Question

Assuming the nitrogen molecule is moving with r.m.s. velocity at $400\text{K}$, the de-broglie wavelength of nitrogen molecule is close to :
(Given : nitrogen molecule weight : $4.64\times {10}^{-26}\text{kg},$ Boltzman constant : $1.38\times {10}^{-23}\text{J/K},$ Planck constant :$6.63\times {10}^{-34}\text{J-s}$)

• A. $0.24\mathring{\text{A}}$
• B. $0.20\mathring{\text{A}}$
• C. $0.34\mathring{\text{A}}$
• D. $0.44\mathring{\text{A}}$

Answer 1

The correct option is A $0.24\mathring{\text{A}}$

RMS speed of gas molecule,

$V_{\text{rms}}=\sqrt{\frac{3kT}{m}}$

de- Broglie wavelength, $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$

$\therefore \lambda =\frac{h}{\sqrt{2m\times \frac{1}{2}m{V}_{\text{rms}}^2}}=\frac{h}{m{V}_{rms}}$

$\Rightarrow \lambda =\frac{h}{m\times \sqrt{\frac{3kT}{m}}}=\frac{h}{\sqrt{3mkT}}$

Substituting the respective values, we get

$\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{3\times 4.64\times {10}^{-26}\times 1.38\times {10}^{-23}\times 400}}=0.24\mathring{\text{A}}$

Hence, option $\left(\text{A}\right)$ is correct.