Dual Nature of Matter

JEE Main level, single correct, medium

28 of 200

Question 28 of 200

The de Broglie wavelength of a molecule in a gas at room temperature (300K)\left(300K\right) is λ1{\lambda }_{1}. If the temperature of the gas is increased to 600K600K, then the de Broglie wavelength of the same gas molecule becomes

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Answer: (C) 12λ1\dfrac{1}{\sqrt{2}}{\lambda }_{1}

The root mean squared velocity of a gas is given by
v=3RTMv=\sqrt{\dfrac{3RT}{M}}
⇒v∝T\Rightarrow v\propto \sqrt{T}
Let
T1=300KT2=600K{T}_{1}=300K {T}_{2}=600K
Taking velocity ratios at the given temperatures,
v1v2=T1T2=300600=12\dfrac{{v}_{1}}{{v}_{2}}=\sqrt{\dfrac{{T}_{1}}{{T}_{2}}}=\sqrt{\dfrac{300}{600}}=\sqrt{\dfrac{1}{2}}
The de Broglie wavelength is given by λ=hmv\lambda =\dfrac{h}{mv}. So,
λ∝1v\lambda \propto \dfrac{1}{v}
The ratio of the wavelengths is
λ1λ2=(v2v1)=21\dfrac{{\lambda }_{1}}{{\lambda }_{2}}=\left(\dfrac{{v}_{2}}{{v}_{1}}\right)=\dfrac{\sqrt{2}}{1}
⇒λ2=(12λ1)\Rightarrow {\lambda }_{2}=\left(\dfrac{1}{\sqrt{2}}{\lambda }_{1}\right)

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