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(D) The average translational kinetic energy of a gas molecule is given by $KE = \frac{3}{2} k_B T$. The kinetic energy gained by an electron accelera

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$77.3 	imes 10^3 \,K$

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(D) The average translational kinetic energy of a gas molecule is given by $KE = \frac{3}{2} k_B T$. The kinetic energy gained by an electron accelerated through a potential difference $V$ is $KE = eV$. Given that these two energies are equal: $\frac{3}{2} k_B T = eV$ Substituting the values $V = 10 \,V$, $e = 1.6 	imes 10^{-19} \,C$, and $k_B = 1.38 	imes 10^{-23} \,JK^{-1}$: $T = \frac{2eV}{3k_B} = \frac{2 	imes 1.6 	imes 10^{-19} 	imes 10}{3 	imes 1.38 	imes 10^{-23}}$ $T = \frac{32 	imes 10^{-19}}{4.14 	imes 10^{-23}} \approx 7.73 	imes 10^4 \,K = 77.3 	imes 10^3 \,K$.

If the average translational kinetic energy of a molecule in a gas is equal to the kinetic energy of an electron accelerating from rest through $10 \,V$, then the temperature of the gas molecule is (Boltzmann constant $= 1.38 \times 10^{-23} \,JK^{-1}$)

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