Q 11-12-152JEE MainJEE Main 2018 (15 Apr, Shift 2)Easy
The value closest to the thermal velocity of a Helium atom at room temperature $(300\ \text{K})$ in $\text{m s}^{-1}$ is: $\left[k_B = 1.4\times10^{-23}\ \text{J/K};\ m_\text{He} = 7\times10^{-27}\ \text{kg}\right]$
Answer: (D) $1.3\times10^3$
$$v_\text{rms} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3\times1.4\times10^{-23}\times300}{7\times10^{-27}}} = \sqrt{1.8\times10^6} \approx 1.3\times10^3\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics