Light of wavelength 300 nm is used in an experiment on photoelectric effect with lithium having a work function of 2.7 eV. The maximum kinetic energy of ejected photo-electrons will be

1.43 eV

The correct answer is Option (4) → 1.43 eV

Energy of incident photons: $E = \frac{hc}{\lambda}$

Given $\lambda = 300\ \text{nm} = 300 \times 10^{-9}\ \text{m}$, $h = 6.626 \times 10^{-34}\ \text{Js}$, $c = 3 \times 10^8\ \text{m/s}$

$E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{300 \times 10^{-9}} \approx 6.626 \times 10^{-19}\ \text{J}$

Convert to eV: $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$

$E \approx \frac{6.626 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 4.14\ \text{eV}$

Maximum kinetic energy of photoelectrons: $K_\text{max} = E - \phi = 4.14 - 2.7 = 1.44\ \text{eV}$

Final Answer: $K_\text{max} \approx 1.44\ \text{eV}$