At what temperature does the average translational KE of a molecule in a gas becomes equal to the K.E. of an electron accelerated through a PD of $$3$$V?
Correct option is C. $$23,200$$ K
$$\dfrac{3}{2}k_BT=KE$$
$$\dfrac{3}{2}\times k_B\times T=eV$$
$$k_B=1.38 × 10^{-23}\ m^2 kg s^{-2} K^{-1}$$
$$e = 1.6\times 10^{-19}$$
$$\dfrac{3}{2}\times 1.38\times 10^{-23}\times T=1.6\times 10^{-19}\times 3$$
$$T=\dfrac{1.6\times 10^{-19}\times 3\times 2}{1.38\times 10^{-23}\times 3}$$
$$T=2.31\times 10^4$$
(temp) $$T=23188.40\approx 23,200\ K$$
Answer C.