When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to:

According to Einstein’s photo-electric equation, the maximum kinetic energy of the emitted photo-electrons is given by

$$K_{\text{max}} = h\nu - \phi,$$

where $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the incident radiation and $$\phi$$ is the work function of the metal.

Since frequency $$\nu$$ and wavelength $$\lambda$$ are related by $$\nu = \dfrac{c}{\lambda}$$ (with $$c$$ the speed of light), we may write

$$K_{\text{max}} = \dfrac{hc}{\lambda} - \phi.$$

Let us call the first wavelength $$\lambda_1 = 500 \text{ nm}$$ and the corresponding maximum kinetic energy $$K_1$$. For the second case $$\lambda_2 = 200 \text{ nm}$$ and the kinetic energy is given to be three times larger, so $$K_2 = 3K_1$$.

Writing Einstein’s equation for each case, we have

$$K_1 = \dfrac{hc}{\lambda_1} - \phi \quad\quad (1)$$ $$K_2 = \dfrac{hc}{\lambda_2} - \phi. \quad\quad (2)$$

Because $$K_2 = 3K_1$$, substituting this into equation (2) gives

$$3K_1 = \dfrac{hc}{\lambda_2} - \phi. \quad\quad (3)$$

Now we subtract equation (1) from equation (3) to eliminate $$\phi$$:

$$3K_1 - K_1 = \left(\dfrac{hc}{\lambda_2} - \phi\right) - \left(\dfrac{hc}{\lambda_1} - \phi\right).$$

This simplifies to

$$2K_1 = hc\left(\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1}\right).$$

So the single-energy $$K_1$$ is

$$K_1 = \dfrac{hc}{2}\left(\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1}\right).$$

We next substitute numerical values. A convenient constant for photo-electric problems is

$$hc = 1240 \ \text{eV·nm}.$$

The reciprocal wavelengths are

$$\dfrac{1}{\lambda_2} = \dfrac{1}{200 \text{ nm}} = 0.005 \ \text{nm}^{-1},$$ $$\dfrac{1}{\lambda_1} = \dfrac{1}{500 \text{ nm}} = 0.002 \ \text{nm}^{-1}.$$

Thus

$$\dfrac{1}{\lambda_2} - \dfrac{1}{\lambda_1} = 0.005 - 0.002 = 0.003 \ \text{nm}^{-1}.$$

Now we find $$K_1$$:

$$K_1 = \dfrac{1240 \ \text{eV·nm}}{2} \times 0.003 \ \text{nm}^{-1}.$$

$$K_1 = 620 \ \text{eV·nm} \times 0.003 \ \text{nm}^{-1} = 1.86 \ \text{eV}.$$

With $$K_1$$ known, we return to equation (1) to extract the work function:

$$\phi = \dfrac{hc}{\lambda_1} - K_1.$$

The photon energy for $$\lambda_1 = 500 \text{ nm}$$ is

$$\dfrac{hc}{\lambda_1} = \dfrac{1240 \ \text{eV·nm}}{500 \ \text{nm}} = 2.48 \ \text{eV}.$$

Therefore

$$\phi = 2.48 \ \text{eV} - 1.86 \ \text{eV} = 0.62 \ \text{eV}.$$

This value is closest to $$0.61 \ \text{eV}$$ among the given options.

Hence, the correct answer is Option D.