At what temperature the average translational \mathrm{KE} of the molecules of a gas will become equal to the \mathrm{KE} of an electron accelerated from rest through 1 \mathrm{~V} potential difference?
A \quad 10^{6} \mathrm{~K}
B 2.34 \times 10^{4} \mathrm{~K}
c 7.73 \times 10^{3} \mathrm{~K}
D none of these
At what temperature the average translational \mathrm{KE} of the molecules of a gas will become equal to the \mathrm{KE} of an electron accelerated from rest through 1 \mathrm{~V} potential difference?
A \quad 10^{6} \mathrm{~K}
B 2.34 \times 10^{4} \mathrm{~K}
c 7.73 \times 10^{3} \mathrm{~K}
D none of these

Correct option is
C 7.73 \times 10^{3} \mathrm{~K}

    \[\frac{3}{2} \mathrm{KT}=1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\]

Or”

    \[\mathrm{T}=\frac{2 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}}=7730 \mathrm{~K}=7.73 \times 10^{3} \mathrm{~K}\]