Question

If \(K_1\) and \(K_2\) are maximum kinetic energies of photoelectrons emitted when lights of wavelengths \(\lambda_1\) and \(\lambda_2\), respectively incident on a metallic surface and \(\lambda_1 = 3\lambda_2\), then

• A. \(K_1 > \frac{K_2}{3}\)
• B. \(K_1 < \frac{K_2}{3}\)
• C. \(K_1 = 2K_2\)
• D. \(K_2 = 2K_1\)

Hint:

For the photoelectric effect, the kinetic energy is directly related to the frequency, which is inversely proportional to the wavelength.

Answer 1

The Correct Option is B

The kinetic energy of the emitted photoelectron is given by: \[ K = h\nu - \phi \] Where \(\nu = \frac{c}{\lambda}\). Since \(\lambda_1 = 3\lambda_2\), this implies that \(\nu_1 = \frac{1}{3} \nu_2\). Thus, the kinetic energy \(K_1\) will be less than \(\frac{K_2}{3}\) for the respective wavelengths. \bigskip