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

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

Option: 1

2\lambda _1


Option: 2

\frac{1}{\sqrt{2}} \lambda_1


Option: 3

\sqrt{2} \lambda_1


Option: 4

\frac{1}{2} \lambda_1


Answers (1)

best_answer

\begin{aligned} & \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mK}(\mathrm{T})}} \\ & \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}} \\ & \lambda_2=\lambda_1 \sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\ & =\lambda_1 \sqrt{\frac{300}{600}}=\frac{\lambda_1}{\sqrt{2}} \end{aligned}

Posted by

Divya Prakash Singh

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