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(D) The average translational kinetic energy of a gas molecule is given by $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant $(1.38 	imes 10^{-

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$7.7 	imes 10^3\, K$

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(D) The average translational kinetic energy of a gas molecule is given by $E = \frac{3}{2}kT$,where $k$ is the Boltzmann constant $(1.38 	imes 10^{-23}\, J/K)$ and $T$ is the absolute temperature.The energy gained by an electron accelerated through a potential difference of $V = 1\, V$ is $E = eV$,where $e = 1.6 	imes 10^{-19}\, C$.Equating the two energies:$\frac{3}{2}kT = eV$Solving for $T$:$T = \frac{2eV}{3k}$Substituting the values:$T = \frac{2 	imes 1.6 	imes 10^{-19}}{3 	imes 1.38 	imes 10^{-23}}$$T = \frac{3.2 	imes 10^{-19}}{4.14 	imes 10^{-23}}$$T \approx 0.7729 	imes 10^4\, K = 7.7 	imes 10^3\, K$.

The temperature at which the average translational kinetic energy of a molecule is equal to the energy gained by an electron in accelerating from rest through a potential difference of $1\, V$ is:

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