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Photon Theory

Dual Nature of Radiation and Matter

Physics · Class 12

1. Einstein's Photon Theory

Light shows dual nature – particle as well as wave.
Energy of light is localised in small bundles called quanta of light (photons).
Photons are electrically neutral ⇒ not deflected by electric or magnetic fields.
A photon has both energy and momentum.
Photons have zero rest mass and always travel with speed c (in vacuum).
Energy of a photon depends only on frequency of light (not on intensity).
In a photon–particle collision, total energy and total momentum are conserved.

2. Energy & Momentum of a Photon

E = hν = hcλ = pc

E (eV) = 12400 eV·Åλ (in Å) = 1240 eV·nmλ (in nm)

p = hλ = hνc = Ec

Wave equation : c = νλ (or v = fλ)

Symbol	Meaning / value
h	Planck's constant = 6.63 × 10⁻³⁴ J·s = 4.14 × 10⁻¹⁵ eV·s
c	speed of light = 3 × 10⁸ m/s
ν (or f)	frequency ; λ → wavelength
1 eV	1.6 × 10⁻¹⁹ J (eV and J both are units of energy)
1 Å , 1 nm	10⁻¹⁰ m , 10⁻⁹ m

h in eV·s : 6.63 × 10⁻³⁴1.6 × 10⁻¹⁹ = 4.14 × 10⁻¹⁵ eV·s

e.g. λ = 620 nm ⇒ E = 1240620 = 2 eV = 3.2 × 10⁻¹⁹ J

3. Number of Photons per second (N)

Monochromatic (single wavelength) source of light power P, wavelength λ, frequency ν :

N = PE = Phν = Pλhc

If efficiency of source is η% then useful light power :

P = η100 × (power of source)

e.g. bulb 100 W, 80% ⇒ P = 80100 × 100 = 80 W

Intensity of light (I) :

Power per unit area (energy flux) ; unit W/m² = J/(s·m²) ; P = I A.
For a point source spreading light uniformly over a sphere of radius r :

I = P4πr² ⇒ I ∝ 1r²

4. Photons incident on a Surface per sec

Light beam of cross-section area A, intensity I, wavelength λ falls on a surface :

N = PE = IAhν = IAλhc

Q. A 10 W source emits light of λ = 620 nm. Find photons emitted per second.
E = 1240/620 = 2 eV = 3.2 × 10⁻¹⁹ J
N = PE = 103.2 × 10⁻¹⁹ = 3.1 × 10¹⁹ photons/s

Key Points :

hc = 12400 eV·Å = 1240 eV·nm — use for quick eV calculations.
Smaller λ (higher ν) ⇒ more energy per photon.
For the same power, longer λ ⇒ more photons per second (N ∝ λ).

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Radiation Pressure

5. Spectrum of Light

E = hcλ = 12400 eV·Åλ (in Å) ⇒ λ more → E less

Radiation	Photon energy	Wavelength
Radio waves	≤ 1.24 μeV	≥ 1 m (10⁶ μm)
Microwaves	1.24 μeV – 1.24 meV	1 m – 1 mm
Infrared	1.24 meV – 1.77 eV	1 mm – 7000 Å
Visible	1.77 eV – 3.1 eV	7000 Å – 4000 Å
Ultraviolet	3.1 eV – 124 eV	4000 Å – 100 Å
X-rays	124 eV – 124 keV	100 Å – 0.1 Å
Gamma rays	124 keV – 124 MeV	≤ 0.1 Å

6. Nature of Surface

Light falling on a surface is partly reflected, partly transmitted (refracted) and partly absorbed.

P = P_r + P_t + P_a ⇒ 1 = P_rP + P_tP + P_aP

r + t + a = 1

r = reflectivity, t = transmittivity, a = absorptivity

Surface	r	a	t
Ideal mirror	1	0	0
Ideal transparent glass slab	0	0	1
Ideal black body	0	1	0

e.g. r = 0.6, t = 0.3 ⇒ a = 0.1 → 60% reflected, 30% transmitted, 10% absorbed.
For a metallic plate we take t = 0 ⇒ a + r = 1.

7. Radiation Force & Pressure

A light beam falling on a surface exerts a force on it = radiation force.
Force (⊥ to surface) per unit area = radiation pressure : P = F_⊥/A

Normal incidence (t = 0, reflectivity r, a = 1 − r) :

Photons incident per sec : N = IAλhc

Reflected : N_r = N r ; Absorbed : N_a = N(1 − r)

F_r = N_r·2hλ = IAλ rhc·2hλ = 2IArc

F_a = N_a·hλ = IA(1 − r)c

F = F_r + F_a = 2IArc + IA(1 − r)c

F = IA(1 + r)c , P = FA = I(1 + r)c

Surface	Force	Pressure
Perfect absorber (r = 0)	IA/c	I/c
Perfect reflector (r = 1)	2IA/c	2I/c

Oblique incidence (angle θ with the normal) :

P = I(1 + r)c cos²θ

(one cosθ because beam spreads over area A/cosθ, one cosθ for the normal component of momentum)

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Matter Waves

8. Radiation Force on Bodies

Parallel beam of intensity I on a body ⇒ for an absorbing body, F = I × (projected area ⊥ beam) / c.

Sphere (radius R)

Cylinder (R, h)

Cone (base radius R)

Body	Projected area	Force
Sphere (any r)	πR²	IπR²/c
Cylinder (r = 0)	2Rh	I(2Rh)/c
Cone, apex to beam (r = 0)	πR²	IπR²/c

For a sphere, F = IπR²/c for any value of r (light reflected from a sphere spreads in all directions, so on average it adds no extra push along the beam).
Cylinder & cone values are for absorbing surfaces ; for reflecting ones the result changes (e.g. cylinder : F = (2IRh/c)(1 + r/3)).

9. de Broglie Wavelength (Matter Waves)

Light has both wave and particle nature ⇒ de Broglie proposed that matter must also have both natures.
Waves associated with material particles are called matter waves.
Verified when electrons were observed to diffract (a wave phenomenon) – Davisson & Germer experiment.

Particle of mass m, speed v : K = ½mv² = p²2m ⇒ p = √(2mK)

λ = hp = hmv = h√(2mK)

10. Graphs of λ

p vs λ : pλ = h

1/p vs λ : line

K vs λ : K = h²/(2mλ²)

1/K vs λ : parabola

11. Common Particles

1 amu = 1.67 × 10⁻²⁷ kg

Particle	Charge	Mass
α-particle	+2e	4 amu	₂⁴He²⁺
β-particle	−e	9.1 × 10⁻³¹ kg	electron
γ-ray	no charge	no rest mass	photon
Proton	+e	1 amu	₁¹H⁺
Deuteron	+e	2 amu	₁²H⁺

12. Accelerating Voltage & K.E.

A charge q moving through potential difference V changes its K.E. by qV.
Accelerating voltage : K_f = K_i + qV ; Retarding voltage : K_f = K_i − qV

If initial K.E. is not given, take K_i = 0 ⇒ K = qV :

λ = hp = h√(2mK) = h√(2mqV)

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de Broglie & Waves

13. de Broglie λ of an Electron

λ = 6.63 × 10⁻³⁴√(2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × V) m

λ_e = √150V Å = 12.27√V Å

(V = accelerating voltage in volt)

14. λ for Other Particles

λ = h√(2mqV) ⇒ λ ∝ 1√(mq) for the same V

Particle	m, q	λ
Electron	m_e, e	12.27/√V Å
Proton	1 amu, e	0.286/√V Å
Deuteron	2 amu, e	0.286/(√2·√V) = 0.202/√V Å
α-particle	4 amu, 2e	0.286/(√8·√V) = 0.101/√V Å
C⁺ ion	12 amu, e	0.286/(√12·√V) Å
Neutron	1 amu, 0	0.286/√E Å (E = K.E. in eV)

Q. Compare λ of a proton accelerated through 50 V and a neutron of K.E. 50 eV.
Proton : K = qV = 50 eV ⇒ λ = 0.286√50 = 0.040 Å
Neutron : λ = 0.286√50 = 0.040 Å (same mass, same K.E. ⇒ same λ)

15. λ of Gas Particles

Gas particle at temperature T : K = 32kT

λ = h√(2mK) = h√(2m × 3kT/2)

λ = h√(3mkT)

k = Boltzmann constant = 1.38 × 10⁻²³ J/K ; T in kelvin

Q. Find λ of an O₂ molecule at 27 °C.
T = 300 K, m = 32 amu = 32 × 1.67 × 10⁻²⁷ = 5.34 × 10⁻²⁶ kg
λ = 6.63 × 10⁻³⁴√(3 × 5.34 × 10⁻²⁶ × 1.38 × 10⁻²³ × 300)
= 6.63 × 10⁻³⁴2.58 × 10⁻²³ ≈ 2.57 × 10⁻¹¹ m = 0.257 Å

16. Particle vs Photon

	Photon	Particle
λ & p	λ = h/p (valid for both)
λ & energy E	λ = hc/E	λ = h/√(2mE), E = K.E.

17. Waves (Quick Recap)

y = A sin(ωt − kx + φ) , f = ω2π , v = fλ

y = A sin(10¹⁴t + φ) ⇒ f = 10¹⁴2π Hz
y = A sin 2π(10¹⁴t + φ) ⇒ ω = 2π × 10¹⁴ ⇒ f = 10¹⁴ Hz
Sum of two waves A₁sin(ω₁t − k₁x + φ₁) + A₂sin(ω₂t − k₂x + φ₂) ⇒ frequencies ω₁2π and ω₂2π.
Product A sin(ω₁t − …) sin(ω₂t − …) : use 2 sinA sinB = cos(A − B) − cos(A + B) ⇒ two waves of frequencies

f′ = ω₁ + ω₂2π , f″ = ω₁ − ω₂2π

Standing wave between two fixed ends :

l = nλ2 (n = integer = no. of loops)

18. Bohr's Quantization Condition

Electron moves only in orbits whose circumference is an integral multiple of its de Broglie wavelength :

2πr = nλ = nhmv

mvr = nh2π

Q. A particle of mass m moves inside a narrow tube of length L ; its matter wave forms a standing wave with nodes at both ends and N loops. Find (a) λ, (b) momentum, (c) energy.
L = Nλ2 ⇒ (a) λ = 2L/N
(b) p = hλ = Nh2L (c) K = p²2m = N²h²8mL²

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Photoelectric Effect

19. Photoelectric Effect

When light of suitable wavelength falls on a clean metal surface, electrons are ejected instantaneously (time lag ≲ 10⁻⁹ s). These are photoelectrons.
Red light (low ν) → no electron ejected, however intense ; violet / UV → electrons ejected. Brighter violet light → more electrons.
One photon is absorbed by one electron completely (1 electron ↔ 1 photon) and converted into its energy.

E = hν = hcλ = 12400λ (Å) eV ; N = IAhν = IAλhc

20. Efficiency (η%)

Number of electrons emitted per 100 incident photons (usually very small, ~0.001%).

N_e = η100 N

21. Work Function (φ)

Minimum energy required to eject an electron from the surface of a metal.
It depends only on the material (not on light). Minimum for caesium.

22. Threshold Wavelength & Frequency

λ_th (cut-off wavelength) = maximum wavelength for which electrons are ejected.
ν_th = minimum frequency for which electrons are ejected.

φ = hν_th = hcλ_th = 12400 eV·Åλ_th (Å) , c = ν_thλ_th

Condition for emission : E ≥ φ ⇔ ν ≥ ν_th ⇔ λ ≤ λ_th

e.g. φ = 2 eV ⇒ minimum photon energy needed = 2 eV
λ_th = 124002 = 6200 Å ; light with λ > 6200 Å cannot eject electrons.

23. Maximum K.E. of Photoelectron

K_max = E − φ = hν − φ = ½mv_max²

Electrons below the surface lose some energy before escaping ⇒ ejected electrons have 0 ≤ K ≤ K_max.

24. Photoelectric Experiment

Light falls on metal plate C (emitter) ; electrons move to plate A (collector) ; ammeter reads photocurrent i.
Intensity I is changed by changing distance from source (I ∝ 1/r²) ; ν (or λ) is changed by changing the source (e.g. green → blue).

Voltage	K at collector
Accelerating V (A at +)	K_max + eV (slowest : 0 + eV)
Retarding V (A at −)	K_max − eV (slow ones turn back)

25. Saturation Current (i_s)

As accelerating voltage increases, photocurrent increases, but after a certain value it becomes constant (all emitted electrons are collected) = saturation current.

i_s = ΔqΔt = N_e e = η100 N e

i_s = N_ee = η100·IAhν·e

26. Stopping Potential (V_s)

As retarding voltage increases, photocurrent decreases. At a particular retarding voltage (stopping potential) even the fastest electron is stopped ⇒ current = 0.

K_max = eV_s

eV_s = hν − hν_th = hcλ − hcλ_th

e.g. Stopping potential = 10 V ⇒ K_max = 10 eV.
φ = 2 eV, E = 5 eV ⇒ K_max = 3 eV ⇒ V_s = 3 V ; electrons come out with any K from 0 to 3 eV.

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Photoelectric Effect

27. Observations of the Experiment

(1) Saturation current vs intensity (ν constant)

i_s = η100·IAhν·e ⇒ ν = const ⇒ i_s ∝ I

(2) Stopping potential vs frequency

eV_s = hν − φ ⇒ V_s = heν − φe

Slope = tanθ = he = 4.14 × 10⁻¹⁵ V·s — same for all metals (parallel lines).
x-intercept = ν_th ; y-intercept = −φ/e.

(3) Photocurrent vs tube voltage

same ν, I₂ > I₁

same photon flux, ν₂ > ν₁

Changing I (same ν) changes only i_s ; stopping potential stays the same.
Higher ν ⇒ larger V_s (more negative cut-off).

(4) Photoelectric effect is instantaneous

No time lag (~10⁻⁹ s) between incidence of light and emission, even for very weak light.

28. What depends on what?

Quantity	Intensity I	Frequency ν
Photon energy E	no	yes (E = hν)
Work function φ	no	no (material only)
K_max , V_s	no	yes
Saturation current i_s	yes	yes (i_s ∝ I/ν)

29. If ν is Doubled

K_max = eV_s = hν − φ (form y = mx + c, not y = mx)

So K_max is NOT ∝ ν and V_s is NOT ∝ ν. If ν becomes n times (n > 1), K_max and V_s become more than n times.

e.g. E₁ = 5 eV, φ = 2 eV ⇒ K₁ = 5 − 2 = 3 eV
ν₂ = 2ν₁ ⇒ E₂ = 10 eV ⇒ K₂ = 10 − 2 = 8 eV (> 2 × 3 eV)
Analogy : earn 10,000, spend 6,000 (fixed) → save 4,000 ; earn 20,000 → save 14,000 (more than double).

30. Failures of Wave Theory

Intensity problem : wave theory says K.E. of electrons should increase with intensity ; experimentally K_max = hν − φ does not depend on intensity.
Frequency problem : wave theory says emission should occur at every frequency ; experimentally no emission for ν < ν_th, whatever the intensity.
Time-delay problem : wave theory predicts a time lag for energy to accumulate ; experimentally emission is instantaneous (~10⁻⁹ s).

31. Isolated Metal Sphere (Solved)

Q. Light of λ = 4000 Å falls on an isolated metal sphere of radius R, φ = 2.5 eV. At steady state find (1) potential of the sphere, (2) charge on it.
Sphere loses electrons ⇒ becomes positive (+q, potential V). Steady state when even the fastest electron cannot escape : K_max − eV = 0 ⇒ V = V_s.
E = 12400/4000 = 3.1 eV ; K_max = 3.1 − 2.5 = 0.6 eV ⇒ eV_s = 0.6 eV
(1) V = 0.6 V
(2) V = kqR ⇒ q = VRk = 0.6 R9 × 10⁹ = 6.7 × 10⁻¹¹ R coulomb (R in m)

Formula Revision :

E = hν = hc/λ = 12400/λ(Å) eV ; p = h/λ = E/c
N = P/E = IAλ/hc ; radiation pressure P = I(1 + r)cos²θ/c
λ = h/p = h/√(2mK) = h/√(2mqV) ; λ_e = 12.27/√V Å ; gas : h/√(3mkT)
φ = hν_th = hc/λ_th ; K_max = hν − φ = eV_s
i_s = (η/100)(IA/hν)e ; V_s–ν slope = h/e