The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is 1. If the temperature of the gas is increased to 600 K, the de Broglie wavelength becomes:

Given: Initial temperature is T1 = 300 K and final temperature is T2 = 600 K. Initial de Broglie wavelength is 1. Find: The new de Broglie wavelength 2. The provided the solution applies Wien’s displacement law, which is unrelated to de Broglie wavelength for gas molecules. For a gas molecule, = hp, p T so 1 T Hence, 2 1 = T1T2 Substituting the values, 2 1 = 300600 = 1 2 Therefore, 2 = 1 2 So the correct option based on the question and the hint is B. The provided the solution concludes /2, which is a mismatch because it uses an incorrect physical law for this question.

The de Broglie wavelength of a molecule in a gas at room temperat… — Solution | JEE Physics

MCQEasyJEE 2023de Broglie Relation

The de Broglie wavelength of a molecule in a gas at room temperature ( $300 \, \text{K}$ ) is $\lambda_1$ . If the temperature of the gas is increased to $600 \, \text{K}$ , the de Broglie wavelength becomes:

A
$2\lambda_1$
B
$\lambda_1/\sqrt{2}$
C
$\sqrt{2}\lambda_1$
D
$\lambda_1/2$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Initial temperature is $T_1 = 300 \, \text{K}$ and final temperature is $T_2 = 600 \, \text{K}$ . Initial de Broglie wavelength is $\lambda_1$ .

Find: The new de Broglie wavelength $\lambda_2$ .

The provided the solution applies Wien’s displacement law, which is unrelated to de Broglie wavelength for gas molecules. For a gas molecule,

\lambda = \frac{h}{p}, \qquad p \propto \sqrt{T}

\lambda \propto \frac{1}{\sqrt{T}}

Hence,

\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}}

Substituting the values,

\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{300}{600}} = \frac{1}{\sqrt{2}}

Therefore,

\lambda_2 = \frac{\lambda_1}{\sqrt{2}}

So the correct option based on the question and the hint is B. The provided the solution concludes $\lambda/2$ , which is a mismatch because it uses an incorrect physical law for this question.

Common mistakes

Using Wien’s displacement law here is incorrect because the question is about de Broglie wavelength of a gas molecule, not thermal radiation. Use $\lambda = h/p$ with molecular kinetic energy related to temperature instead.
Assuming $\lambda \propto 1/T$ is wrong for matter waves in a gas. Since $p \propto \sqrt{T}$ , the correct relation is $\lambda \propto 1/\sqrt{T}$ .
Doubling the temperature does not halve the de Broglie wavelength directly. First form the ratio $\lambda_2/\lambda_1 = \sqrt{T_1/T_2}$ , then substitute the temperatures.

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