Question

Light of wavelength 600 nm is incident on a metal surface with work function 2 eV. The maximum kinetic energy of photoelectrons is:

• A. 0.07 eV
• B. 0.7 eV
• C. 2.07 eV
• D. 4.14 eV

Hint:

\(E(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})}\).

Answer 1

The Correct Option is A

Photon energy: \[ E = \frac{1240}{600} \approx 2.07\ \text{eV} \] \[ K_{\max} = E - \phi = 2.07 - 2 = 0.07\ \text{eV} \]