Paper: CSIR-NET June 2013 · Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Quantum Chemistry – General · Marks: 2 · Difficulty: Medium
For the particle-in-a-box problem in $(0, \mathrm{L})$ the value of $\left\langle\hat{\mathrm{x}}^{3}\right\rangle$ in the $\mathrm{n} \rightarrow \infty$ limit would be
(a)$\frac{\mathrm{L}^{3}}{6}$
(b)$\frac{\mathrm{L}^{3}}{3}$
(c)$\frac{\mathrm{L}^{3}}{4}$
(d)$\frac{\mathrm{L}^{4}}{4}$
Answer
Answer: C ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
As n → ∞ the probability density becomes uniform (1/L): ⟨x³⟩ = (1/L)∫₀^L x³ dx = L³/4.