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

Question

Dr. Abubacker · Accepted Answer

(a) According to Richardson-Dushman equation
$J = AT^{2}e^{- b/T}$
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 $(\log_{e}A)$ on $\log_{e}\frac{J}{T^{2}}$ axis

Q. 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

Explanation

Reviewed by Dr. Abubacker

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