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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A particle having electric charge 3 × 10⁻¹⁹ C and mass 6 × 10⁻²⁷ kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is α × 10⁻¹² m. The value of α is ________. (Take Planck's constant = 6.6 × 10⁻³⁴ J⋯)

Numerical Answer Type:
Enter a numerical value Answer: 10 to 10 +4 marks

Solution & Explanation

Related Formula
λ = (h)/(p) = h√(2mK)

K = qV

Core Logic

For a charged particle accelerated through a potential difference V, the de Broglie wavelength is:

λ = h√(2mqV)
Step 1: Value Substitution
λ = 6.6 × 10⁻³⁴ 2 × 6 × 10⁻²⁷ × 3 × 10⁻¹⁹ × 1.21 λ = 6.6 × 10⁻³⁴ 36 × 10⁻⁴⁶ × 1.21
Step 2: Arithmetic Evaluation

Evaluate the square root:

√(36) × √(1.21) × 10⁻⁴⁶ = 6 × 1.1 × 10⁻²³ = 6.6 × 10⁻²³ λ = 6.6 × 10⁻³⁴6.6 × 10⁻²³ λ = 10⁻¹¹ m
Step 3: Final Conclusion

We need to express this in the form α × 10⁻¹² m:

10⁻¹¹ m = 10 × 10⁻¹² m

Therefore, α = 10.

Pattern Recognition

The expression inside the radical always evaluates cleanly in JEE. 2 × 6 × 3 = 36 and 1.21 = (1.1)² are engineered to perfectly cancel the 6.6 in the numerator.

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_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%)