Q99 · CSIR-NET Chemistry, December 2012

Paper: CSIR-NET December 2012 · Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Harmonic Oscillator · Marks: 2 · Difficulty: Medium

The energy of a harmonic oscillator in its ground state is $1 / 2 \mathrm{h} \omega^{\prime}$. According to the virial theorem, the average kinetic (T) and potential (V) energies of the above are
(a)$\mathrm{T}=1 / 4 \mathrm{h} \omega^{\prime} \mathrm{V}=1 / 4 \mathrm{h} \omega^{\prime}$
(b)$\mathrm{T}=1 / 8 \mathrm{h} \omega^{\prime} \mathrm{V}=3 / 8 \mathrm{h} \omega^{\prime}$
(c)$\mathrm{T}=\mathrm{h} \omega^{\prime} \mathrm{V}=1 / 2 \mathrm{h} \omega^{\prime}$
(d)$\mathrm{T}=3 / 8 \mathrm{h} \omega^{\prime} \mathrm{V}=1 / 8 \mathrm{h} \omega^{\prime}$
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
Virial theorem for a quadratic potential: ⟨T⟩ = ⟨V⟩ = ½E = ¼hν.

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