From Energy Quantum to Particle
The photoelectric effect gave evidence to the strange fact that light, in interaction with matter, behaved as if it was made of quanta or packets of energy, each of energy $h\nu$. Einstein's explanation of that effect rested on a single quantum of radiation being absorbed by a single electron. But a packet of energy alone is not yet a particle.
The decisive step, as NCERT §11.7 records, was Einstein's result that the light quantum can also be associated with a momentum, $h\nu/c$. A definite value of both energy and momentum is a strong sign that the light quantum can be associated with a particle. This particle was later named the photon. The particle-like behaviour of light was confirmed in 1924 by A. H. Compton's experiment on the scattering of X-rays from electrons.
So the logic runs in two stages: the photoelectric effect establishes that radiant energy comes in quanta of $h\nu$; the assignment of momentum $h\nu/c$ to each quantum elevates that quantum to a genuine particle. The photon is therefore the particle answer to the wave description of light.
A photon carries a definite energy $h\nu$ and a definite momentum $h\nu/c$, and always travels at the speed of light $c$. The pairing of energy and momentum is what makes it a particle.
It is worth being precise about what "particle" means here. A photon is not a tiny ball of light; it is a localised packet carrying a definite, indivisible quantum of energy and momentum that interacts with matter all-at-once rather than depositing energy gradually like a continuous wave. When a single photon meets a single electron in the photoelectric effect, the whole quantum $h\nu$ is handed over in one event — there is no fraction of a photon. This whole-quantum exchange is exactly why the photoelectric effect is instantaneous and why the ejected electron's maximum kinetic energy depends on frequency rather than on how bright the beam is.
Energy of a Photon
Each photon of radiation of frequency $\nu$ carries an energy equal to one quantum:
$$E = h\nu = \dfrac{hc}{\lambda}$$
Here $h = 6.63 \times 10^{-34}\ \text{J s}$ is Planck's constant, $c = 3 \times 10^{8}\ \text{m s}^{-1}$ is the speed of light, and $\lambda$ the wavelength. The two forms are connected by $c = \nu\lambda$. The energy rises with frequency and falls with wavelength: a high-frequency (short-$\lambda$) photon is more energetic than a low-frequency (long-$\lambda$) one. In atomic and nuclear physics the energy is often expressed in electron volts, with $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$.
A practical consequence of $E = hc/\lambda$ is that the product $hc$ acts as a fixed conversion constant: $hc \approx 1.99 \times 10^{-25}\ \text{J m}$, or about $1240\ \text{eV nm}$ with energy in electron volts and wavelength in nanometres. So a $600\ \text{nm}$ visible photon carries $\approx 1240/600 \approx 2.07\ \text{eV}$, while a $0.1\ \text{nm}$ X-ray photon carries about $12400\ \text{eV}$ — four orders of magnitude more for the same kind of "particle". This is why X-ray and gamma photons ionise matter readily while long-wavelength radio photons are individually feeble. The energy is a property of the single photon, decided by the frequency of the light alone.
Momentum of a Photon
The feature that completes the particle picture is momentum. Einstein associated each light quantum with momentum
$$p = \dfrac{h\nu}{c} = \dfrac{h}{\lambda}$$
The second form follows directly from $c = \nu\lambda$ and shows that a photon's momentum is set entirely by its wavelength. The same result can be read off the energy: since $E = pc$ for a photon, $p = E/c = h\nu/c$. NCERT later notes that this is exactly the de Broglie relation $\lambda = h/p$ applied to a photon, so the photon's wavelength as a particle equals the wavelength of the radiation of which it is the quantum.
The relation $E = pc$ is the photon's signature. For a slow massive particle, energy and momentum are tied by the quadratic $E_k = p^2/2m$; for a massless photon they are simply proportional, with $c$ as the proportionality constant — and that linear tie is what lets a massless object carry momentum at all. The momentum is real and measurable: light striking a surface exerts radiation pressure, and the electron recoil in Compton scattering is the direct fingerprint of photon momentum transferred in a collision. NIOS Ch.25 records the same expression $p = h\nu/c$, confirming it is standard rather than particular to one textbook.
| Property | Photon | Notes |
|---|---|---|
| Energy | E = hν = hc/λ | Set by frequency / wavelength |
| Momentum | p = hν/c = h/λ | Follows from $E = pc$ |
| Speed | c | Always the speed of light |
| Rest mass | Zero | Yet carries momentum |
| Charge | Zero (neutral) | Not deflected by E or B fields |
| Relation E and p | E = pc | Energy–momentum tie for a photon |
Zero rest mass does not mean zero momentum
Students reach for $p = mv$, see that a photon has zero rest mass, and wrongly conclude its momentum is zero. The photon still carries momentum $p = h\nu/c = h/\lambda$, drawn from its energy through $p = E/c$. Compton's X-ray scattering experiment confirmed this momentum experimentally. Equally, do not confuse $E = h\nu$ (per photon) with $E = hc/\lambda$ — they are the same quantity written two ways via $c = \nu\lambda$.
Photon: rest mass $= 0$, yet $p = h/\lambda \neq 0$ and $E = pc$.
The Photon Picture: Five Properties
NCERT §11.7 condenses the entire photon model of electromagnetic radiation into five statements. They are reproduced faithfully below, because NEET statement-type questions are drawn almost verbatim from them.
| # | Property of the photon |
|---|---|
| (i) | In the interaction of radiation with matter, radiation behaves as if it is made up of particles called photons. |
| (ii) | Each photon has energy $E = h\nu$, momentum $p = h\nu/c$, and speed $c$. |
| (iii) | All photons of a given frequency $\nu$ (wavelength $\lambda$) have the same energy $E = h\nu = hc/\lambda$ and momentum $p = h\nu/c = h/\lambda$, whatever the intensity. |
| (iv) | Photons are electrically neutral and are not deflected by electric and magnetic fields. |
| (v) | In a photon–particle collision, total energy and total momentum are conserved; the number of photons may change (a photon can be absorbed or created). |
Property (v) deserves emphasis. Energy and momentum are conserved in every photon–electron collision, exactly as in mechanics. But unlike colliding billiard balls, the photons themselves are not conserved in number: one may be absorbed entirely, or a new one created. This is why photon collisions are governed by conservation laws on energy and momentum alone, never on photon count.
Unpacking those statements one property at a time makes the model easier to apply under exam pressure:
- Energy is quantised: a photon's energy $E = h\nu = hc/\lambda$ is fixed by frequency, indivisible, and exchanged whole in a single interaction.
- It carries momentum: $p = h\nu/c = h/\lambda$, equivalently $p = E/c$, so a massless packet still pushes on whatever absorbs or scatters it.
- It travels at speed $c$: in vacuum every photon, of every frequency, moves at the same speed $c = 3 \times 10^{8}\ \text{m s}^{-1}$ — there are no slow or fast photons in free space.
- Zero rest mass: the photon has no rest mass, which is precisely why it can move at $c$ and why $E = pc$ rather than $E = p^2/2m$ applies.
- Electrically neutral: photons carry no charge, so electric and magnetic fields cannot deflect them, unlike electron or proton beams.
- Energy fixed, number flexible: raising the intensity at fixed frequency raises the number of photons per second, never the energy $h\nu$ of any one photon.
- Conservation in collisions: in a photon–electron collision total energy and total momentum are conserved, but the photon count may rise or fall as photons are absorbed or created.
The same energy quantum $h\nu$ powers the work-energy balance for ejected electrons in Einstein's Photoelectric Equation.
Intensity and Photon Flux
A common misreading is that brighter light means more energetic photons. NCERT is explicit: by increasing the intensity of light of a given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon still having the same energy. Photon energy is independent of intensity. Intensity measures the photon-number flux, not the energy per photon.
For fixed frequency, doubling the intensity doubles the photons arriving per second; the energy $h\nu$ of each photon is unchanged. Number of photons per second $\propto$ intensity.
This connects directly to the photoelectric experiments: the photocurrent is proportional to intensity precisely because intensity fixes how many photons (and hence how many absorbing electrons) participate per second, while the maximum kinetic energy of the ejected electrons is governed by the energy $h\nu$ of each photon.
We can make the link between intensity and photon flux quantitative using the laser values already in hand. Intensity $I$ is power per unit area, $I = P/A$, and since each photon carries energy $E = h\nu$, the photon flux — the number of photons crossing unit area per second — is
$$\Phi = \dfrac{I}{h\nu} = \dfrac{P}{A\,h\nu}.$$
Take the laser of NCERT Example 11.1, emitting $P = 2.0 \times 10^{-3}\ \text{W}$ at $\nu = 6.0 \times 10^{14}\ \text{Hz}$ (photon energy $h\nu = 3.98 \times 10^{-19}\ \text{J}$). Spread over $A = 1.0\ \text{cm}^{2} = 1.0 \times 10^{-4}\ \text{m}^{2}$, the intensity is $I = P/A = 20\ \text{W m}^{-2}$ and the photon flux is $\Phi = I/h\nu = 20/3.98 \times 10^{-19} \approx 5.0 \times 10^{19}$ photons per second per square metre. Double the intensity to $40\ \text{W m}^{-2}$ and the flux doubles to $1.0 \times 10^{20}\ \text{m}^{-2}\,\text{s}^{-1}$ — yet each photon still carries the same $3.98 \times 10^{-19}\ \text{J}$. That is the whole content of "intensity changes how many, not how much".
Photon vs Classical Particle vs Matter Wave
The photon sits between two pictures NEET also tests. A classical particle such as a moving electron has rest mass, travels at any speed below $c$, and ties energy to momentum through $E_k = p^2/2m$. The de Broglie matter-wave view, developed on the Wave Nature of Matter — de Broglie page, assigns such a particle a wavelength $\lambda = h/p$. The photon is the mirror image: the particle aspect of a wave, massless, locked to speed $c$, with $E = pc$.
| Feature | Photon | Classical particle (e.g. electron) | de Broglie matter wave |
|---|---|---|---|
| Rest mass | Zero | Non-zero ($m$) | Non-zero ($m$); wave is associated with it |
| Speed | Always $c$ in vacuum | Any $v < c$ | Particle moves at $v$; wavelength $\lambda = h/p$ |
| Energy–momentum | E = pc | E_k = p²/2m | p = h/λ |
| Momentum | $p = h\nu/c = h/\lambda$ | $p = mv$ | $p = h/\lambda$ |
| Charge | Neutral; not deflected by E or B | May be charged; deflected by E or B | Carried by the particle |
| Origin | Particle aspect of EM radiation | Newtonian mechanics | Wave aspect of matter (de Broglie) |
The shared entry is telling: both the photon and a de Broglie matter wave obey $p = h/\lambda$. That single relation is the bridge of the whole "dual nature" chapter — light, classically a wave, gains a particle momentum $h/\lambda$, while matter, classically a particle, gains a wave of the same form. The Davisson–Germer experiment later supplied the experimental proof for the matter-wave half of that symmetry.
Worked Examples
Monochromatic light of frequency $6.0 \times 10^{14}\ \text{Hz}$ is produced by a laser. The power emitted is $2.0 \times 10^{-3}\ \text{W}$. (a) What is the energy of a photon in the beam? (b) How many photons per second, on average, are emitted by the source?
(a) Each photon has energy $$E = h\nu = (6.63 \times 10^{-34}\ \text{J s})(6.0 \times 10^{14}\ \text{Hz}) = 3.98 \times 10^{-19}\ \text{J}.$$
(b) If $N$ photons are emitted per second, the power $P = NE$, so $$N = \dfrac{P}{E} = \dfrac{2.0 \times 10^{-3}\ \text{W}}{3.98 \times 10^{-19}\ \text{J}} = 5.0 \times 10^{15}\ \text{photons per second}.$$
Find the momentum of the photon in NCERT Example 11.1 above, using $c = 3 \times 10^{8}\ \text{m s}^{-1}$.
The photon momentum is $$p = \dfrac{h\nu}{c} = \dfrac{E}{c} = \dfrac{3.98 \times 10^{-19}\ \text{J}}{3 \times 10^{8}\ \text{m s}^{-1}} \approx 1.33 \times 10^{-27}\ \text{kg m s}^{-1}.$$ The energy $E = h\nu$ from part (a) is reused, since $p = E/c$ for any photon.
A monochromatic source emits light of wavelength $\lambda = 600\ \text{nm}$ at a power of $P = 3.3 \times 10^{-3}\ \text{W}$. How many photons does it emit per second? (Take $h = 6.6 \times 10^{-34}\ \text{J s}$, $c = 3 \times 10^{8}\ \text{m s}^{-1}$.)
Each photon has energy $E = hc/\lambda$, and the number per second is $N = P/E = P\lambda/(hc)$: $$N = \dfrac{P\lambda}{hc} = \dfrac{(3.3 \times 10^{-3})(6.0 \times 10^{-7})}{(6.6 \times 10^{-34})(3 \times 10^{8})} = 1.0 \times 10^{16}\ \text{photons per second}.$$ Using $E = hc/\lambda$ rather than $E = h\nu$ avoids first converting wavelength to frequency — the answer is identical because the two energy forms are the same quantity.
Find the momentum of a photon of wavelength $\lambda = 0.50\ \text{nm}$ (an X-ray photon). (Take $h = 6.63 \times 10^{-34}\ \text{J s}$.)
Momentum depends only on wavelength through $p = h/\lambda$: $$p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}\ \text{J s}}{0.50 \times 10^{-9}\ \text{m}} \approx 1.33 \times 10^{-24}\ \text{kg m s}^{-1}.$$ This is a thousand times the momentum of the visible photon found above, mirroring its thousand-fold shorter wavelength — momentum scales as $1/\lambda$.
Particle Nature of Light — The Photon
- Photoelectric effect → energy quanta of $h\nu$; assigning momentum $h\nu/c$ makes the quantum a particle, the photon. Compton scattering (1924) confirmed it.
- Photon energy: $E = h\nu = hc/\lambda$. Photon momentum: $p = h\nu/c = h/\lambda$, equivalently $p = E/c$.
- Every photon travels at speed $c$, has zero rest mass, and is electrically neutral — undeflected by electric and magnetic fields.
- For fixed frequency, intensity sets the number of photons per second; each photon's energy $h\nu$ is independent of intensity.
- In photon–particle collisions, total energy and total momentum are conserved, but the number of photons may change.