Q 12-11-144JEE MainJEE Main 2020 (6 Sep, Shift 2)Medium
Assuming the nitrogen molecule is moving with r.m.s. velocity at $400$ K, the de Broglie wavelength of the nitrogen molecule is close to: (Given: nitrogen molecule weight: $4.64\times10^{-26}$ kg, Boltzmann constant: $1.38\times10^{-23}\ \text{J K}^{-1}$, Planck constant: $6.63\times10^{-34}$ J s)
Answer: (A) $0.24$ Å
$$v_{\text{rms}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3\times1.38\times10^{-23}\times400}{4.64\times10^{-26}}} \approx 597\ \text{m/s}$$
$$\lambda = \frac{h}{mv} = \frac{6.63\times10^{-34}}{4.64\times10^{-26}\times597} \approx 2.4\times10^{-11}\ \text{m} = 0.24\ \text{Å}$$
Solution by Sreeraj P, M.Sc Physics