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.


1 Answer 1


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 $U\neq U_{0}$.

  • $\begingroup$ 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, 2013 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.