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

To determine the angular frequency of the incident light when the produced photoelectrons have zero kinetic energy, we use the photoelectric effect equation:

\(E = h \nu\) ← (1) 

Since the photoelectrons have zero kinetic energy, the energy of the incident light \(E\) is equal to the work function \(\phi\) of the metallic plate:

\(h \nu = \phi\) ← (2)

Where:

• \(h = 6.63 \times 10^{-34}\,\text{Js}\) is Planck's constant,
• \(\nu\) is the frequency of the light,
• \(\phi = 110 \times 10^{-20}\,\text{J}\) is the work function.

We are required to find the angular frequency \(\omega\), where \(\omega = 2\pi\nu\).

From equation (2), solve for \(\nu\):

\(\nu = \frac{\phi}{h}\)

Substitute the given values:

\(\nu = \frac{110 \times 10^{-20}}{6.63 \times 10^{-34}}\)

\(\nu \approx 1.66 \times 10^{14}\,\text{Hz}\)

To find the angular frequency \(\omega\):

\(\omega = 2\pi \nu\)

\(\omega = 2 \times 3.1416 \times 1.66 \times 10^{14}\)

\(\omega \approx 1.04 \times 10^{16}\,\text{rad/s}\)

Therefore, the correct answer is \(1.04 \times 10^{16}\,\text{rad/s}\).