MCQ
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
- ✓
- D
✓
Answer
Correct option: C.
$J=A T^2 e^{-b / T} \Rightarrow \frac{J}{T^2} \propto e^{-b / T}$i.e. $\frac{J}{T^2}$ will vary exponentially with $\frac{1}{T}$, having negative slope.
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