Question

Photons of \(5.5\ \mathrm{eV}\) energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy \(4.0\ \mathrm{eV}\). The stopping voltage required for these electrons is

• A. \(5.5\ \mathrm{V}\)
• B. \(1.5\ \mathrm{V}\)
• C. \(9.5\ \mathrm{V}\)
• D. \(4.0\ \mathrm{V}\)

Hint:

Stopping potential is independent of light intensity and depends only on frequency.

Answer 1

The Correct Option is D

The problem given involves the concept of the photoelectric effect, in which photons fall on a metal surface causing the emission of photoelectrons. The kinetic energy of these emitted electrons can be used to determine the stopping voltage required to prevent their motion.

Here's a step-by-step breakdown of the solution:

1. According to the photoelectric effect, the energy of the incident photons (\(\)) is converted into the work function of the metal (\(<\phi>\)) plus the maximum kinetic energy of the emitted electrons (\(K_{\text{max}}\)). This is expressed mathematically as: \(E_{\text{photon}} = \phi + K_{\text{max}}\).
2. Given:
   • Photon energy, \(E_{\text{photon}} = 5.5 \ \mathrm{eV}\)
   • Maximum kinetic energy of electrons, \(K_{\text{max}} = 4.0\ \mathrm{eV}\)
3. Replace the known values in the formula to find the work function: \(\phi = E_{\text{photon}} - K_{\text{max}}\).
4. Substitute the values: \(\phi = 5.5 \ \mathrm{eV} - 4.0 \ \mathrm{eV} = 1.5 \ \mathrm{eV}\).
5. The stopping voltage (\(\)) required to stop the most energetic photoelectron is equal to the maximum kinetic energy of the electrons divided by the charge of an electron (\(e\)), due to the formula: \(V_0 = \frac{K_{\text{max}}}{e}\).
6. In electronvolts, this voltage is equivalent to the maximum kinetic energy directly, because: \(V_0 = K_{\text{max}} = 4.0 \ \mathrm{V}\).

Therefore, the stopping voltage required is \(4.0\ \mathrm{V}\). This matches with the provided correct answer. Thus, the correct option is :

• \(4.0\ \mathrm{V}\).