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.
c
(c)$J = A{T^2}{e^{ - b/T}}$ ==> $\frac{J}{{{T^2}}} \propto {e^{ - b/T}}$
i.e. $\frac{J}{{{T^2}}}$ will vary exponentially with $\frac{1}{T}$, having negative slope.
(c)$J = A{T^2}{e^{ - b/T}}$ ==> $\frac{J}{{{T^2}}} \propto {e^{ - b/T}}$
i.e. $\frac{J}{{{T^2}}}$ will vary exponentially with $\frac{1}{T}$, having negative slope.
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