# Assuming the nitrogen molecule is moving with

Question:

Assuming the nitrogen molecule is moving with r.m.s. velocity at $400 \mathrm{~K}$, the de-Broglie wavelength of nitrogen molecule is close to:

(Given: nitrogen molecule weight : $4.64 \times 10^{-26} \mathrm{~kg}$, Boltzman constant : $1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$, Planck constant: $6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ )

1. $0.34 A 2.$0.24 A

3. $0.20 A 4.$0.44 A

Correct Option: , 2

Solution:

$\mathrm{v}_{\text {rms }}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}$

$\mathrm{m} \rightarrow$ mass of one molecule $($ in $\mathrm{kg})=$

$\frac{\text { molar mass }}{\text { NA }}$

de-Broglie wavelenth,.

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$

given, $\mathrm{V}=\mathrm{V}_{\mathrm{rms}}$

$\lambda=\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}}$

$\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{KTm}}}$

$=\sqrt{\sqrt{3 \times 1.38 \times 10^{-23} \times 400 \times\left(\frac{28 \times 10^{-3}}{6.023 \times 10^{-23}}\right)}}$

$\lambda=\frac{6.63 \times 10^{-11}}{2.77}=2.39 \times 10^{-11} \mathrm{~m}$

\$\lambda=0.24 A