Question

Q 44 Calculate at what temperature the
average translation K.E. of the molecules
of gas will become equal to the K.E. of
an electron accelerated from rest
through 2 V potential difference,
Ops: A.
15460 K
B.
14380 K
c.
15850 K
D.
14870 K​

Answer 1

Given :

Potential difference = 2V

Charge of the electron = 1.6×10⁻¹⁹C

To Find :

The temperature of the molecules of the gas = ?

Solution :

• The average translational kineti energy of gas is

            $K.E=\frac{3}{2} KT\rightarrow Equation(1)$

• The kinetic energy of an accelerated electron from rest is

            $K.E = qV\rightarrow Equation(2)$

• As the both kinetic energies are equal, By equating equations 1 & 2

            $\frac{3}{2} KT = qV$  

            $\frac{3}{2}\times1.38\times10^{-23} \times T=1.6\times10^{-19}\times2$

            $T = 1.5460\times 10^{4}$

            $T=15460K$

The temperature of the gas molecules is 15460K