

Etrial=ϕ*H^ϕdτϕ*ϕdτ

Substitute the values in the above expression as follows.

Etrial = ϕ * H ^ ϕdτ ϕ * ϕdτ= e k r 2 ( 2 2μ ( 1 r 2 r ( r 2 r ) ) e 2 4π ε 0 r ) e k r 2 dτ e 2k r 2 dτ= 22μ0 e k r 2 ( 1 r 2 r ( r 2 e k r 2 r ))dr0 e 2 4π ε 0 re 2k r 2 dr0e2k r 2dr=22μ0ek r 2( 1 r 2 r( r 2 ( 2kr ) e k r 2 ))dr0e24πε0re2kr2dr0e2kr2dr

The above equation can be simplified as given below.

= 22μ0 e k r 2 ( 2k r 2 ( 3 r 2 e kr 2k r 4 e kr ))dr0 e 2 4π ε 0 re 2k r 2 dr0e2k r 2dr=2μ0e2k r 2(32k r 2)dre24πε00e 2k r 2 rdr0e2kr2dr=32kμ22k2μ0r2e2k r 2dre24πε00e 2k r 2 rdr0e2kr2dr=32kμ22kμ×( π 8 k 3 )1/212( π 2k)1/2( e 24π ε 0)0e2k r 2dr4πr20e2kr2dr4πr2

Solve the integral as follows.

=32kμ22kμ×(π8k3)1/212(π2k)1/2(e24πε0)(14k)(π8k3)1/2=32kμ22μe2πε0(2kπ)1/2

The value of k can be calculated by differentiating the above expression with respect to k as follows.

Etrial=32kμ22μe2πε0(2kπ)1/2Etrialk=32μ12k1/2e2πε0(2π)1/2=0k=2e44π3ε02×μ294=e4μ218π3ε024

Substitute this value in the expression for trial energy as given below.

Etrial=32kμ22μe2πε0(2kπ)1/2=32μ(e4μ218π3ε024)22μe2πε0(2π)1/2(e4μ218π3ε024)1/2=e4μ6π3ε02222μe2πε0(2π)1/2(e4μ218π3ε024)1/2

Conclusion

The variation theory treatment of H using ϕ=ekr2 as an unnormalized trial function has been performed and specific expression for k is e4μ218π3ε024.

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