The de Broglie wavelength of a molecule in a gas at room temperature ( 300 K ) is $\lambda_{1}$. If the temperature of the gas is increased to 600 K , then the de Broglie wavelength of the same gas molecule becomes

(A) $\frac{1}{\sqrt{2}} \lambda_{1}$

(C) $\frac{1}{2} \lambda_{1}$

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Modern Physics

The de Broglie wavelength of a molecule in a gas at room temperature ( 300 K ) is

λ_{1}

. If the temperature of the gas is increased to 600 K , then the de Broglie wavelength of the same gas molecule becomes

(A) $\frac{1}{2} λ_{1}$ ✓ Correct
(B) $2 λ_{1}$
(C) $\frac{1}{2} λ_{1}$
(D) $2 λ_{1}$

Answer: (A)

Solution

The de Broglie wavelength is

λ = \frac{h}{m v}

, and the thermal speed obeys

v \propto T

, so

λ \propto \frac{1}{T}

. Doubling the temperature gives

\frac{λ _{2}}{λ _{1}} = \frac{T _{1}}{T _{2}} = \frac{300}{600} = \frac{1}{2}

. The new wavelength is

\frac{1}{2} λ_{1}

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option

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