MCQ
For a thermionic emitter (metallic) if $J$ represents the current density and $T$ is its absolute temperature then the correct curve between $\log _e \frac{J}{T^2}$ and $\frac{1}{T}$ is
- ✓
- B
- C
- D
✓
Answer
Correct option: A.
According to Richardson-Dushman equation $J=A T^2 e^{-b / T}$ Taking log of this equation $\log _e \frac{J}{T^2}=\log _e A-\frac{b}{T}$i.e. graph between $\log _e \frac{J}{T^2}$ and $\frac{1}{T}$ will be a straight line having negative slope and positive intercept $(\log A)$ on $\log _e \frac{J}{T^2}$ axis.
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