If the thermionic current density is J and em | vedclass

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(C) The Richardson-Dushman equation for thermionic emission is given by $J = AT^2 e^{-b/T}$,where $A$ and $b$ are constants.Dividing both sides by $T^

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(C) The Richardson-Dushman equation for thermionic emission is given by $J = AT^2 e^{-b/T}$,where $A$ and $b$ are constants.Dividing both sides by $T^2$,we get $\frac{J}{T^2} = A e^{-b/T}$.Taking the natural logarithm on both sides,we get $\ln\left(\frac{J}{T^2}\right) = \ln(A) - \frac{b}{T}$.This equation is of the form $y = mx + c$,where $y = \ln\left(\frac{J}{T^2}\right)$,$x = \frac{1}{T}$,$m = -b$ (slope),and $c = \ln(A)$ (intercept).However,if we plot $\frac{J}{T^2}$ directly against $\frac{1}{T}$,the relationship is $\frac{J}{T^2} = A e^{-b/T}$.As $\frac{1}{T}$ increases,the exponent $-b/T$ becomes more negative,causing the value of $\frac{J}{T^2}$ to decrease exponentially towards zero.Therefore,the curve is an exponential decay curve.

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

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