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 of an Electron.

Year 2026 2025 2024 Total
Questions 7 16 8 31
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An electron of mass 'm' with an initial velocity v=v₀ i(v₀>0) enters an electric field E=-E₀ k If the initial de Broglie wavelength is λ₀, the value after time t would be :-

Solution & Explanation

Related Formula

The de Broglie wavelength relation matching a moving particle momentum is given by:

λ = (h)/(p) = hm| v|
Core Logic

Calculate the electric acceleration force acting component on the charge :

a = q Em = (-e)(-E₀ k)m = eE₀m k

Applying kinematics to find velocity at time t:

v(t) = v₀ i + ( eE₀tm) k
Step 1: Calculating Velocity Magnitude and Final Wavelength

Find the magnitude of the updated velocity vector:

| v| = v₀² + ( eE₀tm)² = v₀ 1 + e²E₀²t²m²v₀²

Substitute this into the wavelength equation:

λ' = hmv₀ 1 + e²E₀²t²m²v₀²

Since initial wavelength matches λ₀ = hmv₀, the expression simplifies to :

λ' = λ₀ 1 + e²E₀²t²m²v₀²
Pattern Recognition

The perpendicular field increases the particle's overall velocity and momentum. Since wavelength is inversely proportional to momentum, it must decrease, which rules out options with a plus sign in the numerator.

Chapter Mix

Class 12 Physics: 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_2025_24_jan_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%)