Question

Light of energy E falls normally on a metal of work function \(\frac{E}{3} \). The kinetic energies (K) of the photo electrons are

• A. \(k=\frac{2E}{3}\)
• B. \(k=\frac{E}{3}\)
• C. \(O≤K≤\frac{2E}{3}\)
• D. \(O≤K≤\frac{E}{3}\)

Answer 1

The Correct Option is C

The energy of the incident light is \( E \), and the work function of the metal is \( \frac{E}{3} \). According to the photoelectric equation: \[ K = E - \text{Work function} \] Substitute the given work function: \[ K = E - \frac{E}{3} = \frac{3E}{3} - \frac{E}{3} = \frac{2E}{3} \] Thus, the kinetic energy of the photoelectrons can range from \( 0 \) to \( \frac{2E}{3} \). Thus, the solution is \( 0 \leq K \leq \frac{2E}{3} \). 

Answer 2

The photoelectric equation is given by: \[ E = W + K \] where: - \( E \) is the energy of the incident light, - \( W \) is the work function of the metal, and - \( K \) is the kinetic energy of the emitted photoelectron. The work function of the metal is \( W = \frac{E}{3} \), so the energy of the photoelectron is: \[ K = E - W = E - \frac{E}{3} = \frac{2E}{3} \] Thus, the maximum kinetic energy \( K_{\text{max}} \) of the photoelectron is \( \frac{2E}{3} \). Since the energy of the incident light can vary, the kinetic energy of the photoelectrons can range from 0 to \( \frac{2E}{3} \). Therefore, the kinetic energy \( K \) satisfies the condition: \[ 0 \leq K \leq \frac{2E}{3} \] Thus, the correct answer is option (C).