Given, Work function, $W_0 = 2.3 \text{ eV} = 2.3 \times 1.6 \times 10^{-19} \text{ J} = 3.68 \times 10^{-19} \text{ J}$ Incident light wavelength, $\lambda = 2500 \text{ \AA} = 2500 \times 10^{-10} \text{ m} = 2.5 \times 10^{-7} \text{ m}$ Planck's constant, $h = 6.634 \times 10^{-34} \text{ J s}$ Speed of light, $c = 3 \times 10^8 \text{ m/s}$ Mass of electron, $m = 9.11 \times 10^{-31} \text{ kg}$ According to Einstein's photoelectric equation, $$ \begin{aligned} E &= hf = \frac{hc}{\lambda} \\ &= \frac{6.634 \times 10^{-34} \times 3 \times 10^8}{2.5 \times 10^{-7}} \\ &= 7.9608 \times 10^{-19} \text{ J} \end{aligned} $$ Maximum kinetic energy of the emitted photo-electron, $$ \begin{aligned} K_{max} &= E - W_0 \\ &= 7.9608 \times 10^{-19} - 3.68 \times 10^{-19} \\ &= 4.2808 \times 10^{-19} \text{ J} \end{aligned} $$ We know that $K_{max} = \frac{1}{2}mv^2$. Therefore, $$ \begin{aligned} \frac{1}{2}mv^2 &= 4.2808 \times 10^{-19} \\ v^2 &= \frac{2 \times 4.2808 \times 10^{-19}}{9.11 \times 10^{-31}} \\ v^2 &= 9.398 \times 10^{11} \\ v &= \sqrt{9.398 \times 10^{11}} \\ v &\approx 9.69 \times 10^5 \text{ m/s} \end{aligned} $$ The velocity of the emitted photo-electron is approximately $9.69 \times 10^5 \text{ m/s}$.