The de Broglie wavelength of a molecule in a | vedclass

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Question

(A) From the Kinetic Theory of Gases ($K$.$T$.$G$.),the root mean square velocity of a gas molecule is given by $v_{RMS} = \sqrt{\frac{3 k_B T}{m}}$.T

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) From the Kinetic Theory of Gases ($K$.$T$.$G$.),the root mean square velocity of a gas molecule is given by $v_{RMS} = \sqrt{\frac{3 k_B T}{m}}$.This implies that $v_{RMS} \propto \sqrt{T}$.The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m v_{RMS}}$.Since $v_{RMS} \propto \sqrt{T}$,we have $\lambda \propto \frac{1}{\sqrt{T}}$.Therefore,the ratio of the wavelengths is $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}}$.Given $T_1 = 300 \ K$ and $T_2 = 600 \ K$,we get $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{300}{600}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus,the new wavelength is $\lambda_2 = \frac{\lambda_1}{\sqrt{2}}$.

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 $..............$.

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