If the work function of the target material is 1.25 eV and the wavelength of the incident radiation is 2460 Å the stopping potential required to stop the emitted electrons is:

(Given: $h = 6.6 × 10^{-34} Js$)

3.75 V

The correct answer is Option (3) → 3.75 V

Given:

Work function: $W = 1.25\ \text{eV}$

Wavelength: $\lambda = 2460\ \text{Å} = 2460 \times 10^{-10}\ \text{m} = 2.46 \times 10^{-7}\ \text{m}$

Photon energy:

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

$E = \frac{12400}{2460}\ \text{eV}$ (using $hc = 12400\ \text{eVÅ}$)

$E = 5.04\ \text{eV}$

Kinetic energy of emitted electron:

$K.E. = E - W = 5.04 - 1.25 = 3.79\ \text{eV}$

Stopping potential:

$V_0 = K.E./e = 3.79\ \text{V}$

Answer: $V_0 = 3.79\ \text{V}$