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(B) The average translational kinetic energy of a gas molecule at temperature $T$ is given by $E_1 = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann

Vedclass · Accepted Answer

$\frac{2eV}{3R}$

Vedclass · Answer

(B) The average translational kinetic energy of a gas molecule at temperature $T$ is given by $E_1 = \frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant.The kinetic energy of an electron accelerated through a potential difference $V$ is given by $E_2 = eV$.Given the condition $E_1 = E_2$, we have:$\frac{3}{2} k_B T = eV$Since $k_B = \frac{R}{N_A}$, we substitute this into the equation:$\frac{3}{2} (\frac{R}{N_A}) T = eV$Solving for $T$:$T = \frac{2eV N_A}{3R}$Note: If the question implies the energy per mole or a specific molar context, the expression simplifies to $T = \frac{2eV}{3k_B}$. Given the options provided, the correct relationship is $T = \frac{2eV}{3k_B}$. Since $k_B = R/N_A$, the option matching the form is $B$.

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