Question

Let the energy of an emitted photoelectron be \( E \) and the wavelength of incident light be \( \lambda_0 \). What will be the change in \( E \) if \( \lambda_0 \) is doubled?

• A. \( E = \frac{E_0}{2} \)
• B. \( E = E_0 \)
• C. \( E = 2E_0 \)
• D. \( E = \frac{E_0}{4} \)

Hint:

The energy of photoelectrons is directly proportional to the frequency of the incident light. Doubling the wavelength reduces the energy by a factor of 4.

Answer 1

The Correct Option is D

Step 1: Energy of Photoelectron.
The energy of a photoelectron is related to the frequency of the incident light: \[ E = h f - \phi \] where \( h \) is Planck’s constant, \( f \) is the frequency, and \( \phi \) is the work function of the metal.
Step 2: Doubling Wavelength.
Since frequency \( f \) is inversely proportional to the wavelength, doubling the wavelength results in reducing the frequency by half. Therefore, the energy of the emitted electron will decrease by a factor of 4.
Step 3: Conclusion.
The correct answer is (D), \( E = \frac{E_0}{4} \).