If the de-Broglie wavelength at a temperature | vedclass

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(A) The de-Broglie wavelength $\lambda$ of a gas molecule at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$.This implies that $\lambda

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$\lambda / 2$

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(A) The de-Broglie wavelength $\lambda$ of a gas molecule at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$.This implies that $\lambda \propto \frac{1}{\sqrt{T}}$.Given $T_1 = 27^oC = (27 + 273) K = 300 K$ and $T_2 = 927^oC = (927 + 273) K = 1200 K$.Using the ratio: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values: $\frac{\lambda}{\lambda_2} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.Therefore,$\lambda_2 = \frac{\lambda}{2}$.

If the de-Broglie wavelength at a temperature of $27^oC$ is $\lambda$,what will be the de-Broglie wavelength at $927^oC$?

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