A full lesson: a cartoon movie to start, then light as packets of energy, the energy and momentum of one photon, the five photon properties, counting photons from the power of a lamp, why "brighter" is not "stronger", when light behaves as a wave and when as photons, NEET shortcuts, problems to solve yourself, and 12 MCQs.
A very bright red lamp fails, a tiny blue torch succeeds. Light arrives in packets (photons) of energy hν. Each glow-star molecule needs one packet that is big enough; a thousand small packets do not add up.
Meera has glow-in-the-dark stars stuck on her bedroom ceiling. To "charge" them, Dadi switches on her big, very bright red night lamp for five minutes. Lights off: nothing. The stars stay dark. Meera then shines her tiny blue key-chain torch on them for ten seconds. Lights off: the stars glow green.
If light were only a smooth wave, the bright red lamp should win easily: it pours far more energy on the ceiling every second. It loses because light is delivered in packets. A red packet is too small to excite one molecule of the glow paint, and a molecule cannot collect two small packets. A blue packet is big enough on its own. Those packets are photons, and this lesson is about them.
"The colour decides the size of each packet. The brightness only decides how many packets arrive. Hold on to this one line and half of this lesson is already done."
"Try it at home with a glow toy: charge it with your phone torch through red cellophane, then through blue cellophane. Which one makes it glow? And would a red lamp ten times brighter help?"
Blue works, red does not. A red lamp ten times brighter only sends ten times more red photons, each still too small, so it still fails. Energy per photon E = hν depends only on the frequency (colour), never on the brightness.
From the photoelectric effect, Einstein concluded that light energy is carried in quanta of E = hν. He also gave each quantum a definite momentum p = hν/c. Something with a definite energy and a definite momentum behaves like a particle, and it was later named the photon. Slide the wavelength and watch the packets change.
h = 6.63 × 10⁻³⁴ J s. Higher frequency (shorter λ) means a bigger packet.
A photon has momentum even though its rest mass is zero. Light can push (radiation pressure).
1 eV = 1.6 × 10⁻¹⁹ J. Red 620 nm: 2.0 eV. Violet 400 nm: 3.1 eV. X-ray 0.1 nm: 12.4 keV.
"Two formulas, one idea. Energy is h nu, momentum is h over lambda, and dividing them gives c. If NEET gives you λ in nanometres, go straight to 1240 over λ for the energy in electron volts."
NCERT lists what we know about photons. Test three of them yourself: send a photon and an electron through electric and magnetic fields, let an X-ray photon collide with an electron, and let an atom swallow a photon.
Every photon of one frequency has exactly the same energy and momentum, whatever the source.
Brighter light of the same colour means more photons crossing each square metre every second, each with the same energy.
Electric and magnetic fields cannot bend a photon's path. An electron in the same fields curves.
In a photon–electron collision, total energy and total momentum are conserved (Compton, 1923).
An atom can absorb a photon (it disappears) or emit a new one. Unlike charge, photon number is not conserved.
A photon always moves at c in vacuum and has zero rest mass. NEET sometimes asks its "equivalent mass" E/c² = h/(cλ).
"In the collision, the photon gives some energy to the electron. Does the photon slow down, or does something else change?"
It never slows down: a photon always moves at c. Losing energy means a smaller hν, so its frequency falls and its wavelength gets longer. That longer-wavelength scattered X-ray is exactly what Compton measured.
A source of power P sends out P joules every second. Each photon carries hc/λ. Divide, and you get the number of photons per second. Choose a source or set your own power and wavelength; each dot on the screen stands for 10¹⁴ photons.
Change λ to metres first: 500 nm = 5 × 10⁻⁷ m. Then E = 6.63 × 10⁻³⁴ × 3 × 10⁸ / 5 × 10⁻⁷ = 3.98 × 10⁻¹⁹ J.
Change mW to W first: 2 mW = 2 × 10⁻³ W, that is 2 × 10⁻³ J every second.
N = 2 × 10⁻³ / 3.98 × 10⁻¹⁹ ≈ 5.0 × 10¹⁵ photons per second. Same power, longer λ: more photons (N ∝ λ).
"Five thousand million million photons every second from a tiny laser! That is why light looks smooth and continuous to us: the packets are tiny and there are too many to notice one by one."
Back to the photoelectric effect, now with photon eyes. Light falls on caesium (work function φ₀ = 2.14 eV, as in NCERT). Turn up the intensity and the frequency separately and watch what each one changes. Think of rain and hail: more raindrops do not dent a car, one hailstone does.
| Light made… | Photons | Electrons |
|---|---|---|
| Brighter | same energy, more per second | more current, same K max and V₀ |
| Bluer, same rate | more energy, same number | same current, larger K max and V₀ |
| Bluer, same P | more energy, fewer (N ∝ λ) | less current, larger K max and V₀ |
The fastest electron has K max = eV₀, and the photon energy is shared as hν = φ₀ + K max.
Example: φ₀ = 2.0 eV, V₀ = 1.5 V → E = 3.5 eV → λ = 1240/3.5 ≈ 354 nm. Add both in eV, or both in joule, never mixed.
"One photon, one electron. That is the whole secret. Intensity is how many photons, so it controls how many electrons. Frequency is how big each photon is, so it controls how fast the electrons come out."
Interference, diffraction and polarisation need the wave picture. The photoelectric effect and Compton scattering need the photon picture. Watch both at once: send photons through two slits one at a time. Each lands as a single dot (particle), but thousands of dots build interference fringes (wave).
| Phenomenon | Picture needed, and why |
|---|---|
| Interference | Wave: needs phase and superposition |
| Diffraction | Wave: bending round edges, adding wavelets |
| Polarisation | Wave: needs a transverse vibration direction |
| Photoelectric | Photon: threshold frequency, instant emission, K max independent of intensity |
| Compton | Photon: a collision with energy and momentum bookkeeping |
| Seeing | Both: the lens focuses a wave; rods and cones absorb photons |
"Light is not sometimes a wave and sometimes a ball. It is one thing with two faces. When it travels, the wave face shows. When it gives energy to matter, the photon face shows."
400 nm → 3.1 eV, 620 nm → 2.0 eV, 1 nm → 1240 eV. Use 12400/λ(Å) if λ is in angstrom.
500 nm → 1.3 × 10⁻²⁷ kg m/s. For E in eV: p = E × 1.6 × 10⁻¹⁹ / (3 × 10⁸).
Quick form: N ≈ 5 × 10¹² × P(mW) × λ(nm). 2 mW, 500 nm → 5 × 10¹⁵ s⁻¹. Equal power: N ∝ λ, so red sources send more photons than blue.
Momentum per second = N × h/λ = P/c. A perfect mirror reverses it, doubling the force.
φ₀ in eV and V₀ in volts: eV₀ in eV is the same number as V₀. Add, then divide.
Rest mass is 0. This is a "mass" only in the sense of E = mc²; a photon at rest does not exist.
"Memorise 1240. With that one number you can turn any wavelength in nanometres into electron volts in five seconds, and NEET answer options are almost always in eV or nm."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute. Take h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, e = 1.6 × 10⁻¹⁹ C.
A laser produces monochromatic light of frequency 6.0 × 10¹⁴ Hz. The power emitted is 2.0 × 10⁻³ W. (a) What is the energy of a photon? (b) How many photons per second, on average, does the source emit?
E = hν, then N = P/E.
Find the energy (in J and keV) and the momentum of an X-ray photon of wavelength 0.1 nm.
E = hc/λ with λ = 10⁻¹⁰ m; p = h/λ.
Light falls on caesium (φ₀ = 2.14 eV). The stopping potential is 0.60 V. Find the wavelength of the light.
Photon energy = eV₀ + φ₀, both in eV. Then λ = 1240/E.
A 10 W laser beam of wavelength 600 nm falls normally on a perfectly absorbing plate. Find the number of photons hitting the plate per second and the force on it. What is the force if the plate is a perfect mirror?
N = Pλ/(hc). Each photon brings momentum h/λ, so F = N h/λ = P/c.
A red lamp (700 nm) and a violet lamp (400 nm) each emit 1 W of light. Which sends more photons per second, and by what factor? Which photons carry more energy each?
N = Pλ/(hc), so at equal P, N ∝ λ. E ∝ 1/λ.
Take h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s and hc = 1240 eV nm. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "energy of a photon of wavelength λ" | E = hc/λ; in eV: 1240/λ(nm) | Forgetting to change nm to m when using SI units |
| "momentum of a photon" | p = h/λ = E/c | Writing p = mv and saying p = 0 because the rest mass is 0 |
| "number of photons emitted per second" | N = Pλ/(hc) | Not changing mW to W; using frequency where λ is given |
| "intensity is doubled" | photon energy same, photon number doubles | Doubling the energy of each photon (or K max) |
| "deflected by electric or magnetic field" | photon is neutral: no deflection | Treating light like a beam of charged particles |
| "conserved in a photon–electron collision" | total energy and total momentum | Claiming the number of photons is conserved |
| "stopping potential V₀, work function φ₀, find λ" | λ = hc/(eV₀ + φ₀) | Subtracting instead of adding; mixing eV and J |
| "force exerted by a light beam" | F = P/c (absorb), 2P/c (reflect) | Forgetting the factor 2 for a mirror |
If a wave (light) can behave like a particle with p = h/λ, can a particle behave like a wave? Lesson 5 turns the same formula around: λ = h/p, de Broglie's matter waves.
Values used: h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J, hc ≈ 1240 eV nm (1243 with these constants), work function of caesium 2.14 eV (NCERT Table 11.1). The glow-paint threshold in the movie (about 2.6 eV) is approximate.
Follow the four steps: understand it, remember it, try it, then prove it with past papers.