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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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When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ____ m. (h = 6.63× 10⁻³⁴J.s,e = 1.6× 10⁻¹⁹C,c = 3× 10⁸m / s)

Solution & Explanation

Related Formula
eVₛ = (hc)/(λ) - φ
Core Logic

For the first light:

e(3.2) = (hc)/(λ) - φ (1)

For the second light (wavelength 2λ):

e(0.7) = (hc)/(2λ) - φ (2)

Photoelectric Effect diagram for Q33 - JEE Main 2026 Evening
Photoelectric Effect diagram for Q33 - JEE Main 2026 Evening

Step 1: Solving the Equation

Subtract Equation (2) from Equation (1):

e(3.2) - e(0.7) = ( (hc)/(λ) - φ ) - ( (hc)/(2λ) - φ ) e(2.5) = (hc)/(2λ)
Step 2: Calculate Wavelength
2.5 = ((hc)/(e)) ((1)/(2λ))

Using (hc)/(e) ≈ 12400 eV·AA:

2.5 = (12400)/(2λ) λ = (12400)/(5) AA λ = 2480 AA = 2.48 × 10⁻⁷ m
Pattern Recognition

When you have two states of photoelectric effect for the same metal, subtracting the two stopping potential equations instantly eliminates the work function φ. Convert hc/e directly to 12400 eV·AA to expedite calculation.

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_24_january_evening

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Rankbit System
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%)