Do photoelectrons move fast enough to use special relativity? (+ homework)

Question

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.

---

NONRELATIVISTIC APPROACH:

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 $A_0 = 1.87eV$.

---

RELATIVISTIC APPROACH:

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?

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.