Q. If the thermionic current density is $J$ and emitter temperature is $T$ then the curve between $\frac{J}{T^{2}}$ and $\frac{1}{T}$ will be
A)
B)
C)
D)
Explanation
(c) $J = AT^{2}e^{- b/T}$
$\Rightarrow \frac{J}{T^{2}} \propto e^{- b/T}$
$i.e.,\frac{J}{T^{2}}$ will vary exponentially with $\frac{1}{T}$, having negative slope
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.