Use Einstein’s photoelectric equation

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

Here,

• $\phi$ = work function of the metal
• $K_{\max}$ = maximum kinetic energy of emitted electron

Relate stopping potential to kinetic energy

The stopping potential is $V_0 = 0.5\,\text{V}$, so

$K_{\max} = eV_0 = 0.5\,\text{eV}$

Calculate energy of incident radiation

Wavelength given:

$\lambda = 250\,\text{nm}$

Photon energy:

$E = \frac{hc}{\lambda}$

Using the standard relation

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

$E = \frac{1240}{250} = 4.96\,\text{eV}$

Find the work function

$\phi = E - K_{\max}$

$\phi = 4.96 - 0.5 = 4.46\,\text{eV}$

This is approximately

$\phi \approx 4.5\,\text{eV}$

Match with the options

The correct option is:

B: $4.5\,\text{eV}$