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I have been trying to solve a problem about photocell and have obtained the wrong result. Can anyone tell me where did I go wrong? So it is easier I will write down all my calculations.

We have a photocell and we shine a light with of $\lambda=550nm$ on a cathode with a surface $4mm^2$. On the surface of the cathode $j=1\times10^3W/m^2$.

(a) What is the work function $A_0$ of the metal from which cathode is made if the stopping voltage equals $U_0=1.2V$?

(b) What current flows through the photocell if we connect it to a voltage (not the stopping voltage) and only $5\%$ of the incoming photons manage to trigger the photoelectric effect.


First I calculated the energy of an incoming photon:

\begin{align} W = \frac{hc}{\lambda} = \frac{1.602\times 10^{-19}Js\cdot 2.99\times10^{8}\tfrac{m}{s}}{550\times10^{-9}m} = 3.602\times10^{-19}J \end{align}

Then I calculated the work function for question (a): \begin{align} W &= W_k + A_0\\ W&= U_0 e_0 + A_0\\ A_0 &= W - U_0 e_0\\ A_0 &= 3.602\times10^{-19}J - 1.2V \cdot 1.602\times10^{-19}As\\ A_0 &= 1.680\times 10^{-19}J\\ A_0 &= 1.05 eV \end{align}

Until now, it went well and my results match with the book, but when I try to calculate the answer for (b) i don't get the right result. This is what I did:

\begin{align} j=\smash{\frac{dP}{dS}} \longrightarrow P &= j \cdot S \\ P&= 1\times10^3 W/m^2 \cdot 4\times10^{-6}m^2\\ P&= 4 \times 10 ^{-3} W \end{align}

This is the power of all of the incoming photons while only $5\%$ manage to cause the photoelectric effect. But the power of the photons who manage to cause the photoelectric effect is smaller so $P'= 0.05 \cdot P = 2\times10^{-4}W$. Using $P'$ i now try to calculate the current flowing in the circuit and i get the wrong result (the right one is $I=89\mu A$):

\begin{align} P' &= UI\\ I &= \frac{P'}{U}\longleftarrow \substack{\text{I am not sure about this step where}\\\text{I inserted $U=1.2V$. Does $U$ effect}\\\text{the current $I$?}}\\ I &= \frac{2\times10^{-4}W}{1.2V}\\ I &= 166.6\overline{6} \mu A \end{align}

I get about twice as much as is the solution but it is unfortunately wrong. A hint or some explanation would be helpful.

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The formula that gives you the right answer is

$I=e\:\left(=1.602*10^{-19}\:\text{C}\right)\cdot\frac{dn_{\text{el}}}{dt}=e\cdot\eta\left(=5\%\right)\cdot\frac{P\left(=\:4\cdot 10^{-3}\:\text{W}\right)}{W=h\cdot\nu}=8.895058301\cdot10^{-5}\:A\approx89.9\mu A,$

where $\frac{dn_{\text{el}}}{dt}$ describes the rate of emitted electrons per time due to the photo effect. You are not supposed to use the voltage $U_{0}$ (as written in part b) of the problem text) since you have now another stop voltage $U\neq U_{0}$.

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  • $\begingroup$ Thank you. I did not know that the current is to be quantized like that. I think this question is important in this manner for all the students like me (self taught). $\endgroup$ – 71GA Jul 13 '13 at 8:21

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