Question

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and the stopping potential is 2V, what is the wavelength of the electromagnetic wave? Given \( hc = 1242 \, \text{eV} \cdot \text{nm} \) where \( h \) is the Planck constant and \( c \) is the speed of light in vacuum.

• A. 400 nm
• B. 600 nm
• C. 200 nm
• D. 300 nm

Hint:

The energy of the photon in the photoelectric effect can be calculated using the equation \( E = \frac{hc}{\lambda} \), where \( \lambda \) is the wavelength.

Answer 1

The Correct Option is D

This problem addresses the photoelectric effect, where incident light on a metal surface causes electron emission. The provided work function and stopping potential are used to determine the incident electromagnetic wave's wavelength.

Given:

• Work function (\( \phi \)) = 2.14 eV
• Stopping potential (\( V_0 \)) = 2 V
• Product of Planck's constant and speed of light (\( hc \)) = 1242 eV·nm

To determine the wavelength (\( \lambda \)) of the electromagnetic wave, follow these calculations:

The total energy (\( E \)) of an incident photon equals the sum of the work function and the kinetic energy of the emitted electrons, which is related to the stopping potential:

1. \(E = \phi + eV_0 = 2.14 \, \text{eV} + 2 \, \text{eV} = 4.14 \, \text{eV}\)

The energy (\( E \)) of the incident photon is also related to its wavelength (\( \lambda \)) by the equation:

1. \(E = \frac{hc}{\lambda}\)

Substitute the known values into the equation and solve for \( \lambda \):

1. \(4.14 = \frac{1242}{\lambda}\)

Rearrange the equation to isolate \( \lambda \):

1. \(\lambda = \frac{1242}{4.14} \approx 300 \, \text{nm}\)

Therefore, the wavelength of the electromagnetic wave is 300 nm.

Comparing this result with the provided options:

• 400 nm
• 600 nm
• 200 nm
• 300 nm

The calculated wavelength matches the option of 300 nm.