Question

Assume that a molecule is moving with the root mean square speed at temperature 300 K. The de Broglie wavelength of nitrogen molecule is (Atomic mass of nitrogen = 14.0076 u, h = $6.63\times {10}^{-27}Js,k_B=1.38\times {10}^{-23}J{K}^{-1},1u=1.66.\times {10}^{-27}kg)$

• A. $2.75\times {10}^{-11}m$
• B. $2.75\times {10}^{-12}m$
• C. $3.24\times {10}^{-11}m$
• D. $3.24\times {10}^{-12}m$

Answer 1

The correct option is A $2.75\times {10}^{-11}m$

Mean Kinetic energy if a molecule:

$\frac{1}{2}m{v}_{rms}^2=\frac{3}{2}{k}_{B}T$

Here, m = $2\times 14.0067=28.02u$

As $\lambda =\frac{h}{m{v}_{rms}}=\frac{h}{\sqrt{3m{k}_{B}T}}$

$=\frac{6.63\times {10}^{-34}}{\sqrt{3\times (28.02\times 1.66\times {10}^{-27})}\times 1.38\times {10}^{-23}\times 300}$

$=2.75\times {10}^{-11}m$