Monochromatic light falling on a metal surface of work function 4.2 eV ejects electrons with kinetic energies from zero to 2.6 eV. What is the energy of the incident photon?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6.8 eV
By Einstein’s photoelectric equation, \( K_{\mathrm{max}}=h\nu-\phi_0 \). The electrons that need only the work function to escape come out with the maximum kinetic energy. More tightly bound electrons come out with less, down to zero. So the top value of the range, 2.6 eV, is \( K_{\mathrm{max}} \).
\( \begin{aligned} h\nu &= K_{\mathrm{max}}+\phi_0 \\ &= 2.6\,\mathrm{eV}+4.2\,\mathrm{eV} \\ &= 6.8\,\mathrm{eV} \end{aligned} \)
NCERT (2026-27):
Einstein’s Photoelectric Equation: Energy Quantum of Radiation – “… In photoelectric effect, an electron absorbs a quantum of energy (hν) of radiation. If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function φ0), the electron is emitted with maximum kinetic energy
\( K_{\mathrm{max}}=h\nu-\phi_0 \)\( \qquad (11.2) \)
More tightly bound electrons will emerge with kinetic energies less than the maximum value.”