The surface of certain metal is first illuminated with light of wavelength $$\lambda_1 = 350$$ nm and then, by a light of wavelength $$\lambda_2 = 540$$ 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 $$= \frac{1240}{\lambda \; in \; nm}$$ eV)

For photo-emission we always begin with Einstein’s photoelectric equation

$$K_{\text{max}} \;=\; E_{\text{photon}} \;-\; \phi$$

where $$K_{\text{max}}$$ is the maximum kinetic energy of the emitted electron, $$E_{\text{photon}}$$ is the incident photon energy and $$\phi$$ is the work function of the metal.

The problem itself supplies the convenient numerical relation

$$E_{\text{photon}}\;(\text{in eV}) \;=\;\frac{1240}{\lambda\;(\text{in nm})}$$

First we evaluate the photon energies for the two given wavelengths.

For $$\lambda_1 = 350\ \text{nm}$$, we have

$$E_1 \;=\;\frac{1240}{350}\;=\;3.542857\ \text{eV}$$

For $$\lambda_2 = 540\ \text{nm}$$, we have

$$E_2 \;=\;\frac{1240}{540}\;=\;2.296296\ \text{eV}$$

Let $$K_1$$ and $$K_2$$ denote the maximum kinetic energies of the photo-electrons corresponding to $$\lambda_1$$ and $$\lambda_2$$ respectively. Hence, from Einstein’s equation,

$$K_1 \;=\;E_1 - \phi$$

$$K_2 \;=\;E_2 - \phi$$

Next we translate the given information about the speeds of the electrons into a relation between the kinetic energies. The kinetic energy of an electron is

$$K \;=\;\frac{1}{2} m v^2$$

The statement “the maximum speed of the photo-electrons in the two cases differ by a factor of 2” means

$$v_1 \;=\;2\,v_2$$

Substituting this into the kinetic-energy formula, we obtain

$$K_1 \;=\;\tfrac12 m (v_1)^2 \;=\;\tfrac12 m (2 v_2)^2 \;=\;4\left(\tfrac12 m v_2^2\right) \;=\;4 K_2$$

Thus

$$K_1 \;=\;4\,K_2$$

We now substitute the expressions for $$K_1$$ and $$K_2$$ obtained from Einstein’s equation:

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

Writing the numerical values of $$E_1$$ and $$E_2$$:

$$3.542857 - \phi \;=\;4\,(2.296296 - \phi)$$

Expanding the right-hand side,

$$3.542857 - \phi \;=\;9.185184 - 4\phi$$

Now we collect the terms containing $$\phi$$ on the left:

$$3.542857 - \phi + 4\phi \;=\;9.185184$$

$$3.542857 + 3\phi \;=\;9.185184$$

Isolating $$\phi$$ gives

$$3\phi \;=\;9.185184 - 3.542857$$

$$3\phi \;=\;5.642327$$

$$\phi \;=\;\frac{5.642327}{3}$$

$$\phi \;\approx\;1.880776\ \text{eV}$$

Rounding to the precision implied by the options, the work function is very close to $$1.8\ \text{eV}$$.

Hence, the correct answer is Option B.