# Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wavelength of nitrogen molecule is close to : (Given :  nitrogen molecule weight : $4.64\times 10^{-26}kg,$ Boltzman constant : $1.38\times 10^{-23} J/K,$ Planck constant :$6.63\times 10^{-34} J.s$) Option: 1   Option: 2 Option: 3   Option: 4

$\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}} \\ \mathrm{m} \rightarrow mass of one molecule (in \mathrm{kg} ) = \frac{\text { molar mass }}{\text { NA }} \\ de-Broglie wavelenth, \lambda=\frac{\mathrm{h}}{\mathrm{mv}} \\ given, \mathrm{v}=\mathrm{v}_{\mathrm{rms}} \\ \lambda=\frac{\mathrm{h}}{\mathrm{m} \sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}} \\ \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{KTm}}} =\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.38 \times 10^{-23} \times 400 \times\left(\frac{28 \times 10^{-3}}{6.023 \times 10^{-23}}\right)}}$

$\begin{array}{l} \lambda=\frac{6.63 \times 10^{-11}}{2.77}=2.39 \times 10^{-11} \mathrm{~m} \\ \ \\ \lambda=0.24 \AA \end{array}$

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