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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 de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is 25% of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

Solution & Explanation

Related Formula

For a photon, the energy (which is its kinetic energy) is:

Eₚ = (h c)/(λₚ) λₚ = (h c)/(Eₚ)

For an electron, the de-Broglie wavelength related to its kinetic energy (Kₑ) is:

λₑ = (h)/(mₑ vₑ) = (h vₑ)/(2 Kₑ)

since Kₑ = (1)/(2) mₑ vₑ² mₑ vₑ = (2 Kₑ)/(vₑ).

Core Logic

Given that the velocity of the electron is 25% of the speed of light:

vₑ = 0.25c = (c)/(4)

Substituting this into the electron's wavelength equation:

λₑ = (h ((c)/(4)))/(2 Kₑ) = (h c)/(8 Kₑ)
Step 1: Equate Wavelengths

We are given that the de-Broglie wavelengths are equal (λₚ = λₑ):

(h c)/(Eₚ) = (h c)/(8 Kₑ)

Cancelling h c from both sides:

(1)/(Eₚ) = (1)/(8 Kₑ) 8 Kₑ = Eₚ
Step 2: Find the Energy Ratio

The ratio of the kinetic energy of the electron to that of the photon (Eₚ) is:

(Kₑ)/(Eₚ) = (1)/(8)

Therefore, the ratio is 1/8.

Pattern Recognition

A direct comparison between massive particle kinetic energy (K = (p²)/(2m) = (p v)/(2)) and photon energy (E = p c) when they share the same momentum p (same wavelength) yields a simple universally applicable rule: KparticleEphoton = (v)/(2c).

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_2024_29_jan_morning

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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%)