Q139 · CSIR-NET Chemistry, June 2011

Paper: CSIR-NET June 2011 · Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Perturbation Theory · Marks: 2 · Difficulty: Medium

The unperturbed energy levels of a system are $\varepsilon_{0}=0, \varepsilon_{1}=2$ and $\varepsilon_{2}=4$. The second order correction to energy for the ground state in pressure of the perturbation V for which $\mathrm{V}_{10}=2, \mathrm{V}_{20}=4$ and $\mathrm{V}_{12}=6$ has been found to be
(a)-6
(b)0
(c)+6
(d)-8
Answer
Answer: A ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
E₀⁽²⁾ = Σ_n |V_n0|²/(ε₀ − ε_n) = 2²/(0 − 2) + 4²/(0 − 4) = −2 − 4 = −6 (V₁₂ does not enter).

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