Question

A certain metal when irradiated by light (\( r = 3.2 \times 10^{16} \, \text{Hz} \)) emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light (\( r = 2.0 \times 10^{16} \, \text{Hz} \)). The \( \nu \) of metal is

• A. \( 1.2 \times 10^{14} \, \text{Hz} \)
• B. \( 8 \times 10^{15} \, \text{Hz} \)
• C. \( 1.2 \times 10^{16} \, \text{Hz} \)
• D. \( 1.2 \times 10^{14} \, \text{Hz} \)

Hint:

The photoelectric effect equation relates the kinetic energy of emitted photoelectrons to the frequency of the incident light and the work function of the material.

Answer 1

The Correct Option is A

By using the photoelectric equation \( E_k = h \nu - \phi \), where \( \nu \) is the frequency of light, and knowing that the kinetic energy doubles, we can solve for the work function \( \phi \), finding that \( \nu = 1.2 \times 10^{14} \, \text{Hz} \).