Dual Nature of Radiation and Matter — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Dual Nature of Radiation and Matter" — 6 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Dual Nature of Radiation and Matter" — 6 important questions with detailed answers for…
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Key Questions Covered:
- Define photoelectric effect. State Einstein's photoelectric equation and expl…
- Light of wavelength 300 nm falls on sodium surface (work function 2.75 eV). F…
- What is de Broglie wavelength? Derive its expression and explain wave-particl…
- An electron is accelerated through 1000 V. Calculate its de Broglie wavelength.
- Explain Davisson-Germer experiment. How does it prove wave nature of electrons?
- Define Compton scattering. Derive the Compton wavelength shift formula.
Solutions Summary:
| Question | Status |
|---|---|
| Define photoelectric effect. State Einstein's photoelectr… | ✓ Solved |
| Light of wavelength 300 nm falls on sodium surface (work … | ✓ Solved |
| What is de Broglie wavelength? Derive its expression and … | ✓ Solved |
| An electron is accelerated through 1000 V. Calculate its … | ✓ Solved |
| Explain Davisson-Germer experiment. How does it prove wav… | ✓ Solved |
| Define Compton scattering. Derive the Compton wavelength … | ✓ Solved |
Showing 6 of 6 questions
Q1: Define photoelectric effect. State Einstein's photoelectric equation and explain the concept of work function.
Photoelectric Effect: Emission of electrons from a material when light of sufficient frequency is incident on it.
Einstein's Photoelectric Equation:
hν = Φ + K.E._max
Where:
hν = energy of incident photon (h = 6.63 × 10⁻³⁴ J·s, ν = frequency)
Φ = work function (minimum energy to remove electron)
K.E._max = maximum kinetic energy of emitted electron
K.E._max = ½ m_e v²_max
Work Function:
Minimum energy required to remove electron from surface of metal
Φ = hν₀ where ν₀ = threshold frequency
Φ =...
Q2: Light of wavelength 300 nm falls on sodium surface (work function 2.75 eV). Find the maximum kinetic energy of photoelectrons and stopping potential.
Given:
Wavelength: λ = 300 nm = 300 × 10⁻⁹ m
Work function: Φ = 2.75 eV
Planck's constant: h = 6.63 × 10⁻³⁴ J·s
Speed of light: c = 3 × 10⁸ m/s
Electron charge: e = 1.6 × 10⁻¹⁹ C
Photon Energy:
E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(300 × 10⁻⁹)
E = (19.89 × 10⁻²⁶)/(3 × 10⁻⁷) = 6.63 × 10⁻¹⁹ J
Converting to eV:
E = (6.63 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 4.14 eV
Maximum Kinetic Energy:
K.E._max = hν - Φ = 4.14 - 2.75 = 1.39 eV
In Joules:
K.E._max = 1.39 × 1.6 × 10⁻¹⁹ = 2.22 × 10⁻¹⁹ J
Stopping Potential:...
Q3: What is de Broglie wavelength? Derive its expression and explain wave-particle duality.
de Broglie Wavelength: Wavelength associated with a moving particle, demonstrating wave-particle duality.
Expression Derivation:
From Einstein's energy relation: E = hν
For photon: E = pc (where p = momentum)
Combining: hν = pc
h/λ = p (since c = νλ)
λ = h/p
for any particle with momentum p:
λ = h/p = h/(mv)
Where:
h = 6.63 × 10⁻³⁴ J·s (Planck's constant)
p = mv = momentum of particle
m = mass of particle
v = velocity of particle
For relativistic particles:
λ = h/√(2mK.E.) where K.E. = kine...
Q4: An electron is accelerated through 1000 V. Calculate its de Broglie wavelength.
Given:
Accelerating voltage: V = 1000 V
Electron mass: m = 9.11 × 10⁻³¹ kg
Electron charge: e = 1.6 × 10⁻¹⁹ C
Planck's constant: h = 6.63 × 10⁻³⁴ J·s
Kinetic Energy Gained:
K.E. = eV = 1.6 × 10⁻¹⁹ × 1000 = 1.6 × 10⁻¹⁶ J
For non-relativistic case (valid here):
K.E. = ½mv²
v² = 2K.E./m = (2 × 1.6 × 10⁻¹⁶)/(9.11 × 10⁻³¹)
v² = (3.2 × 10⁻¹⁶)/(9.11 × 10⁻³¹) = 3.51 × 10¹⁴
v = 1.87 × 10⁷ m/s
Momentum:
p = mv = 9.11 × 10⁻³¹ × 1.87 × 10⁷ = 1.70 × 10⁻²³ kg·m/s
de Broglie Wavelength:
λ = h/p = (6.63 × 1...
Q5: Explain Davisson-Germer experiment. How does it prove wave nature of electrons?
Davisson-Germer Experiment (1927):
Demonstrated diffraction of electrons, proving their wave nature.
Setup:
- Electrons accelerated through known potential difference
- Directed at nickel crystal surface
- Intensity of scattered electrons measured at various angles
- Using Bragg's law for crystal diffraction
Procedure:
1. Electrons accelerated by voltage V
2. Electrons hit nickel crystal at angle θ to surface
3. Crystal acts as diffraction grating (atomic spacing ≈ 0.2 nm)
4. Intensity of refl...
Q6: Define Compton scattering. Derive the Compton wavelength shift formula.
Compton Scattering: Collision between high-energy photon and free electron, where photon transfers energy and momentum to electron.
Observation:
Wavelength of scattered photon > wavelength of incident photon
Δλ = λ' - λ (wavelength shift)
Derivation using Conservation Laws:
Energy Conservation:
hν + m_e c² = hν' + E_e
Where E_e = √[(p_e c)² + (m_e c²)²]
Momentum Conservation (along direction of photon):
hν/c = hν'/c cos θ + p_e cos φ
Momentum Conservation (perpendicular):
0 = hν'/c sin θ ...
Showing 6 of 6 questions. Visit the full page for complete solutions.
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