A metallic surface with a work function of 2 | vedclass

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(C) The emission current $i$ is given by the Richardson-Dushman equation: $i = A T^2 S e^{-\phi / kT}$.Given for the first surface: $\phi_1 = 2 \, eV$

Vedclass · Accepted Answer

$4$

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(C) The emission current $i$ is given by the Richardson-Dushman equation: $i = A T^2 S e^{-\phi / kT}$.Given for the first surface: $\phi_1 = 2 \, eV$,$T_1 = 800 \, K$,$i_1 = 1 \, mA$.Given for the second surface: $\phi_2 = 4 \, eV$,$T_2 = 1600 \, K$,$A_1 = A_2$,$S_1 = S_2$.We observe that the ratio $\frac{\phi_1}{T_1} = \frac{2}{800} = 0.0025$ and $\frac{\phi_2}{T_2} = \frac{4}{1600} = 0.0025$. Thus,$\frac{\phi_1}{T_1} = \frac{\phi_2}{T_2}$.Taking the ratio of the currents: $\frac{i_2}{i_1} = \frac{A T_2^2 S e^{-\phi_2 / kT_2}}{A T_1^2 S e^{-\phi_1 / kT_1}} = \left( \frac{T_2}{T_1} ight)^2 \cdot e^{-(\phi_2/kT_2 - \phi_1/kT_1)}$.Since $\frac{\phi_2}{kT_2} = \frac{\phi_1}{kT_1}$,the exponential term becomes $e^0 = 1$.Therefore,$\frac{i_2}{i_1} = \left( \frac{1600}{800} ight)^2 = (2)^2 = 4$.Hence,$i_2 = 4 	imes i_1 = 4 	imes 1 \, mA = 4 \, mA$.

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