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.
a
(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 $(logeA)$ on ${\log _e}\frac{J}{{{T^2}}}{\rm{axis}}{\rm{.}}$
(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 $(logeA)$ on ${\log _e}\frac{J}{{{T^2}}}{\rm{axis}}{\rm{.}}$
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