What are the speeds of the photoelectrons which are knocked out of the metal during photoeffect? Do i have to use special relativity?

I am having trouble solving a homework using relativity. But if i do it in a nonrelativistic way my result is wrong...

The homework:

We expose a same metal to a light with $\lambda_1=350nm$ and a light with $\lambda_2=540nm$. What is the work function $A_0$ of this metal if the $v_1 = 2 v_2$ relation holds for the maximum speeds - the speeds with which electrons leave the metal.


It was easy doing this in a nonrelativistic maner when kinetic energy is of a form $W_k=\tfrac{1}{2}mv^2$:

\begin{align} W_1 - A_0 = W_{k1} &= \tfrac{1}{2}m{v_1}^2 = 2m{v_2}^2\\ W_2 - A_0 = W_{k2} &= \tfrac{1}{2}m{v_2}^2 \end{align}

The workfunction is the same in both cases so:

\begin{align} A_0 &= W_1 - 2m{v_2}^2 \xleftarrow{~ ~ \text{i insert for $v_2$}~ ~ } {v_2}^2 = \tfrac{2(W_2 - A_0)}{m}\\ A_0 &= W_1 - 4(W_2-A_0)\\ A_0 &= W_1 - 4W_2 + 4A_0\\ A_0 &= \frac{4W_2 - W_1}{3}\\ A_0 &= \frac{4\tfrac{hc}{\lambda_2} - \tfrac{hc}{\lambda_1}}{3}\\ A_0 &= \tfrac{hc}{3}\left(\tfrac{4}{\lambda_2} - \tfrac{1}{\lambda_1}\right)\\ A_0 &= \tfrac{6.626\cdot 10^{-34}Js~~~ \cdot ~~~2.99\cdot10^{8}\tfrac{m}{s}}{3} \left(\tfrac{4}{350\cdot 10^{-9}m} - \tfrac{1}{540\cdot 10^{-9}m}\right)\\ A_0 &= 6.32\cdot 10^{-19}J\\ A_0 &= 3.95 eV \end{align}

The result is wrong as it should be $A_0 = 1.87eV$.


I can't seem to solve this problem in a relativistic way where $W_k = mc^2[\gamma(v)-1]$. Here is what i did do:

\begin{align} W_1 - A_0 & = W_{k1} = mc^2\gamma(v_1) - mc^2\\ W_2 - A_0 & = W_{k1} = mc^2\gamma(v_2) - mc^2 \end{align}

I solve second eq. for $mc^2$ and insert it in the first one to get:

\begin{align} W_1 - A_0 &= mc^2 \gamma(v_1) - \left(mc^2\gamma(v_2) - W_2 + A_0\right)\\ W_1 - A_0 &= mc^2 \gamma(v_1) - mc^2\gamma(v_2) + W_2 - A_0\\ W_1 - W_2 &= mc^2 \gamma(v_1) - mc^2\gamma(v_2)\\ W_1 - W_2 &= mc^2\underbrace{\left[\gamma(v_1) - \gamma(v_2)\right]}_{\llap{\text{how can i get this relation out of a $v_1=2v_2$?}}} \end{align} This is where i am stucked. Can anyone help me to solve this?

  • $\begingroup$ It is always safe to use relativistic mechanics. It's just often more difficult. $\endgroup$ Jul 7, 2013 at 14:32

1 Answer 1


You obviously don't need a relativistic calculation because the rest mass of an electron is around half a MeV and the energies you're dealing with are orders of magnitude less than this.

In fact you got the correct formula:

$$ A = \frac{4W_2 - W_1}{3} $$

but you've made a mistake with the numbers somewhere.

$W_1$ (350nm) = 3.542eV

$W_2$ (540nm) = 2.296eV

Feed these into your equation and you get $A$ = 1.881eV.

  • $\begingroup$ Thanks! Can you tell me what are the aproximate speed magnitudes of an electrons when knocked from the metal? $\endgroup$
    – 71GA
    Jul 7, 2013 at 8:15
  • 1
    $\begingroup$ If the work function is 1.881eV then the electron energy is 0.415eV for the 540nm light or 1.661ev for the 350nm light. The corresponding velocities are about $3.8 \times 10^5$ and $7.6 \times 10^5$ m/s respectively. $\endgroup$ Jul 7, 2013 at 8:22
  • $\begingroup$ $7.6\times10^5 = 0.0025 c$. Thats $0.25\%$ So how much $%$ of the speed of light must speeds be to apply special relativity? Is $5\%$ enough already? $\endgroup$
    – 71GA
    Jul 7, 2013 at 8:27
  • 1
    $\begingroup$ If you're asking for a rough guide, compare the energies in your system with the rest mass of the particles. If they are comparable you can expect relativistic effects to be large, while if they're a lot lower you can expect relativistic effects to be small. Relativistic effects are always present of course, so it's just a matter how small they need to get before you decide you can ignore them. To see how big the effects are just calculate $\gamma$. For example $\gamma = 1.01$, i.e. a 1% correction, corresponds to 0.14$c$. $\endgroup$ Jul 7, 2013 at 8:34
  • $\begingroup$ I'd like to emphasize John's last comment there: there is not magic speed where you flip from Newtonian to Einsteinian mechanics. Relativity is always correct, but you might be able to use Newton if you can tolerate the errors. So the twin questions are "What size errors do I care about?" and "What size errors do I think I make if I use Newton in this case?". $\endgroup$ Jul 7, 2013 at 18:48

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