A metal with work function 2.5 eV is illuminated by light of wavelength 300 nm. What is the maximum kinetic energy of the emitted electrons?

Step-by-step solution

1. Find the photon energy using $E=\frac{hc}{\lambda }$. Since $hc=1240\text{ eV⋅nm}$, $E=\frac{1240}{300}=4.133\text{ eV}$.
2. Subtract the work function: $K_{max}=E-\Phi =4.133-2.5=1.633\text{ eV}$.
3. Round to one decimal place: $K_{max}\approx 1.6\text{ eV}$.
4. Choose the matching option.

Concept tested: photoelectric effect maximum kinetic energy

Published 6 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked