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 $N= 14.0076 \; u$

Question

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

 

Answer 1

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}$

$= \large\frac{h}{\sqrt{3mkT}}$

$= \large\frac{6.63 \times 10^{-34}}{\sqrt{[3 \times 28.0152 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300]}}$

$= .028 \times 10^{-9}\: m$