Question

At what temperature the linear kinetic energy of a gas molecule will be equal to that of an electron accelerated through a potential difference of 10 volt?

• A. 77.3 x 10${}^3$ K
• B. 38.65 x 10${}^3$ K
• C. 19.x10${}^3$ K
• D. 273 K

Answer 1

Answer: (A)

77.3 x 10${}^3$ K

Energy of an electron accelerated through a potential V: $E=eV$

Given,

$E=\frac{3}{2}kT$

$\Rightarrow eV=\frac{3}{2}kT$

$\Rightarrow T=\frac{2eV}{3k}=\frac{2e\times 10}{3\times 1.38\times {10}^{-23}}=\frac{20eV}{3\times 1.38\times {10}^{-23}}=\frac{20\times 1.6\times {10}^{-19}J}{3\times 1.38\times {10}^{-23}}=77.3\times {10}^{3}K$