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: Einstein's photoelectric equation: K_max = hν – φ. Photon energy determines K_max; intensity only affects the number of emitted electrons, not their maximum kinetic energy.

Step 1: Calculate initial K_max.
K_max = 4 eV – 2 eV = 2 eV.

Step 2: Assess intensity change effect.
Doubling intensity doubles electron count but photon energy remains 4 eV, so K_max stays 2 eV.

Final Answer: Remain 2 eV (option c).