Relation between kinetic energy (E) and RMS speed (u) is given by:

u = {(\frac{2 E}{m})}^{1 / 2}

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u = {(\frac{3 E}{2 m})}^{1 / 2}

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u = {(\frac{E}{2 m})}^{1 / 2}

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u = {(\frac{E}{3 m})}^{1 / 2}

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Solution

The correct option is B $u = {(\frac{2 E}{m})}^{1 / 2}$
Since,
$p = \frac{1}{3} \times \frac{m}{v} \times c^{2}$ ( $c =$ rms velocity)

$\Rightarrow \sqrt{\frac{3 P V}{m}} = c$
For 1 molecule, $P V = k T \Rightarrow c = \sqrt{\frac{3 k T}{m}}$

$. \cdot . K E = \frac{3}{2} k T$

$\Rightarrow 2 K E = 3 k T$

$\Rightarrow c = \sqrt{\frac{2 K E}{m}}$

Hence, the correct answer is option $A$ .

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