The surface of certain metal is first illuminated with light of wavelength $\lambda_1 = 350\ \text{nm}$ and then, by a light of wavelength $\lambda_2 = 540\ \text{nm}$. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to (Energy of photon $= \dfrac{1240}{\lambda\ \text{in nm}}\ \text{eV}$)
Answer: (B) $1.8$
Photon energies: $E_1 = \dfrac{1240}{350} = 3.54\ \text{eV}$, $E_2 = \dfrac{1240}{540} = 2.30\ \text{eV}$.
Since $v_1 = 2v_2$, the maximum kinetic energies are in the ratio $4:1$:
$$E_1 - \phi = 4(E_2 - \phi) \Rightarrow 3\phi = 4E_2 - E_1 = 9.19 - 3.54 = 5.65$$
$$\phi \approx 1.9\ \text{eV} \approx 1.8\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics