JEE Main · Physics ↓ Falling

Dual Nature of Radiation and Matter appeared 31 times across 3 years — 3.6% of Physics. This question is from Photoelectric Effect.

Year 2026 2025 2024 Total
Questions 7 16 8 31
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A light wave described by E = 60[ (3 × 10¹⁵t) + (12 × 10¹⁵t)] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately) ____ eV. (h = 6.6 × 10⁻³⁴ J⋯. and e = 1.6 × 10⁻¹⁹C)

Solution & Explanation

Related Formula
Kmax = hνmax - φ₀ v = (ω)/(2π)
Core Logic

The light wave consists of two frequencies governed by ω₁ and ω₂. ω₁ = 3 × 10¹⁵ rad/s ω₂ = 12 × 10¹⁵ rad/s

The maximum kinetic energy of ejected photoelectrons will be determined by the highest frequency photon, which corresponds to ω₂ = 12 × 10¹⁵ rad/s.

Step 1: Calculate Photon Energy

Frequency νmax = (ω₂)/(2π) = 12 × 10¹⁵2 × 3.14 ≈ 1.91 × 10¹⁵ Hz.

Energy of this photon:

Ephoton = hν = (6.6 × 10⁻³⁴) × (1.91 × 10¹⁵) = 1.26 × 10⁻¹⁸ J

Convert this energy to eV:

Emax = 1.26 × 10⁻¹⁸1.6 × 10⁻¹⁹ ≈ 7.87 eV ≈ 7.9 eV
Step 2: Calculate Maximum Kinetic Energy

Using Einstein's photoelectric equation:

Kmax = Emax - φ₀ Kmax = 7.9 - 2.8 = 5.1 eV
Pattern Recognition

When a wave has multiple frequency components (E = E₁ ω₁ t + E₂ ω₂ t), the Kmax is always strictly determined by the highest frequency (highest energy) component. Ignore the lower frequency terms.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Reference Study Guides

Dual Nature Of Radiation And Matter

More Dual Nature of Radiation and Matter Previous-Year Questions — Page 7

Q50 jee_main_2024_31_jan_morning Photoelectric Effect
When a metal surface is illuminated by light of wavelength λ, the stopping potential is 8 V. When the same surface is illuminated by light of wavelength 3λ, stopping potential is 2 V. The threshold wavelength for this surface is:
  • A. 5λ
  • B. 3λ
  • C. 9λ
  • D. 4.5λ

Solution

Related Formula
(hc)/(λ) = φ + Kmax Kmax = eV₀ φ = (hc)/(λ₀)
Core Logic

Using Einstein's Photoelectric equation for the two cases:

Case 1 (Wavelength λ, Stopping Potential 8 V):

(hc)/(λ) = (hc)/(λ₀) + 8e (i)

Case 2 (Wavelength 3λ, Stopping Potential 2 V):

(hc)/(3λ) = (hc)/(λ₀) + 2e (ii)
Step 2: Solving the Equations

Multiply equation (ii) by 4 to eliminate e:

(4hc)/(3λ) = (4hc)/(λ₀) + 8e

Equating this to equation (i):

(hc)/(λ) - (hc)/(λ₀) = (4hc)/(3λ) - (4hc)/(λ₀)

Divide entirely by hc:

(1)/(λ) - (1)/(λ₀) = (4)/(3λ) - (4)/(λ₀) (4)/(λ₀) - (1)/(λ₀) = (4)/(3λ) - (1)/(λ) (3)/(λ₀) = (4 - 3)/(3λ) (3)/(λ₀) = (1)/(3λ) λ₀ = 9λ
Chapter Mix

Class 12 Physics: Dual Nature Of Radiation And Matter

More Dual Nature of Radiation and Matter Questions — jee_main_2026_21_jan_morning

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)