Question

Photons of energy \(4\,\text{eV}\) are incident on a metal surface whose work function is \(2\,\text{eV}\). If the intensity of this incident light is doubled, the maximum kinetic energy of the emitted photoelectrons will:

• A. Double to \(4\,\text{eV}\)
• B. Quadruple to \(8\,\text{eV}\)
• C. Remain \(2\,\text{eV}\)
• D. Become zero

Hint:

In the photoelectric effect, intensity controls the number of photoelectrons, while frequency (or photon energy) controls the maximum kinetic energy. Never confuse these two effects.

Answer 1

The Correct Option is C

Concept: This question is based on Einstein's Photoelectric Equation and the effect of intensity on the photoelectric effect. According to Einstein's photoelectric theory, when light of sufficiently high frequency falls on a metal surface, electrons are emitted from the surface. The maximum kinetic energy of the emitted photoelectrons depends only on the frequency (or energy) of the incident photons and not on the intensity of the light. The photoelectric equation is: where:
• \(K_{\max}\) = maximum kinetic energy of photoelectrons
• \(h\nu\) = energy of the incident photon
• \(\phi\) = work function of the metal The work function is the minimum energy required to remove an electron from the metal surface.

Step 1: Identify the given quantities. From the question, \[ h\nu = 4\,\text{eV} \] and \[ \phi = 2\,\text{eV} \]

Step 2: Apply Einstein's photoelectric equation. Substituting the given values, \[ K_{\max}=4-2 \] \[ K_{\max}=2\,\text{eV} \] Thus, the maximum kinetic energy of the emitted photoelectrons is \(2\,\text{eV}\).

Step 3: Analyze the effect of doubling the intensity. Intensity of light represents the number of photons incident per unit area per unit time. When the intensity is doubled:
• The number of photons striking the surface per second increases.
• Therefore, the number of emitted photoelectrons increases.
• However, the energy of each individual photon remains unchanged at \(4\,\text{eV}\). Since the photon energy remains unchanged, the maximum kinetic energy given by Einstein's equation also remains unchanged.

Step 4: Examine each option.
• Option (a): Double to \(4\,\text{eV}\)
Incorrect. Kinetic energy does not depend on intensity.
• Option (b): Quadruple to \(8\,\text{eV}\)
Incorrect. Increasing intensity cannot increase the energy of individual photons.
• Option (c): Remain \(2\,\text{eV}\)
Correct. Photon energy remains \(4\,\text{eV}\), so maximum kinetic energy remains \(2\,\text{eV}\).
• Option (d): Become zero
Incorrect. The incident photon energy is greater than the work function, so photoelectrons continue to be emitted. Final Conclusion: Doubling the intensity increases only the number of emitted photoelectrons and does not affect their maximum kinetic energy. Therefore, \[ K_{\max}=2\,\text{eV} \] even after doubling the intensity. Hence, the correct answer is \[ \boxed{\text{(c) Remain }2\,\text{eV}} \]