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(D) The root mean square velocity (thermal velocity) of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3k_B T}{m}}$.Given values are:$

Vedclass · Accepted Answer

$1.3 	imes 10^3$

Vedclass · Answer

(D) The root mean square velocity (thermal velocity) of a gas molecule is given by the formula: $v_{rms} = \sqrt{\frac{3k_B T}{m}}$.Given values are:$k_B = 1.4 	imes 10^{-23}\,J/K$$T = 300\,K$$m_{He} = 7 	imes 10^{-27}\,kg$Substituting these values into the formula:$v = \sqrt{\frac{3 	imes 1.4 	imes 10^{-23} 	imes 300}{7 	imes 10^{-27}}}$$v = \sqrt{\frac{1260 	imes 10^{-23}}{7 	imes 10^{-27}}}$$v = \sqrt{180 	imes 10^4}$$v = \sqrt{1.8 	imes 10^6}$$v \approx 1.34 	imes 10^3\,m/s$.Thus,the value closest to the thermal velocity is $1.3 	imes 10^3\,m/s$.

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

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