Exam Notes
Dual Nature Of Radiation Class 12 Physics Revision — Grandmaster Guide
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Last Updated: 2026-04-20
- E = hv
- λ = h/p
- λ = h/(mv)
- E = mc²
- E = hf = hc/λ
- p = h/λ
- λ = h/(γmc)
- v = c/√(1 + (m₀/m)²)
- E = √((pc)² + (mc²)²)
- λ = h/√(2mE)
- K = E - U = (1/2)mv²
- de Broglie wavelength λ = h/√(2mK)
🪤 The 5 Mistakes That Cost Marks
- Not understanding the concept of dual nature of radiation
- Confusing wave and particle properties of radiation
- Not being able to apply de Broglie wavelength formula
- Forgetting to use relativistic mass ∈ calculations
- Not being able to distinguish between photon and particle properties
✏️ 3 Solved PYQs
- Question 1: What is the de Broglie wavelength of an electron with kinetic energy 100 eV? Step 1: First, we need to find the momentum of the electron using the formula K = (1/2)mv² Step 2: Then, we can use the formula λ = h/√(2mK) to find the de Broglie wavelength Step 3: Given that K = 100 eV = 1.6 × 10⁻¹⁷ J and m = 9.1 × 10⁻³¹ kg, we can calculate λ Step 4: λ = h/√(2mK) = (6.626 × 10⁻³⁴)/(2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁷) = 1.2 × 10⁻¹⁰ m
- Question 2: What is the energy of a photon with wavelength 500 nm? Step 1: We can use the formula E = hc/λ to find the energy of the photon Step 2: Given that h = 6.626 × 10⁻³⁴, c = 3 × 10⁸ m/s, and λ = 500 × 10⁻⁹ m, we can calculate E Step 3: E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(500 × 10⁻⁹) = 3.98 × 10⁻¹⁹ J
- Question 3: What is the momentum of a photon with energy 2 eV? Step 1: We can use the formula E = pc to find the momentum of the photon Step 2: Given that E = 2 eV = 3.2 × 10⁻¹⁹ J and c = 3 × 10⁸ m/s, we can calculate p Step 3: p = E/c = (3.2 × 10⁻¹⁹)/(3 × 10⁸) = 1.07 × 10⁻²⁷ kg m/s
🧠 The One Thing Most Students Get Wrong
- Most students get confused between the wave and particle properties of radiation, and they are not able to apply the correct formulas ∈ the correct situations
- For example, they may use the formula E = hv for a particle, or they may use the formula λ = h/p for a photon
- To avoid this mistake, students should make sure to understand the dual nature of radiation and apply the correct formulas ∈ the correct situations
👁️ Ayush's Note
- Always remember that the de Broglie wavelength of a particle is given by λ = h/√(2mK)
- Also, remember that the energy of a photon is given by E = hv = hc/λ
- Make sure to use the correct formulas ∈ the correct situations, and do not get confused between the wave and particle properties of radiation
🔁 Last 5 Minutes Box
- de Broglie wavelength λ = h/√(2mK)
- Energy of a photon E = hv = hc/λ
- Momentum of a photon p = E/c
- Relativistic mass m = γm₀
- Photon properties: E = pc, p = E/c, λ = h/p
📝 Practice MCQs
1. What is the de Broglie wavelength of an electron with kinetic energy 100 eV?
A) 1 × 10⁻¹⁰ m
B) 1.2 × 10⁻¹⁰ m
C) 1.5 × 10⁻¹⁰ m
D) 2 × 10⁻¹⁰ m
Answer: B) 1.2 × 10⁻¹⁰ m.
2. What is the energy of a photon with wavelength 500 nm?
A) 2 × 10⁻¹⁹ J
B) 3.98 × 10⁻¹⁹ J
C) 4 × 10⁻¹⁹ J
D) 5 × 10⁻¹⁹ J
Answer: B) 3.98 × 10⁻¹⁹ J.
3. What is the momentum of a photon with energy 2 eV?
A) 1 × 10⁻²⁷ kg m/s
B) 1.07 × 10⁻²⁷ kg m/s
C) 1.5 × 10⁻²⁷ kg m/s
D) 2 × 10⁻²⁷ kg m/s
Answer: B) 1.07 × 10⁻²⁷ kg m/s.
4. What is the relativistic mass of an electron with kinetic energy 100 eV?
A) 9.1 × 10⁻³¹ kg
B) 1.1 × 10⁻³⁰ kg
C) 1.5 × 10⁻³⁰ kg
D) 2 × 10⁻³⁰ kg
Answer: B) 1.1 × 10⁻³⁰ kg.
5. What is the wavelength of a photon with energy 2 eV?
A) 500 nm
B) 600 nm
C) 700 nm
D) 800 nm
Answer: A) 500 nm.
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This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.
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