# How to determine the work function and electron velocities from data on the photoelectric effect?

A metal received light with wavelength $$300~\mathrm{nm}$$ in experiment A and light with wavelength $$500~\mathrm{nm}$$ in experiment B. In one experiment the electron speed was twice the speed of that in the other experiment. Calculate the velocities and the work function ($$W$$) of the metal.

Because it's the same metal in both experiments we have $$W_\mathrm{A} = W_\mathrm{B}$$, which means that $$E_\mathrm{A} - K_\mathrm{A} = E_\mathrm{B} - K_\mathrm{B}$$.

We know that $$E=hc/\lambda$$ and $$K=mv^2/2$$. Because $$\lambda_\mathrm{A} < \lambda_\mathrm{B}$$, we have $$E_\mathrm{A} > E_\mathrm{B}$$ and $$K_\mathrm{A} < K_\mathrm{B}$$ which means $$v_\mathrm{A} < v_\mathrm{B}$$ and $$v_\mathrm{B} = 2v_\mathrm{A}$$.

Then I made the substituions in my initial equality but when I try to calculate $$v_\mathrm{A}$$ I obtained a negative number inside my root.

• Hint: if $E_\mathrm{A} - K_\mathrm{A} = E_\mathrm{B} - K_\mathrm{B}$ and $E_\mathrm{A} > E_\mathrm{B}$ then $K_\mathrm{A} > K_\mathrm{B}$. Given two numbers you have to subtract more from a bigger one to make them equal. Or, from physical perspective, the bigger is the energy of a photon you shoot, the bigger is the kinetic energy of an electron ejected. Feb 20 '16 at 15:51

Consider the difference between the kinetic energy in case A and B: $$K_A-K_B=(E_A-W)-(E_B-W)=E_A-E_B$$ where the work function cancels out because it is the same metal in each experiment. The right hand side of this is easy to evaluate using $$hc/\lambda$$, which gives $$E_A-E_B=hc\left(\frac{1}{\lambda_A}-\frac{1}{\lambda_B}\right)=(\pu{1.9878\times10^{-25}J\cdot m})\cdot(\pu{1.333\times10^6m^{-1}})=\pu{2.65\times10^{-19}J}$$

To solve for the left hand side, we simplify to a single variable by recognizing that $$v_a=2v_b$$ (as Wildcat said, the right hand side is positive and the only way the left hand side will be is if A has the larger velocity): $$K_A-K_B=\frac{m}{2}(v_A^2-v_B^2)=\frac{m}{2}(4v_B^2-v_B^2)=\frac{3}{2}mv_B^2$$ With an electron mass of $$\pu{9.109\times10^{-31}kg}$$, this gives $$v_B=\pu{4.404\times10^5 \frac{m}{s}}$$ and $$v_A=\pu{8.809\times10^5 \frac{m}{s}}$$. The work function can then be solved by evaluating either $$K_A$$ or $$K_B$$ and solving for the difference $$E_x-K_x=W$$ where $$x$$ is either A or B.