1. Use Einstein’s photoelectric equation

For a metal surface, $K_{\max}=h\nu-\phi=E-\phi$ where $\phi$ is the work function.

Also, $K_{\max}=\frac{1}{2}mv_{\max}^2$

Given that the maximum speeds differ by a factor of $2$, we use the fact that kinetic energy is proportional to $v^2$.

2. Find photon energies for the two wavelengths

Using $E(\text{in eV})=\frac{1240}{\lambda(\text{in nm})}$

For $\lambda_1=350\,\text{nm}$, $E_1=\frac{1240}{350}=3.543\,\text{eV}$

For $\lambda_2=540\,\text{nm}$, $E_2=\frac{1240}{540}=2.296\,\text{eV}$

3. Relate the speeds to kinetic energies

Let the corresponding maximum speeds be $v_1$ and $v_2$.

Since the shorter wavelength has higher energy, it gives larger speed, so $v_1=2v_2$

Then, $\frac{K_1}{K_2}=\frac{\frac12 mv_1^2}{\frac12 mv_2^2}=\left(\frac{v_1}{v_2}\right)^2=2^2=4$

Thus, $K_1=4K_2$

But $K_1=E_1-\phi,\qquad K_2=E_2-\phi$

So, $(E_1-\phi)=4(E_2-\phi)$

4. Substitute values and solve for $\phi$

$3.543-\phi=4(2.296-\phi)$

$3.543-\phi=9.184-4\phi$

$3\phi=9.184-3.543=5.641$

$\phi=\frac{5.641}{3}=1.880\,\text{eV}$

So the work function is approximately $\phi\approx 1.8\,\text{eV}$

5. Check options

The closest option is:

A: 1.8

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.