Kinetic Theory

JEE Main level, single correct, medium

112 of 168

Question 112 of 168

The value closest to the thermal velocity of a Helium atom at room temperature (300 K)(300 \mathrm{~K}) in ms−1\mathrm{ms}^{-1} is:
[kB=1.4×10−23 J/K;mHe=7×10−27 kg]\left[\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{~J} / \mathrm{K} ; \mathrm{m}_{\mathrm{He}}=7 \times 10^{-27} \mathrm{~kg}\right]

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Answer: (D) 1.3×1031.3 \times 10^3

v=3kBT/mHe=3×1.4×10−23×3007×10−27=1.34×103 m/s\begin{aligned} v &=\sqrt{3 k_{\mathrm{B}} T / m_{\mathrm{He}}}=\sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{7 \times 10^{-27}}} \\ &=1.34 \times 10^3 \mathrm{~m} / \mathrm{s} \end{aligned}

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