Dual nature of radiation and matter

NEET

Dual Nature of Radiation & Matter

1. Photoelectric Effect

Emission of electrons from a metal surface when light of suitable frequency falls on it.

Laws:
1. Instantaneous process (Time lag < 10-9 s).
2. Saturation current ∝ Intensity of incident light.
3. Stopping Potential depends on Frequency, NOT Intensity.
4. No emission below Threshold Frequency (ν0).
Photoelectric Effect Experimental Setup

Einstein's Photoelectric Equation

Kmax = hν - φ0
Kmax = eV0 (Where V0 is stopping potential).
φ0 = hν0 (Work Function).
Graphs: Current vs V, V0 vs Frequency

(Asked in NEET 2017, 2019, 2021)

2. Particle Nature of Light (Photon)

- Energy of photon E = hν = hc/λ.
- Momentum p = E/c = h/λ.
- Rest mass of photon is zero.
- Photons are electrically neutral.

3. Matter Waves (de Broglie Hypothesis)

Moving material particles exhibit wave-like properties.

de Broglie Wavelength:
λ = h/p = h/mv = h/√(2mK)
For Electron accelerated by V volts:
λ = 12.27 / √V Å.

(Asked in NEET 2016, 2018, 2020, 2023)

4. Davisson-Germer Experiment

- Proved the wave nature of electrons (Electron Diffraction).
- Constructive interference observed at 54V and scattering angle 50°.

Numericals: Dual Nature

Q1. Find energy of a photon of wavelength 4000 Å in eV.
Solution:
E (eV) = 12400 / λ (Å).
E = 12400 / 4000 = 12.4 / 4 = 3.1 eV.
Q2. Work function of a metal is 2.14 eV. Find threshold frequency. (h = 6.6 × 10-34)
Solution:
φ0 = hν0.
2.14 eV = 2.14 × 1.6 × 10-19 J.
ν0 = (2.14 × 1.6 × 10-19) / (6.6 × 10-34).
ν0 = (3.424 / 6.6) × 1015 = 0.518 × 1015 = 5.18 × 1014 Hz.
Q3. Light of 5 eV falls on metal with work function 3 eV. Find max K.E. and stopping potential.
Solution:
Kmax = E - φ0 = 5 - 3 = 2 eV.
Stopping Potential V0 = Kmax / e = 2 V.
Q4. Find ratio of de Broglie wavelength of proton and alpha particle accelerated through same potential.
Solution:
λ ∝ 1 / √(mq).
λp / λα = √(mαqα) / √(mpqp).
mα = 4mp, qα = 2qp.
Ratio = √(4 × 2) = √8 = 2√2 : 1.
Q5. 100W bulb emits light of 540nm. Calculate number of photons emitted per second. (Efficiency 100%)
Solution:
P = nE = n hc/λ.
E = (6.63 × 10-34 × 3 × 108) / (540 × 10-9) = 3.68 × 10-19 J.
n = P/E = 100 / (3.68 × 10-19) = 27.1 × 1019.
n ≈ 2.7 × 1020 /s.
Q6. Find momentum of a photon of frequency 1.5 × 1013 Hz.
Solution:
p = hν / c.
p = (6.63 × 10-34 × 1.5 × 1013) / (3 × 108).
p = (9.945 / 3) × 10-29 = 3.31 × 10-29 kg m/s.
Q7. If frequency of incident light is doubled, what happens to stopping potential? (Assuming ν > ν0)
Solution:
eV0 = hν - φ.
eVnew = h(2ν) - φ = 2hν - φ = 2(eV0 + φ) - φ = 2eV0 + φ.
So, Vnew > 2V0.
It becomes more than double.
Q8. Find de Broglie wavelength of a thermal neutron at 27°C.
Solution:
λ = 30.8 / √T Å (Approx formula for neutron) or h/√(3mkT).
T = 300 K.
Using λ = 0.286 / √E where E in eV. E = 3/2 kT = 0.04 eV.
λ ≈ 1.4 Å. (Standard value at 300K is 1.4 Å).
Q9. An electron has de Broglie wavelength 1 Å. Find its velocity.
Solution:
λ = h / mv → v = h / mλ.
v = 6.63 × 10-34 / (9.1 × 10-31 × 10-10).
v = (6.63 / 9.1) × 107 ≈ 7.3 × 106 m/s.
Q10. Cut-off voltage is 1.5 V. What is MAX kinetic energy of electrons?
Solution:
Kmax = e × V0.
Kmax = e × 1.5 = 1.5 eV.

Important Formulae

1. Photoelectric Effect

Types of Energy:

E = hν = hc/λ (in J)
E (eV) = 12400 / λ (Å)

Einstein's Equation:

Kmax = E - φ0 = h(ν - ν0)

Stopping Potential:

V0 = (h/e)ν - (φ0/e)
2. Matter Waves (de Broglie)

Momentum p:

λ = h/p = h/√(2mK)

For Electron:

λ = 12.27 / √V Å
3. Photon properties

Power (P):

P = nE (n = photons/sec)

Force (Absorbing):

F = IA/c = P/c

20 NEET Golden Facts

  • 1. Photons: Travel with speed of light c in vacuum. Speed changes in medium. Energy/Frequency remains same.
  • 2. Work Function (φ0): Minimum energy required to eject electron. Cs (Cesium) has lowest φ0.
  • 3. 1 eV: Energy gained by electron accelerated through 1 Volt. 1 eV = 1.6 × 10-19 J.
  • 4. Slope of V0 vs ν graph: Slope = h/e (Universal constant).
  • 5. Intensity: Increasing intensity increases Number of Photoelectrons (Saturation Current). No effect on Kmax.
  • 6. Frequency: Increasing frequency increases Kmax and Stopping Potential. No effect on Current.
  • 7. Momentum of Photon: p = E/c. Matter momentum p = mv.
  • 8. de Broglie λ: λ ∝ 1/v and λ ∝ 1/√K.
  • 9. For same K.E.: λe > λp > λα (Lighter particle has longer wavelenth).
  • 10. For same potential V: λ ∝ 1/√(mq).
  • 11. Efficiency: Photoelectric efficiency is very low (< 1%). Not every photon ejects an electron.
  • 12. Specific Charge: e/m of photon is not defined (m=0).
  • 13. Stopping Potential: Is negative potential at anode to stop most energetic electron.
  • 14. Force: Reflecting surface exerts 2× Force compared to absorbing surface (Change in momentum 2p).
  • 15. Neutral particle: Neutron also has de Broglie wavelength. λ = 0.286 / √E Å (Thermal).
  • 16. Effective Mass of Photon: m = E/c2 = hν/c2.
  • 17. Threshold Wavelength: λ0 = hc/φ0. Light λ must be < λ0.
  • 18. Heisenberg Uncertainty: Δx Δp ≥ h/4π.
  • 19. Davisson Germer: Crystal acts as 3D diffraction grating. Braggs Law 2d sinθ = nλ.
  • 20. Gas molecule: λ = h / √(3mkT).
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