Question

The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is: (hc = 1240 eV·nm)

• A. 3.09 eV
• B. 1.42 eV
• C. 1.51 eV
• D. 1.68 eV

Hint:

In photoelectric effect, the energy of the incoming photon is the sum of the kinetic energy of the emitted electron and the work function.

Answer 1

The Correct Option is C

The energy of the photon is given by: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] Substitute \( \lambda = 400 \, \text{nm} \), \( hc = 1240 \, \text{eV} \cdot \text{nm} \) to find the photon energy. Then use the photoelectric equation: \[ E_{\text{photon}} = E_{\text{kinetic}} + \text{Work function} \] Solving gives the work function as 1.51 eV.
Final Answer: \[ \boxed{1.51 \, \text{eV}} \]