Paper: CSIR-NET June 2014 · Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Particle In A Box · Marks: 2 · Difficulty: Medium
For a particle of mass m confined in a box of length L, assume $\Delta \mathrm{x}=\mathrm{L}$. Assume further that $\Delta \mathrm{p}(\min )= \left\langle\mathrm{p}^{2}\right\rangle^{1 / 2}$. Use the uncertainity principle to obtain an estimate of the energy of the particle. The value will be
(a)$\frac{\mathrm{h}^{2}}{8 \mathrm{mL}^{2}}$
(b)$\frac{\hbar^{2}}{8 \mathrm{mL}^{2}}$
(c)$\frac{\mathrm{h}^{2}}{32 \mathrm{mL}^{2}}$
(d)$\frac{\mathrm{h}^{2}}{2 \mathrm{mL}^{2}}$
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Δx Δp ≥ ħ/2 with Δx = L gives p_min ≈ ħ/2L; E ≈ p²/2m = ħ²/(8mL²).