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  4. The value closest to the thermal velocity of a Helium atom at room temperature (300 K) in ms-1 is: [kB = 1.4 × 10-23 J/K; mHe = 7 × 10-27 kg ]

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Q. The value closest to the thermal velocity of a Helium atom at room temperature $(300\, K)$ in $ms^{-1}$ is :$ [k_B = 1.4 \times 10^{-23} \, J/K; m_{He} = 7 \times 10^{-27} \, kg ]$

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Solution:

We know that $v_{ rms }=\sqrt{\frac{3 k T}{m}}$.
Given, $k=1.4 \times 10^{-23} \,J / K ; T=300 \,K ; m=7 \times 10^{-27}\, kg$. Therefore,
$v_{ rms } =\sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{7 \times 10^{-27}}}=\sqrt{\frac{3 \times 300 \times 14 \times 10^{-23}}{7 \times 10 \times 10^{-27}}}=\sqrt{\frac{3^{2} \times 10^{2} \times 2 \times 10^{4}}{10}} $
$=3 \times 10 \times 10^{2} \times \sqrt{\frac{2}{10}} $
$=3 \times \sqrt{10} \times 10^{2} \times \sqrt{2}=3 \times \sqrt{2} \times \sqrt{5} \times 10^{2} \times \sqrt{2}=3 \times 2 \times \sqrt{5} \times 10^{2}$
$v_{ rms } =6 \times 10^{2} \times \sqrt{5}=13.41 \times 10^{2}$
or $ v_{ rms }=1.34 \times 10^{3}$