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Q.

The de Broglie wavelength of a molecule in a mono atom gas at room temperature (300 K) is $λ_{1} .$ If the gas is subjected to an adiabatic process so that volume becomes 1/8^th , then the de Broglie wavelength of the same gas molecule becomes

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a

$\sqrt{2} λ_{1}$

b

$\frac{1}{4} λ_{1}$

c

$\frac{1}{\sqrt{2}} λ_{1}$

d

$2 λ_{1}$

answer is B.

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Detailed Solution

$T V^{r - 1} = c o n s \tan t$
$300 V^{\frac{2}{3}} = T_{2} {(\frac{V}{8})}^{\frac{2}{3}}$
$\begin{array}{l} T_{2} = 1200 K \\ λ \propto \frac{1}{\sqrt{T}} \Rightarrow \frac{λ_{2}}{λ_{1}} = \sqrt{\frac{300}{1200}} \\ λ_{2} = \frac{λ_{1}}{4} \end{array}$

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The de Broglie wavelength of a molecule in a mono atom gas at room temperature (300 K) is λ1. If the gas is subjected to an adiabatic process so that volume becomes 1/8th , then the de Broglie wavelength of the same gas molecule becomes