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: $6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}$ )
Correct Option: , 2
$\mathrm{v}_{\text {rms }}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}$
$\mathrm{m} \rightarrow$ 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
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