Question:medium

If the energy of incident radiation is increased by $25\%$, the kinetic energy of the photoelectrons emitted increases from 0.6 eV to 0.9 eV. The work function of the metal is:

Updated On: Mar 27, 2026
  • 0.6 eV
  • 1.2 eV
  • 1.5 eV
  • 1 eV
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The Correct Option is A

Solution and Explanation

The photoelectric effect is governed by the equation: \( KE = hf - \phi \). Here, \( KE \) represents the kinetic energy of the photoelectrons, \( hf \) is the energy of the incident photons, and \( \phi \) is the work function of the metal. Initially, the kinetic energy is \( KE_1 = 0.6 \, \text{eV} \). When the incident radiation's energy increases by 25%, the new photon energy is \( hf_2 = 1.25hf_1 \). Consequently, the kinetic energy rises to \( KE_2 = 0.9 \, \text{eV} \).

Applying the photoelectric effect equation to both situations yields:

1. Initial state: \( 0.6 = hf_1 - \phi \)
2. Increased energy state: \( 0.9 = 1.25hf_1 - \phi \)

Solving these two equations simultaneously allows for the determination of the work function \( \phi \). Subtracting equation 1 from equation 2:

\( (0.9 - 0.6) = (1.25hf_1 - hf_1) \)

This simplifies to:

\( 0.3 = 0.25hf_1 \)

Solving for \( hf_1 \):

\( hf_1 = \frac{0.3}{0.25} = 1.2 \, \text{eV} \)

Substituting \( hf_1 = 1.2 \, \text{eV} \) back into equation 1:

\( 0.6 = 1.2 - \phi \)

Solving for \( \phi \):

\( \phi = 1.2 - 0.6 = 0.6 \, \text{eV} \)

Therefore, the work function of the metal is 0.6 eV.

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