For a thermionic emitter (metallic),if J repr | vedclass

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(A) According to the Richardson-Dushman equation,the current density $J$ is given by $J = A T^2 e^{-b/T}$,where $A$ and $b$ are constants.Taking the n

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(A) According to the Richardson-Dushman equation,the current density $J$ is given by $J = A T^2 e^{-b/T}$,where $A$ and $b$ are constants.Taking the natural logarithm on both sides:$\log_e J = \log_e A + 2 \log_e T - \frac{b}{T}$Rearranging the terms to isolate $\log_e \frac{J}{T^2}$:$\log_e \frac{J}{T^2} = \log_e A - \frac{b}{T}$This equation is of the form $y = mx + c$,where $y = \log_e \frac{J}{T^2}$,$x = \frac{1}{T}$,$m = -b$ (slope),and $c = \log_e A$ (intercept).Since the slope $m = -b$ is negative,the graph is a straight line with a negative slope and a positive intercept on the $y$-axis.Therefore,the correct curve is represented by the graph with a downward-sloping straight line.

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:

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