Question

For a photocell, the work function is \(\phi\) and the stopping potential is \(V_s\). The wavelength of the incident radiation is

• A. \(\dfrac{hc}{\phi + eV_s}\)
• B. \(\dfrac{\phi + eV_s}{hc}\)
• C. \(\dfrac{\phi - eV_s}{hc}\)
• D. \(\dfrac{hc}{\phi - eV_s}\)

Hint:

Always add work function and maximum kinetic energy to get the photon energy in photoelectric effect problems.

Answer 1

The Correct Option is A

Step 1: Write Einstein’s photoelectric equation.
The energy of the incident photon is given by:
\[ E = h\nu = \phi + K_{\text{max}} \]
Step 2: Relate maximum kinetic energy to stopping potential.
\[ K_{\text{max}} = eV_s \]
Step 3: Substitute in the photoelectric equation.
\[ h\nu = \phi + eV_s \]
Step 4: Express frequency in terms of wavelength.
\[ \nu = \frac{c}{\lambda} \]
Step 5: Solve for wavelength.
\[ \lambda = \frac{hc}{\phi + eV_s} \]
Step 6: Conclusion.
The wavelength of the incident radiation is \(\dfrac{hc}{\phi + eV_s}\).