Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is:

For photoelectric emission, we always begin with Einstein’s photoelectric equation, which states:

$$h\nu \;=\; \phi \;+\; K_{\text{max}}$$

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

First we calculate the photon energy. The wavelength of the radiation is given as $$\lambda = 250\ \text{nm} = 250 \times 10^{-9}\ \text{m}.$$ The frequency $$\nu$$ is related to the wavelength by $$\nu = \dfrac{c}{\lambda},$$ so the photon energy is

$$E_{\text{photon}} = h\nu = \dfrac{hc}{\lambda}.$$

Substituting the values $$h = 6.626 \times 10^{-34}\ \text{J\,s},\; c = 3.00 \times 10^{8}\ \text{m/s},\; \lambda = 250 \times 10^{-9}\ \text{m},$$ we obtain

$$E_{\text{photon}} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{250 \times 10^{-9}}\ \text{J}.$$

Now we multiply the two numbers in the numerator:

$$6.626 \times 3.00 = 19.878,$$ so

$$E_{\text{photon}} = \dfrac{19.878 \times 10^{-26}}{250 \times 10^{-9}}\ \text{J}.$$

Express the denominator in scientific notation: $$250 \times 10^{-9} = 2.5 \times 10^{-7}.$$ Hence

$$E_{\text{photon}} = \dfrac{19.878 \times 10^{-26}}{2.5 \times 10^{-7}}\ \text{J} = \left(\dfrac{19.878}{2.5}\right) \times 10^{-19}\ \text{J} = 7.951 \times 10^{-19}\ \text{J}.$$

To convert this energy to electron-volts we use $$1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}.$$ Thus

$$E_{\text{photon}} = \dfrac{7.951 \times 10^{-19}}{1.602 \times 10^{-19}}\ \text{eV} \approx 4.96\ \text{eV}.$$

Next we compute the maximum kinetic energy of the emitted photoelectrons. It is stopped by a potential of $$V_s = 0.5\ \text{V},$$ and the stopping potential is related to kinetic energy by the formula

$$K_{\text{max}} = eV_s,$$

where $$e = 1.602 \times 10^{-19}\ \text{C}$$ is the elementary charge. Therefore

$$K_{\text{max}} = (1.602 \times 10^{-19}\ \text{C})(0.5\ \text{V}) = 0.801 \times 10^{-19}\ \text{J}.$$

Converting this to electron-volts as before:

$$K_{\text{max}} = \dfrac{0.801 \times 10^{-19}}{1.602 \times 10^{-19}}\ \text{eV} = 0.500\ \text{eV}.$$

Now we apply Einstein’s equation in electron-volts:

$$\phi = E_{\text{photon}} - K_{\text{max}} = 4.96\ \text{eV} - 0.50\ \text{eV} = 4.46\ \text{eV}.$$

On rounding to the usual significant figures, the work function is $$\phi \approx 4.5\ \text{eV}.$$

Hence, the correct answer is Option C.