Explanation

The light consists of two angular frequencies:

\[ \omega_1 = 3 \times 10^{15}, \quad \omega_2 = 12 \times 10^{15} \]

Frequency is:

\[ f = \frac{\omega}{2\pi} \]

Maximum kinetic energy depends on highest frequency component.

\[ f = \frac{12 \times 10^{15}}{2\pi} \]

Photon energy:

\[ E = hf = \frac{6.6 \times 10^{-34} \times 12 \times 10^{15}}{2\pi} \]

≈ 7.9 × 10⁻¹⁹ J

Convert to eV:

\[ \frac{7.9 \times 10^{-19}}{1.6 \times 10^{-19}} ≈ 4.9 \text{ eV} \]

\[ K_{max} = 4.9 - 2.8 = 5.1 \text{ eV} \]

Correct Answer: 5.1 eV

Related Theory (Comprehensive Modern Physics Master Section)

The photoelectric effect is one of the most important and conceptually deep topics in Modern Physics for JEE Main and JEE Advanced. It established the particle nature of light and led to the development of quantum theory.

1. Einstein’s Photoelectric Equation

\[ hf = \phi + K_{max} \]

This equation states that energy of incident photon is used partly to overcome work function (φ) and remaining appears as kinetic energy of emitted electron.

2. Work Function (φ)

Work function is minimum energy required to remove an electron from metal surface. It is material dependent and measured in eV.

3. Threshold Frequency

\[ f_0 = \frac{\phi}{h} \]

If incident frequency is less than threshold frequency, no emission occurs regardless of intensity.

4. Angular Frequency Relation

Given wave in form sin(ωt), where:

\[ \omega = 2\pi f \]

Students commonly forget this 2π factor, leading to incorrect energy calculation.

5. Multi-Frequency Radiation

When light contains multiple frequencies, each frequency acts independently. The maximum kinetic energy corresponds to highest frequency component because:

\[ K_{max} = h f_{max} - \phi \]

Intensity does NOT change kinetic energy. It changes number of electrons emitted.

6. Classical Theory Failure

Classical wave theory predicted:

• Energy depends on intensity.
• Time delay before emission.

Experiments showed:

• No time delay.
• Kinetic energy depends only on frequency.

This contradiction led Einstein to propose photon model.

7. Graphical Interpretation

Plot of Kmax vs frequency is straight line:

Slope = h
Intercept on frequency axis = threshold frequency

This graph is extremely important for JEE.

8. Stopping Potential

\[ eV_0 = K_{max} \]

Thus kinetic energy can be experimentally measured using stopping potential.

9. Advanced JEE Insights

• If multiple frequencies present, highest frequency governs Kmax.
• If one frequency below threshold and one above, emission still occurs due to higher one.
• Photoelectric current proportional to intensity.

10. Energy Unit Conversion

1 eV = 1.6 × 10⁻¹⁹ J

Careful unit conversion is crucial in JEE numericals.

11. Common Mistakes

• Ignoring 2π factor.
• Using lower frequency instead of higher.
• Forgetting to subtract work function.
• Mixing joules and eV units.

12. Why Only Highest Frequency Matters?

Each photon interacts with one electron. Electron absorbs one photon completely. Thus energy transfer is quantized and independent of other photons.

This quantum behavior is foundation of modern quantum mechanics.

Exam Relevance

Photoelectric effect appears every year in JEE Main. It is scoring and concept-based. Angular frequency-based questions are common trap questions.

Mastering Einstein equation and unit conversions ensures full marks in this chapter.

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