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

Question

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.

Answer 1

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.