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  • 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 :

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 0.24\; \AA
 
Option: 2 0.20\; \AA
Option: 3 0.34\; \AA  
Option: 4 0.44\; \AA

Answers (1)

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\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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Deependra Verma

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