JEE Challenger
More from Dual Nature of Radiation and Matter

Dependence of Allowed Energy on Box Length in Quantum Well

Comprehension Passage

Paragraph for Question Nos. 56 to 58

When a particle is restricted to move along xx-axis between x=0x = 0 and x=ax = a, where aa is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x=0x = 0 and x=ax = a. The wavelength of this standing wave is related to the linear momentum pp of the particle according to the de Broglie relation. The energy of the particle of mass mm is related to its linear momentum as E=p22mE = \frac{p^2}{2m}. Thus, the energy of the particle can be denoted by a quantum number 'nn' taking values 1,2,3,…1, 2, 3, \dots (n=1n = 1, called the ground state) corresponding to the number of loops in the standing wave.

Use the model described above to answer the following three questions for a particle moving in the line x=0x = 0 to x=ax = a. Take h=6.6×10−34 Jsh = 6.6\times 10^{-34}\text{ Js} and e=1.6×10−19 Ce = 1.6\times 10^{-19}\text{ C}.

The allowed energy for the particle for a particular value of nn is proportional to

Options

A

a−2a^{-2}

Correct
B

a−3/2a^{-3/2}

C

a−1a^{-1}

D

a2a^2

Step-by-Step Solution

According to the passage, the standing wave formed by the particle in the region x=0x = 0 to x=ax = a has nodes at both ends. For a state with nn loops, the length of the region aa is related to the de Broglie wavelength λ\lambda by:
a=n(λ2)a = n \left(\frac{\lambda}{2}\right)

This gives the wavelength:
λ=2an\lambda = \frac{2a}{n}

Using the de Broglie relation, the linear momentum pp of the particle is:
p=hλ=nh2ap = \frac{h}{\lambda} = \frac{nh}{2a}

The energy EE of the particle is given by:
E=p22m=12m(nh2a)2=n2h28ma2E = \frac{p^2}{2m} = \frac{1}{2m} \left(\frac{nh}{2a}\right)^2 = \frac{n^2 h^2}{8ma^2}

For a particular value of the quantum number nn, the allowed energy is:
E∝1a2=a−2E \propto \frac{1}{a^2} = a^{-2}

Therefore, the correct option is A.