Question

Assuming the nitrogen molecule is moving with root mean square velocity at $400\mathrm{K}$, the$\mathrm{de}-\mathrm{Broglie}$ wavelength of nitrogen molecule is close to -

(given: nitrogen molecule weight: $4.64\times {10}^{-26}\mathrm{kg}$, Boltzmann constant: $1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$, Planck's constant: $6.63\times {10}^{-34}\mathrm{Js}$)-

• A. $0.44$ $\stackrel{\circ }{\mathrm{A}}$
• B. $0.34\stackrel{\circ }{\mathrm{A}}$
• C. $0.20\stackrel{\circ }{\mathrm{A}}$
• D. $0.24\stackrel{\circ }{\mathrm{A}}$

Answer 1

The correct option is D

$0.24\stackrel{\circ }{\mathrm{A}}$

Step 1: Formula for the de-Broglie wavelength - $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{mv}}$ where, $\mathrm{h}$ is the Planck's constant, $\mathrm{m}$ is the mass of the molecule, $\mathrm{v}$ is the velocity

Formula for the root mean square velocity - ${\mathrm{\nu }}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{KT}}{\mathrm{m}}}$ where, $\mathrm{K}$ is the Boltzmann constant, $\mathrm{T}$ is the temperature and $\mathrm{m}$ is the mass of the molecule

Step 2: Putting the formula of ${\mathrm{\nu }}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{KT}}{\mathrm{m}}}$ in place of $\mathrm{\nu }$ in the formula of de-Broglie wavelength, we get-

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

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

Step 3: Calculation of the de-Broglie wavelength of nitrogen molecule-

$\mathrm{h}=6.63\times {10}^{-34}\mathrm{Js}$

$\mathrm{K}=1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$

$\mathrm{m}=4.64\times {10}^{-26}\mathrm{kg}$

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

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

$\mathrm{\lambda }=0.24\stackrel{\circ }{\mathrm{A}}$

So, the de-Broglie wavelength of the nitrogen molecule is $0.24\stackrel{\circ }{\mathrm{A}}$

Hence, option D is the correct answer.