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Q.

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 1200 K, then the de-Broglie wavelength of the same gas molecule becomes

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a

$\frac{1}{2} λ_{1}$

b

$\sqrt{2} λ_{1}$

c

$\frac{1}{\sqrt{2}} λ_{1}$

d

$2 λ_{1}$

answer is D.

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Detailed Solution

From K.T.G
$v_{R M S} = \sqrt{\frac{3 k_{B} T}{m}} v_{R M S} \propto T a n d \frac{h}{m v_{R M S}} = λ i . e ., λ \propto \frac{1}{\sqrt{T}} \frac{λ_{2}}{λ_{1}} = \sqrt{\frac{T_{1}}{T_{2}}} = \sqrt{\frac{300}{1200}} = \frac{1}{2} λ_{2} = \frac{λ_{1}}{2}$

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The de-Broglie wavelength of a molecule in a gas at room temperature (300 K) is λ1. If the temperature of the gas is increased to 1200 K, then the de-Broglie wavelength of the same gas molecule becomes