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What is the de Broglie wavelength of a nitrogen molecule in air at $300\; K?$ Assume molecule is moving with the rms speed of molecules at this temp. At. mass of $N= 14.0076 \; u$

$\begin {array} {1 1} (a)\;028 \times 10^9\:m & \quad (b)\;0.0028\: nm \\ (c)\;0.028 \times 10^{-9}\: m & \quad (d)\;2.8\: nm \end {array}$


1 Answer

Ans : (c)
$T=300\: K$
molecular mass of $N_2=2 \times 14.0076\: u = 28.0152 \times 1.66 \times 10^{-27} kg$
$= 28.0152 \times 1.66 \times 10^{-27} kg$
So, $\large\frac{1}{2} mv^2_{rms} = \large\frac{3}{2} kT$
$ \Rightarrow V_{rms} = \sqrt{ \bigg( \large\frac{3kt}{m} \bigg)}$
Hence, de Broglie wavelength, $ \lambda = h/mv_{rms}$
$= \large\frac{h}{\sqrt{3mkT}}$
$= \large\frac{6.63 \times 10^{-34}}{\sqrt{[3 \times 28.0152 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300]}}$
$= .028 \times 10^{-9}\: m$


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