Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Variation Method · Exam: NET JUNE 2016 · Marks: 2 · Difficulty: Medium
Given a trial wave function $\psi=\mathrm{C}_{1} \phi_{1}+\mathrm{C}_{2} \phi_{2}$ and the Hamiltonian matrix elements. $\int \phi_{1}^{*} \mathrm{H} \phi_{1} \mathrm{dv}=0$, $\int \phi_{1}^{*} \mathrm{H} \phi_{2} \mathrm{dv}=2.5, \int \phi_{2}^{*} \mathrm{H} \phi_{2} \mathrm{dv}=12.0$ the variationally determined ground state energy is
(a)-0.52
(b)-0.50
(c)12.50
(d)12.52
Answer
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