# The temperature at which average translational K.E. of a molecule is equal to the K.E. of an...

## Question:

The temperature at which average translational K.E. of a molecule is equal to the K.E. of an electron accelerated from rest through a potential difference of 1 V, is:

(a) T = 7729 K

(b) T = 8879 K

(c) T = 7.72 K

(d) T = 772.9 K

## Average Translational Kinetic energy:

The average translational kinetic energy of the gas is directly proportional to the temperature and it is constant for all the gases at the same temperature.

{eq}\displaystyle{ \\ }{/eq}

Given Data:

Potential difference {eq}V = 1\ V {/eq}

Magnitude of charge of the electron {eq}(q) = 1.6\times 10^{-19} \ C {/eq}

Since the average translational K.E. of a molecule is equal to the K.E. of an accelerated electron, we can write it mathematically as:

{eq}K.E. = qV {/eq}

This energy has to be equal to the average kinetic energy, therefore:

{eq}\color{red}{K.E. = \dfrac{3}{2}kT} \\ qV = \dfrac{3}{2}kT \\ (1.6 \times 10^{-19})(1) = \dfrac{3}{2} (1.381 \times 10^{-23}) (T) \\ \color{blue}{T = 7.72 \times 10^{3} \ K} {/eq}

Therefore the correct option would be (a).