From Energy Quantum to Genuine Particle

The photoelectric effect showed that light, in interaction with matter, behaves as if made of quanta of energy hνh\nu. But is a 'quantum of energy' really a particle? A particle should carry a definite momentum too.

Einstein supplied the answer: the light quantum carries momentum

p=hνcp = \frac{h\nu}{c}

A definite energy and a definite momentum — that's the calling card of a particle. This particle of light was later named the photon.

The clinching experimental proof came in 1924 from A. H. Compton's scattering of X-rays from electrons: the X-ray photon bounces off an electron exactly like a billiard-ball collision, conserving energy and momentum. Particle behaviour, caught in the act.

Nobel scoreboard

  • Einstein — Nobel Prize 1921, for contributions to theoretical physics and the photoelectric effect.
  • Millikan — Nobel Prize 1923, for the elementary charge and the photoelectric effect.

Photon properties energy momentum and collisions

The Photon Picture: Five Properties to Memorise

NCERT summarises the photon picture of electromagnetic radiation in five statements — learn them as a set:

  1. In interaction with matter, radiation behaves as if made of particles — photons.
  2. Each photon has energy E=hν=hcλE = h\nu = \dfrac{hc}{\lambda}, momentum p=hνc=hλp = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}, and speed c, the speed of light.
  3. All photons of a given frequency (or wavelength) are identical in energy and momentum, whatever the intensity. Increasing intensity only increases the number of photons crossing a given area per second — never the energy per photon.
  4. Photons are electrically neutral, and are not deflected by electric or magnetic fields.
  5. In a photon-particle collision (say photon-electron), total energy and total momentum are conserved — but the number of photons may not be conserved: a photon may be absorbed, or a new one created.

Key Point: Property 5 is subtle and exam-worthy: energy and momentum conservation hold, photon number conservation does not. Light can be born and can die; electrons cannot (in these processes).

[JEE Tip] Two workhorse formulas: E=hcλ=1240 eV nmλ(nm)E = \dfrac{hc}{\lambda} = \dfrac{1240 \text{ eV nm}}{\lambda(\text{nm})} and p=hλ=Ecp = \dfrac{h}{\lambda} = \dfrac{E}{c}. For photon-counting problems, the number emitted per second by a source of power P is N=PhνN = \dfrac{P}{h\nu} — NCERT Example 11.1 exactly.

[NEET Important] A photon has zero rest mass, travels only at c, and its 'effective mass' questions (m=E/c2m = E/c^2) occasionally appear — but its rest mass is strictly zero. Neutrality (no deflection in E or B fields) distinguishes photon beams from cathode rays instantly.

Solved Examples

Example 1: Photons from a laser (NCERT Example 11.1)

Monochromatic light of frequency 6.0×10146.0 \times 10^{14} Hz is produced by a laser of power 2.0×10−32.0 \times 10^{-3} W. (a) What is the energy of each photon? (b) How many photons per second does the source emit?

Solution:

  1. (a) Photon energy: E=hν=(6.63×10−34)(6.0×1014)=3.98×10−19E = h\nu = (6.63 \times 10^{-34})(6.0 \times 10^{14}) = 3.98 \times 10^{-19} J ≈ 2.49 eV.
  2. (b) Counting: power P = N × E per second, so N=PE=2.0×10−33.98×10−19N = \dfrac{P}{E} = \dfrac{2.0 \times 10^{-3}}{3.98 \times 10^{-19}}.
  3. Answer: N≈5.0×1015N \approx 5.0 \times 10^{15} photons per second.
  4. Takeaway: even a feeble milliwatt beam delivers quadrillions of photons a second — which is why light looks continuous.

Example 2: Energy and momentum of one photon (NCERT Exercise 11.4 style)

A helium-neon laser emits at 632.8 nm with power 9.42 mW. Find each photon's energy and momentum, and the photons arriving per second at a target.

Solution:

  1. Energy: E=hcλ=(6.63×10−34)(3×108)632.8×10−9=3.14×10−19E = \dfrac{hc}{\lambda} = \dfrac{(6.63 \times 10^{-34})(3 \times 10^8)}{632.8 \times 10^{-9}} = 3.14 \times 10^{-19} J.
  2. Momentum: p=hλ=6.63×10−34632.8×10−9=1.05×10−27p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{632.8 \times 10^{-9}} = 1.05 \times 10^{-27} kg m/s.
  3. Rate: N=PE=9.42×10−33.14×10−19=3.0×1016N = \dfrac{P}{E} = \dfrac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}} = 3.0 \times 10^{16} photons/s.
  4. Takeaway: EE needs hc/λhc/\lambda; momentum needs only h/λh/\lambda — no factor of c.

Example 3: A hydrogen atom matching photon momentum (NCERT Exercise 11.4c)

How fast must a hydrogen atom (m = 1.66×10−271.66 \times 10^{-27} kg) travel to have the momentum of a 632.8 nm photon?

Solution:

  1. Photon momentum (from Example 2): p=1.05×10−27p = 1.05 \times 10^{-27} kg m/s.
  2. Set equal: mv=p⇒v=1.05×10−271.66×10−27mv = p \Rightarrow v = \dfrac{1.05 \times 10^{-27}}{1.66 \times 10^{-27}}.
  3. Answer: v ≈ 0.63 m/s — walking pace!
  4. Takeaway: photon momenta are tiny; an atom matches one at less than 1 m/s. This is why radiation pressure is so gentle.

Example 4: X-ray production limits (NCERT Exercise 11.1)

Electrons accelerated through 30 kV strike a target. Find (a) the maximum frequency and (b) the minimum wavelength of the X-rays produced.

Solution:

  1. Concept (inverse photoelectric effect): the entire electron energy eV converts, at best, into one photon: hνmax=eVh\nu_{max} = eV.
  2. (a) νmax=eVh=(1.6×10−19)(3×104)6.63×10−34=7.24×1018\nu_{max} = \dfrac{eV}{h} = \dfrac{(1.6 \times 10^{-19})(3 \times 10^4)}{6.63 \times 10^{-34}} = 7.24 \times 10^{18} Hz.
  3. (b) λmin=cνmax=3×1087.24×1018≈0.0414\lambda_{min} = \dfrac{c}{\nu_{max}} = \dfrac{3 \times 10^8}{7.24 \times 10^{18}} \approx 0.0414 nm (or use λmin=1240 eV nm30000 eV=0.0413\lambda_{min} = \dfrac{1240 \text{ eV nm}}{30000 \text{ eV}} = 0.0413 nm).
  4. Takeaway: the photoelectric effect run backwards — electron energy → photon energy — sets the hard short-wavelength limit of an X-ray tube.

Example 5: Which beam has more photons? [NEET Conceptual]

A 1 W red beam (700 nm) and a 1 W violet beam (400 nm): which delivers more photons per second, and by what factor?

Solution:

  1. Rate: N=Phc/λ=PλhcN = \dfrac{P}{hc/\lambda} = \dfrac{P\lambda}{hc} — at equal power, N ∝ λ\lambda.
  2. Ratio: NredNviolet=700400=1.75\dfrac{N_{red}}{N_{violet}} = \dfrac{700}{400} = 1.75.
  3. Answer: the red beam delivers 1.75 times more photons per second — each red photon is weaker, so more are needed to carry the same power.
  4. Takeaway: equal power never means equal photon count; the longer wavelength always wins on numbers.

Example 6: Photon in fields [Board Conceptual]

A beam passes undeflected through strong electric and magnetic fields. Can you conclude it is a photon beam?

Solution:

  1. Photons are electrically neutral — never deflected by E or B fields. Consistent.
  2. But neutrality alone doesn't clinch it: neutrons or any neutral particles also pass undeflected.
  3. Conclusion: undeflected passage is necessary but not sufficient evidence; contrast with cathode rays, which E and B fields visibly bend (Section 1).

Example 7: Photon number is not conserved [JEE Conceptual]

In a photon-electron collision, which of these are conserved: total energy, total momentum, number of photons, number of electrons?

Solution:

  1. Total energy: conserved. Total momentum: conserved. (Standard collision rules.)
  2. Photon number: NOT necessarily conserved — the photon may be absorbed (vanishing) or a new photon created.
  3. Electron number: conserved in these processes.
  4. Takeaway: NCERT's property (v) verbatim — a favourite true/false discriminator.

Example 8: Intensity vs photon energy [NEET Conceptual]

A fixed-wavelength source is turned up from 1 W to 10 W. What happens to (a) each photon's energy and momentum, (b) the photon flux, (c) the beam's total momentum delivered per second?

Solution:

  1. (a) unchanged — E = hc/λhc/\lambda and p = h/λh/\lambda depend only on wavelength.
  2. (b) photon flux (photons/s) rises tenfold: N = Pλ/hcP\lambda/hc ∝ P.
  3. (c) momentum delivered per second = N × p also rises tenfold — radiation pressure scales with intensity.
  4. Takeaway: intensity is a count dial, not an energy-per-photon dial — the photon picture's core message.

Example 9: Compton's role [Board Conceptual]

What did Compton's 1924 experiment demonstrate, and why was it decisive for the photon concept?

Solution:

  1. Compton scattered X-rays off electrons and analysed the scattered radiation.
  2. The results matched a particle-particle collision: a photon of energy hνh\nu and momentum hν/ch\nu/c striking an electron, with energy and momentum conserved.
  3. Decisive because momentum transfer is a particle signature — energy quanta alone (photoelectric effect) could conceivably be a property of absorption; Compton showed the quanta travel as particles.

Example 10: Photons per second from a sodium lamp [JEE Numerical]

A 60 W sodium lamp radiates at an effective wavelength of 589 nm. How many photons leave it per second?

Solution:

  1. Photon energy: E=1240589≈2.11E = \dfrac{1240}{589} \approx 2.11 eV =3.37×10−19= 3.37 \times 10^{-19} J.
  2. Rate: N=PE=603.37×10−19N = \dfrac{P}{E} = \dfrac{60}{3.37 \times 10^{-19}}.
  3. Answer: N≈1.8×1020N \approx 1.8 \times 10^{20} photons per second.
  4. Takeaway: household sources emit ~102010^{20} photons a second — the graininess of light is hidden under sheer numbers.