Dual Nature of Radiation and Matter
This chapter deals with the particle nature of light (photoelectric effect) and the wave nature of matter (de Broglie waves).
1. Electron Emission
The phenomenon of emission of electrons from a metal surface.
- Work Function (Φ0): Minimum energy required to eject an electron from the metal surface. Unit: eV.
- Types: Thermionic emission, Field emission, Photoelectric emission.
2. Photoelectric Effect
Emission of electrons when light of suitable frequency falls on a metal surface.
Observations:
- Photoelectric current ∝ Intensity of incident radiation.
- Max K.E. of photoelectrons depends on frequency, not intensity.
- There exists a Threshold Frequency (ν0) below which no emission occurs.
- The process is instantaneous.
3. Einstein's Photoelectric Equation
Based on the concept of Photons (energy packets).
Kmax = hν - Φ0
eV0 = hν - hν0
eV0 = hν - hν0
- h = Planck's constant (6.63 × 10-34 J s).
- V0 = Stopping potential.
4. Particle Nature of Light: The Photon
- Photons travel at the speed of light c.
- Energy E = hν = hc/λ.
- Momentum p = hν/c = h/λ.
- Photons are electrically neutral.
5. Wave Nature of Matter
De Broglie suggested that moving particles behave like waves.
De Broglie Wavelength (λ): λ = h / p = h / (mv)
For an electron accelerated through potential V:
λ ≈ 12.27 / √V Å
6. Davisson and Germer Experiment
Experimentally confirmed the wave nature of electrons by observing diffraction of electron beam by a nickel crystal.
7. Heisenberg Uncertainty Principle
It is impossible to determine both position and momentum of a subatomic particle simultaneously with absolute accuracy.
Δx Δp ≥ h / 4π
Numericals: Dual Nature of Radiation and Matter
Q1: Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV
electrons.
Given:
Formula: E = eV = hνmax; λmin = c / νmax
Solution:
(a) νmax = eV / h = (1.6 × 10-19 × 3 × 104) / (6.63 × 10-34) ≈ 7.24 × 1018 Hz.
(b) λmin = 3 × 108 / (7.24 × 1018) ≈ 0.041 nm.
Answer: (a) 7.24 × 1018 Hz (b) 0.041 nm
Given:
V = 30 kV = 3 × 104 VFormula: E = eV = hνmax; λmin = c / νmax
Solution:
(a) νmax = eV / h = (1.6 × 10-19 × 3 × 104) / (6.63 × 10-34) ≈ 7.24 × 1018 Hz.
(b) λmin = 3 × 108 / (7.24 × 1018) ≈ 0.041 nm.
Answer: (a) 7.24 × 1018 Hz (b) 0.041 nm
Q2: The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014
Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic
energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted
photoelectrons?
Given:
Formula: Kmax = hν - Φ0; Kmax = eV0; Kmax = 1/2 mv2
Solution:
(a) hν = (6.63 × 10-34 × 6 × 1014) / (1.6 × 10-19) eV ≈ 2.48 eV.
Kmax = 2.48 - 2.14 = 0.34 eV.
(b) V0 = 0.34 V.
(c) vmax = √(2 × Kmax / m) = √(2 × 0.34 × 1.6 × 10-19 / 9.1 × 10-31) ≈ 345 km/s.
Answer: (a) 0.34 eV (b) 0.34 V (c) 3.45 × 105 m/s
Given:
Φ0 = 2.14 eV, ν = 6 × 1014 HzFormula: Kmax = hν - Φ0; Kmax = eV0; Kmax = 1/2 mv2
Solution:
(a) hν = (6.63 × 10-34 × 6 × 1014) / (1.6 × 10-19) eV ≈ 2.48 eV.
Kmax = 2.48 - 2.14 = 0.34 eV.
(b) V0 = 0.34 V.
(c) vmax = √(2 × Kmax / m) = √(2 × 0.34 × 1.6 × 10-19 / 9.1 × 10-31) ≈ 345 km/s.
Answer: (a) 0.34 eV (b) 0.34 V (c) 3.45 × 105 m/s
Q3: Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power
emitted is 2.0 × 10-3 W. (a) What is the energy of a photon in the light beam? (b) How many
photons per second, on an average, are emitted by the source?
Given:
Formula: E = hν; n = P / E
Solution:
(a) E = 6.63 × 10-34 × 6 × 1014 = 3.98 × 10-19 J.
(b) n = 2 × 10-3 / (3.98 × 10-19) ≈ 5 × 1015 photons/s.
Answer: (a) 3.98 × 10-19 J (b) 5 × 1015 photons/s
Given:
ν = 6e14 Hz, P = 2e-3 WFormula: E = hν; n = P / E
Solution:
(a) E = 6.63 × 10-34 × 6 × 1014 = 3.98 × 10-19 J.
(b) n = 2 × 10-3 / (3.98 × 10-19) ≈ 5 × 1015 photons/s.
Answer: (a) 3.98 × 10-19 J (b) 5 × 1015 photons/s
Q4: The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of
frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the
photoelectric emission.
Given:
Formula: V0 = h(ν - ν0) / e
Solution:
V0 = (6.63 × 10-34 × (8.2 - 3.3) × 1014) / (1.6 × 10-19)
V0 ≈ 2.03 V.
Answer: 2.03 V
Given:
ν0 = 3.3e14 Hz, ν = 8.2e14 HzFormula: V0 = h(ν - ν0) / e
Solution:
V0 = (6.63 × 10-34 × (8.2 - 3.3) × 1014) / (1.6 × 10-19)
V0 ≈ 2.03 V.
Answer: 2.03 V
Q5: What is the de Broglie wavelength associated with an electron, accelerated through a potential
difference of 100 volts?
Given:
Formula: λ = 12.27 / √V Å
Solution:
λ = 12.27 / √100 = 12.27 / 10 = 1.227 Å.
Answer: 1.227 Å (or 0.123 nm)
Given:
V = 100 VFormula: λ = 12.27 / √V Å
Solution:
λ = 12.27 / √100 = 12.27 / 10 = 1.227 Å.
Answer: 1.227 Å (or 0.123 nm)
Q6: Work function of sodium is 2.3 eV. Find threshold wavelength.
Answer: 540 nm
Answer: 540 nm
Q7: Momentum of photon of wavelength 400 nm.
Answer: 1.66 × 10-27 kg m/s
Answer: 1.66 × 10-27 kg m/s
Q8: If intensity of light is doubled, what happens to stopping potential? (Answer: No change)
Answer: Remains same
Answer: Remains same
Q9: Rest mass of photon. (Answer: Zero)
Answer: 0
Answer: 0
Q10: De-broglie wavelength of a ball of mass 0.12 kg moving with speed 20 m/s.
Answer: 2.76 × 10-34 m
Answer: 2.76 × 10-34 m
Q11: Energy of 1 eV in Joules.
Answer: 1.6 × 10-19 J
Answer: 1.6 × 10-19 J
Q12: Slope of graph between V0 and ν.
Answer: h / e
Answer: h / e
Q13: Maximum wavelength of light that can eject electrons from metal with Φ=3eV.
Answer: 414 nm
Answer: 414 nm
Q14: Ratio of λ of electron and alpha particle accelerated through same potential V.
Answer: √(mα qα / me qe)
Answer: √(mα qα / me qe)
Q15: Intercept of V0 vs ν graph on frequency axis.
Answer: Threshold frequency (ν0)
Answer: Threshold frequency (ν0)
Q16: Number of photoelectrons depends on? (Answer: Intensity)
Answer: Intensity
Answer: Intensity
Q17: Stopping potential for frequency equal to threshold frequency. (Answer: Zero)
Answer: 0
Answer: 0
Q18: Planck's constant units.
Answer: J-s or kg-m2/s
Answer: J-s or kg-m2/s
Q19: Matter waves are? (Answer: Non-electromagnetic waves)
Answer: Probabilistic waves associated with matter
Answer: Probabilistic waves associated with matter
Q20: Ratio of λ of photon and λ of electron with same momentum.
Answer: 1 : 1
Answer: 1 : 1
Content coming soon...
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App