Find the % difference (approximate) in de Bro | vedclass

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.Since the kinetic energy of a gas molecule is $K.E. = \frac{3}{2}kT$,w

Vedclass · Accepted Answer

$30$

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.Since the kinetic energy of a gas molecule is $K.E. = \frac{3}{2}kT$,we have $\lambda \propto \frac{1}{\sqrt{T}}$.Let the initial temperature be $T_1 = T$ and the final temperature be $T_2 = 2T$.Then,$\lambda_1 = \frac{k'}{\sqrt{T}}$ and $\lambda_2 = \frac{k'}{\sqrt{2T}} = \frac{\lambda_1}{\sqrt{2}}$.The percentage change is given by $\frac{\lambda_2 - \lambda_1}{\lambda_1} 	imes 100 = \left( \frac{1}{\sqrt{2}} - 1 ight) 	imes 100$.Using $\frac{1}{\sqrt{2}} \approx 0.707$,the change is $(0.707 - 1) 	imes 100 = -29.3 \%$.The approximate magnitude of the percentage difference is $30 \%$.

Find the $\%$ difference (approximate) in de Broglie wavelength if the temperature of a gas becomes double.

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