Q. The ratio of work functions and temperatures of two emitters are 1 : 2, then the ratio of current densities obtained by them will be
A) 4:1
B) 2:1
C) 1:2
D) 1:4
Explanation
(d) $\frac{J_{1}}{J_{2}} = \frac{AT_{1}^{2}e^{- 1W_{1}/kT_{1}}}{AT_{2}^{2}e^{- W_{2}/kT_{2}}} = \left( \frac{T_{1}}{T_{2}} \right)^{2}e^{- W_{1}/kT_{1} + W_{2}/kT_{2}}$
$= \left( \frac{T_{1}}{T_{2}} \right)^{2}e^{- W_{1}/kT_{1}\left( 1 - \frac{W_{2}}{W_{1}} \times \frac{T_{1}}{T_{2}} \right)}$
$= \left( \frac{1}{2} \right)^{2}e^{- W_{1}/kT_{1}(1 - 2/1 \times 1/2)} = \frac{1}{4}e^{0} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.