JEE Main · Physics ↓ Falling

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

Year 2026 2025 2024 Total
Questions 7 16 8 31
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The ratio of de Broglie wavelength of a deutron with kinetic energy E to that of an alpha particle with kinetic energy 2E, is n : 1. The value of n is ____. (Assume mass of proton = mass of neutron)

Numerical Answer Type:
Enter a numerical value Answer: 2 to 2 +4 marks

Solution & Explanation

Related Formula
λ = (h)/(p) = h√(2m · KE)
Core Logic

Set up the ratio of wavelengths for the deuteron and the alpha particle based on their given kinetic energies and masses. Let mass of proton/neutron be m. Deuteron mass is 2m, Alpha particle mass is 4m.

Step 1: Ratio setup
(λd)/(λ_α) = √((m_α · KE_α)/(md · KEd))
Step 2: Value Substitution

Substitute m_α = 4m, md = 2m, KE_α = 2E, and KEd = E:

= √((4m · 2E)/(2m · E)) = √((8mE)/(2mE)) = √(4) = 2

So, the ratio is 2 : 1. Therefore, n = 2.

Pattern Recognition

Memorize mass ratios for common particles: proton (m), deuteron (2m), alpha (4m). Substitute directly into inverse-sqrt formula for λ.

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_28_january_morning

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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%)