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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

In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:

Solution & Explanation

Related Formula

For photoelectric emission to occur: f ≥ f₀ where f is incident frequency and f₀ is the threshold frequency.

Core Logic

Initially, the frequency is f₁ = 1.5 f₀. Since f₁ > f₀, emission happens. Then, the frequency is halved: f₂ = (1.5 f₀)/(2) = 0.75 f₀.

Step 1: Check Threshold Condition

Since f₂ = 0.75 f₀, we see that f₂ < f₀. The incident light no longer has enough energy per photon to overcome the work function, regardless of how intense the light is.

Step 2: Conclusion

Because the threshold condition fails, emission completely stops. The number of photoelectrons emitted is zero.

Pattern Recognition

Always check the frequency threshold first in photoelectric questions. Intensity adjustments are irrelevant traps if f < f₀. No emission occurs below threshold frequency.

Chapter Mix

Class 12 Physics: Dual Nature of Radiation and Matter

Reference Study Guides

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_2024_31_jan_evening

Practice all Dual Nature of Radiation and Matter previous-year questions →

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%)