Q. A metallic surface with work function of $2\ eV$, on heating to a temperature of 800 $K$ gives an emission current of $1\ mA$. If another metallic surface having the same surface area, same emission constant but work function $4\ eV$ is heated to a temperature of $1600\ K$, then the emission current will be
A) 1 $mA$
B) 2 $mA$
C) 4 $mA$
D) None of these
Explanation
(c) The emission current $i = AT^{2}Se^{- \phi/kT}$
For the two surfaces $A_{1} = A_{2},\ S_{1} = S_{2},\ T_{1} = 800\ K,$
$T_{2} = 1600\ K,\ \phi_{1}/T_{1} = \phi_{2}/T_{2}$
Therefore, $\frac{i_{2}}{i_{1}} = \left( \frac{T_{2}}{T_{1}} \right)^{2} = (2)^{2} = 4 \Rightarrow i_{2} = 4i_{1} = 4\ mA$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.