The de Broglie wavelength of a molecule in a gas at room temperature (300K) is λ _1. If the temperature of the gas is increased to 600K, then the de…
JEE Main 2023 — Physics Modern Physics
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The de Broglie wavelength of a molecule in a gas at room temperature (300K) is λ1. If the temperature of the gas is increased to 600K, then the de Broglie wavelength of the same gas molecule becomes
Official previous-year question
Held on 10 Apr 2023 · Verified 6 Jul 2026.
Options
A
21λ1
B
2λ1
C
21λ1
D
2λ1
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Solution
The root mean squared velocity of a gas is given by
v=M3RT
⇒v∝T
Let
T1=300KT2=600K
Taking velocity ratios at the given temperatures,
v2v1=T2T1=600300=21
The de Broglie wavelength is given by λ=mvh. So,
λ∝v1
The ratio of the wavelengths is
λ2λ1=(v1v2)=12
⇒λ2=(21λ1)
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