Question

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

• A. $1.3\times 10^4$
• B. $1.3\times 10^3$
• C. $1.3\times 10^5$
• D. $1.3\times 10^2$

Answer 1

Answer: (B)

We know that $v_{rms}=\sqrt{\frac{3kT}{m}}$.
Given, $k=1.4\times 10^{-23}J/K;T=300K;m=7\times 10^{-27}kg$. 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 $v_{rms}=1.34\times 10^3$