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(D) The average kinetic energy of a gas molecule is given by $K_{avg} = \frac{3}{2}kT$,where $k$ is the Boltzmann constant and $T$ is the temperature

Vedclass · Accepted Answer

$7.7 	imes 10^3 \, K$

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

At what temperature is the average kinetic energy equal to the energy gained by an electron accelerated from rest through a potential difference of $1 \, V$?

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