Question

Light is incident on a metallic plate having work function \(110 \times 10^{-20}\,\text{J}\). If the produced photoelectrons have zero kinetic energy, then the angular frequency of the incident light is _________ rad/s. (\(h = 6.63 \times 10^{-34}\,\text{J·s}\))}

• A. \(1.66 \times 10^{16}\)
• B. \(1.04 \times 10^{13}\)
• C. \(1.66 \times 10^{15}\)
• D. \(1.04 \times 10^{16}\)

Hint:

For zero kinetic energy of photoelectrons, the incident light frequency equals the threshold frequency.

Answer 1

The Correct Option is D

According to Einstein’s photoelectric equation, \[ h\nu = \phi + K_{\max}. \]
Step 1: Apply the given condition.
Since the photoelectrons have zero kinetic energy, \[ K_{\max} = 0. \] Thus, \[ h\nu = \phi. \]
Step 2: Substitute the given values.
\[ \nu = \frac{\phi}{h} = \frac{110 \times 10^{-20}}{6.63 \times 10^{-34}} = 1.66 \times 10^{15}\ \text{Hz}. \]
Step 3: Convert frequency to angular frequency.
Angular frequency is given by \[ \omega = 2\pi\nu. \] \[ \omega = 2\pi \times 1.66 \times 10^{15} \approx 1.04 \times 10^{16}\ \text{rad/s}. \]
Final Answer: \[ \boxed{1.04 \times 10^{16}\ \text{rad/s}} \]