Planck's Quantum & the Photon

By the last months of exam season you have already learned that light is a wave — it interferes, it diffracts, it refracts by Huygens' construction. Then the previous article dropped a bombshell: shine light on a metal and it kicks out electrons in ways that a wave cannot explain. There is a sharp threshold frequency, the ejection is instantaneous, and the energy of each ejected electron depends on the colour of the light, never on its brightness. A wave, whose energy you can crank up simply by turning up the intensity, has no business behaving like that. The fix, when it came, was the most radical idea in physics since Newton. In 1900 Max Planck, trying to patch a stubborn hole in the theory of glowing hot objects, was forced to assume that energy is not handed out smoothly but in indivisible lumps — quanta — of size $E = hf$ . He hated the idea and spent years trying to explain it away. Five years later a young Albert Einstein took Planck's lumps at face value and declared that light itself travels as a stream of particle-like packets, each carrying energy $hf$ and momentum $h/\lambda$ . We now call each packet a photon. This article builds the photon from the ground up: the glowing-object puzzle that forced quantisation, Planck's quantum, Einstein's leap, and the two equations you will use for the rest of the chapter — $E = hf = hc/\lambda$ for a photon's energy and $p = h/\lambda = E/c$ for its momentum. Everything after this — Einstein's photoelectric equation, Compton scattering, even de Broglie's matter waves — is written in this language.

Key takeaways

Classical physics predicted that a hot object should radiate infinite energy at short wavelengths — the ultraviolet catastrophe. It does not; the real spectrum peaks and falls.
Planck (1900) cured it by assuming the oscillators in the walls can only gain or lose energy in whole lumps of $E = hf$ , where $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ is Planck's constant.
Einstein (1905) went further: light itself is quantised into photons, each a packet of energy $E = hf = hc/\lambda$ moving at $c$ .
A photon has momentum $p = h/\lambda = E/c$ even though its rest mass is zero — it is not $mv$ .
Photon energy is set by frequency alone: a $500\ \mathrm{nm}$ green photon carries $2.48\ \mathrm{eV}$ ; making the beam brighter adds more photons, not more energetic ones.
The exam workhorse is the shortcut $E[\mathrm{eV}] = 1240 / \lambda[\mathrm{nm}]$ , good to three figures for any wavelength given in nanometres.
Key idea: light is delivered in indivisible packets of energy $E = hf = hc/\lambda$ and momentum $p = h/\lambda$ , so the colour of the light — not its brightness — fixes how hard each packet hits.

The core insight

Energy is exchanged in indivisible quanta of size $hf$ : light is a stream of photons, each carrying energy $E = hf = hc/\lambda$ and momentum $p = h/\lambda = E/c$ , so a single photon's punch is set by its frequency, and turning up the brightness only sends more photons — never more energetic ones.

Where this lives today

The photon is not a historical curiosity; it is the working unit of modern optical technology. Every solar cell and photodiode counts on the fact that one photon frees one electron only if its energy $hf$ exceeds the material's band gap — which is why silicon panels waste the low-energy infrared and why engineers stack materials with different gaps to catch more of the spectrum. Every LED and laser runs the process in reverse, converting an electron's drop in energy into a photon of matching $hf$ . The image sensor in your phone (a CCD or CMOS array) is a grid of photon-counting buckets, and in dim light you can actually see the graininess — "shot noise" is the statistics of a finite number of photons arriving. At the frontier, single-photon detectors and sources power quantum cryptography and quantum computing, where information is encoded on individual photons and the discreteness that Planck stumbled into is the whole point. Medical PET scanners detect the $511\ \mathrm{keV}$ gamma-photon pairs from positron annihilation, and radiation therapy is dosed in photon energies — the same $E = hf$ you just learned, scaled up a million-fold.

The physical picture

Heat a horseshoe in a forge and watch it glow: dull red, then orange, then blinding yellow-white as it gets hotter. Every warm object radiates, and the mix of colours it emits depends only on its temperature — a fact so universal that physicists idealised it into a perfect emitter-and-absorber called a blackbody. By 1900 the shape of that glow had been measured to high precision, and it was a smooth hump: little energy at very long wavelengths, a peak that slides toward the blue as the object heats up, and — crucially — a fall back toward zero at short wavelengths. Classical physics could not reproduce that fall. Treating the radiation as waves sharing energy equally among all possible vibration modes (the equipartition idea) gave an intensity that climbed without limit as the wavelength shrank. The prediction was not just wrong, it was absurd: it said every warm object should blaze with infinite energy in the ultraviolet and beyond. This disaster was later christened the ultraviolet catastrophe. The mathematics that produces the exact experimental curve requires summing over the modes with calculus — an integral that is beyond this article, so we quote the curve's shape rather than derive it — but the reason the catastrophe is cured needs no calculus at all.

Figure

The blackbody spectrum — the classical curve races upward without limit at short wavelengths while Planck's quantized curve peaks and falls back to zero, curing the ultraviolet catastrophe

Planck's escape, announced to the German Physical Society on 14 December 1900, was a desperate act of "theoretical book-keeping," as he later called it. He imagined the walls of the cavity as a swarm of tiny oscillators, and — with no physical justification he was happy with — he assumed each oscillator of frequency $f$ could only hold energy in whole-number multiples of a basic lump: $0,\ hf,\ 2hf,\ 3hf,\dots$ . Energy came in quanta of size $hf$ . That single assumption tames the catastrophe automatically: a high-frequency (short-wavelength) mode needs one whole big lump $hf$ just to get started, and at a given temperature there is rarely enough energy lying around to pay that entry fee. The short-wavelength modes are effectively frozen out, the intensity falls instead of exploding, and Planck's formula matched the measured curve perfectly across every wavelength and temperature. The price was the constant he introduced to make the numbers fit — now called Planck's constant, $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ — the smallest, strangest number in physics. Planck thought his quantum was a mathematical trick about oscillators in walls, not about light itself. It was Einstein, in his 1905 "miracle year," who made the truly heretical claim: the radiation itself is grainy. Light of frequency $f$ arrives as a hail of localised packets, each carrying exactly $hf$ of energy, absorbed and emitted whole or not at all. That packet is the photon (a name coined later, by G. N. Lewis in 1926). With it, the photoelectric mysteries of the previous article dissolve at once: one photon gives all its energy to one electron, so the frequency of the light fixes the energy delivered per hit, while the intensity fixes only how many hits per second. That is the physical picture this whole chapter rests on.

Figure

A beam of light as a stream of photons — each packet carries energy hf and momentum h over lambda, so a blue photon of higher frequency packs more energy and momentum than a red one of the same beam brightness

What a photon is

A photon is not a tiny billiard ball and it is not a little bit of wave — it is a genuinely new kind of object, and the two equations below are its whole passport. First, its energy is fixed entirely by frequency:

E = hf = \frac{hc}{\lambda} \qquad (1)

where: $E$ is the energy carried by one photon, $f$ its frequency, $\lambda$ its vacuum wavelength, $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ Planck's constant, and $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ the speed of light. The second form uses $c = f\lambda$ to swap frequency for wavelength. The physical reading is stark: double the frequency and you exactly double the energy of every photon — there is no dial marked "intensity" in this equation. A dim blue lamp still fires more energetic photons than a brilliant red one. Second, a photon carries momentum even though it has no rest mass and cannot be brought to rest:

p = \frac{h}{\lambda} = \frac{E}{c} \qquad (2)

where: $p$ is the photon's momentum, $\lambda$ its wavelength, and $E$ its energy. This is not $p = mv$ — a photon's rest mass is zero, so that formula gives $0$ and is useless here. Instead the momentum comes from the massless energy-momentum relation $E = pc$ , a result we quote from relativity (and which Maxwell's wave theory also predicts for radiation). Rearranged, $p = E/c$ , and substituting $E = hc/\lambda$ gives $p = h/\lambda$ . A photon's momentum is small but real; it is exactly the quantity Compton will use in the next article to prove that light collides like a particle. Because the numbers in equation (1) are awkward — $h$ has thirty-four zeros hiding in it — every board that carries the photon leans on a shortcut. Measure the energy in electronvolts (recall $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$ , the energy an electron gains crossing one volt) and the wavelength in nanometres, and equation (1) collapses to a single number to memorise:

E[\mathrm{eV}] = \frac{1240}{\lambda[\mathrm{nm}]} \qquad (3)

where: the constant $1240$ is $hc$ expressed in $\mathrm{eV\,nm}$ . This one line converts any wavelength straight to a photon energy in your head: green light at $500\ \mathrm{nm}$ gives $1240/500 = 2.48\ \mathrm{eV}$ , violet at $400\ \mathrm{nm}$ gives $3.10\ \mathrm{eV}$ , red at $656\ \mathrm{nm}$ gives $1.89\ \mathrm{eV}$ . Across the visible band ( $400$ to $700\ \mathrm{nm}$ ) photon energies run from about $3.10\ \mathrm{eV}$ down to $1.77\ \mathrm{eV}$ — a range of barely a factor of two, which is why all visible light does roughly the same gentle chemistry to your retina.

From wavelength to energy: the size of a photon's punch

Where equation (3) really earns its keep is in showing how wildly a photon's energy varies across the electromagnetic spectrum, even though the formula never changes. A radio wave a metre long ( $\lambda = 1\ \mathrm{m} = 10^{9}\ \mathrm{nm}$ ) carries a laughably tiny $1240/10^{9} = 1.24\times10^{-6}\ \mathrm{eV}$ per photon — which is why a radio transmitter pours out photons by the trillion-trillion and none of them can so much as nudge a chemical bond. Go the other way to a gamma ray of wavelength $1\ \mathrm{pm} = 10^{-3}\ \mathrm{nm}$ and each photon carries $1240/10^{-3} = 1.24\times10^{6}\ \mathrm{eV} = 1.24\ \mathrm{MeV}$ — energetic enough to blast electrons clean out of atoms and snap DNA. Same equation, a trillion-fold spread in the punch per packet. The lesson worth carrying out of this chapter is that frequency is energy: the higher up the spectrum you go, the harder each individual photon hits, and that single fact organises everything from why ultraviolet burns your skin while radio does not, to why X-rays image bone.

Derivation APhoton momentum p = h over lambda from its energy

The goal is to show that a photon of energy $E = hf$ must also carry momentum $p = h/\lambda$ , using only algebra and two quoted facts — that a massless particle obeys $E = pc$ , and that any wave satisfies $c = f\lambda$ . This is the single most important derivation in the article, because momentum is what turns "light is grainy in energy" into "light collides like a particle," the idea Compton will need. No calculus appears; every step is a substitution.

Step 1. Start from the energy of the photon, Planck–Einstein equation (1): $E = hf$ . This is the quantum input — energy comes in a lump proportional to frequency, with $h$ the constant of proportionality.

Step 2. Bring in the momentum of a massless particle. Ordinary mechanics ties momentum to mass through $p = mv$ , but a photon has zero rest mass, so that road gives $p = 0$ and is a dead end. The correct relation for a massless particle, quoted from relativity, is $E = pc$ . (Maxwell's electromagnetism independently predicts that a wave carrying energy $E$ carries momentum $E/c$ , so the two pictures agree.) Rearranging, $p = E/c$ .

Step 3. Substitute the photon energy from Step 1 into $p = E/c$ : this gives $p = hf/c$ . Everything on the right is known once you know the frequency of the light.

Step 4. Finally use the wave relation $c = f\lambda$ , i.e. $f/c = 1/\lambda$ , to trade frequency for wavelength: $p = hf/c = h\,(f/c) = h/\lambda$ . So $p = h/\lambda = E/c$ . For $500\ \mathrm{nm}$ green light this is $p = (6.63\times10^{-34})/(500\times10^{-9}) = 1.33\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}$ , and the cross-check $E/c = (3.98\times10^{-19})/(3.00\times10^{8}) = 1.33\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}$ agrees exactly, as it must. Verified: $p(500\ \mathrm{nm}) = 1.33\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}$ by both roads ( $h/\lambda$ and $E/c$ ) — recomputed in verify_hs2-plancks-quantum-and-the-photon.py.

Go deeper · Derivation B · 1 minWhere the 1240 eV·nm shortcut comes from

Derivation BWhere the 1240 eV·nm shortcut comes from

The goal is to justify the exam shortcut $E[\mathrm{eV}] = 1240/\lambda[\mathrm{nm}]$ so you use it with understanding rather than as a magic number — and so you never misapply it to a wavelength that is not in nanometres. The whole trick is to compute the constant $hc$ once, in the mixed units of electronvolts and nanometres, and carry it forever. Only arithmetic is involved.

Step 1. Write the exact energy of a photon in SI units, from equation (1): $E = hc/\lambda$ . With $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ and $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ , the product $hc = 1.99\times10^{-25}\ \mathrm{J\,m}$ . This numerator is the same for every photon; only $\lambda$ in the denominator changes.

Step 2. Convert the numerator's energy unit from joules to electronvolts by dividing by $e = 1.60\times10^{-19}\ \mathrm{C}$ (since $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$ ): $hc = (1.99\times10^{-25})/(1.60\times10^{-19}) = 1.24\times10^{-6}\ \mathrm{eV\,m}$ .

Step 3. Convert the length unit from metres to nanometres. There are $10^{9}\ \mathrm{nm}$ in a metre, so $hc = 1.24\times10^{-6}\ \mathrm{eV\,m} \times 10^{9}\ \mathrm{nm/m} = 1240\ \mathrm{eV\,nm}$ . That is the whole origin of the number: it is Planck's constant times the speed of light, dressed in exam-friendly units.

Step 4. Therefore $E = hc/\lambda$ becomes $E[\mathrm{eV}] = 1240/\lambda[\mathrm{nm}]$ — but only when $\lambda$ is in nanometres and you want $E$ in electronvolts. Test it: $500\ \mathrm{nm}$ gives $1240/500 = 2.48\ \mathrm{eV}$ , matching the direct SI route ( $3.98\times10^{-19}\ \mathrm{J} \div 1.60\times10^{-19}\ \mathrm{J\,eV^{-1}} \approx 2.48\ \mathrm{eV}$ ). Verified: $hc = 1.24\times10^{-6}\ \mathrm{eV\,m} = 1240\ \mathrm{eV\,nm}$ , and $1240/500 = 2.48\ \mathrm{eV}$ — recomputed in verify_hs2-plancks-quantum-and-the-photon.py.

Limiting cases

case	result	meaning
low frequency (radio, $\lambda = 1\ \mathrm{m}$ )	$E = 1.24\times10^{-6}\ \mathrm{eV}$	each photon is feeble; brightness comes from sheer photon count
visible ( $400$ – $700\ \mathrm{nm}$ )	$E = 3.10$ down to $1.77\ \mathrm{eV}$	just enough to drive retinal chemistry, not to ionise
high frequency (gamma, $\lambda = 1\ \mathrm{pm}$ )	$E = 1.24\ \mathrm{MeV}$	one photon can ionise atoms and break bonds
double the frequency	energy and momentum both double	$E = hf$ and $p = h/\lambda$ are strictly linear in $f$
brighter beam, same colour	more photons per second, same $E$ each	intensity sets photon rate, frequency sets photon energy
classical wave limit ( $h \to 0$ )	quantum lumps vanish, spectrum diverges	the ultraviolet catastrophe returns — $h$ is what cures it

Common trapIntensity means more photons, not more energetic photons

The single most expensive misconception in this whole topic is to think that a brighter light delivers "stronger" photons. It does not. The energy of one photon is $E = hf$ — it depends on frequency (colour) and nothing else. Turn up the intensity of a monochromatic beam and you send more photons every second, but each one still carries exactly the same $hf$ . This is precisely why a dim ultraviolet lamp ejects photoelectrons while an intensely bright red lamp ejects none: below the threshold frequency, no single red photon has enough energy to free an electron, and stacking up more of them changes nothing, because photons are absorbed one at a time. Keep "brightness = how many, frequency = how much each" burned into memory and half the traps in this chapter disappear.

Worked examples

Board level — the full profile of a green photon (CBSE and ICSE pattern)

Monochromatic green light has a wavelength of $500\ \mathrm{nm}$ in vacuum. Taking $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ , $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$ and $e = 1.60\times10^{-19}\ \mathrm{C}$ , find (a) its frequency, (b) the energy of one photon in joules and in electronvolts, and (c) the momentum of one photon.

Step 1. Frequency from $c = f\lambda$ : $f = c/\lambda = (3.00\times10^{8})/(500\times10^{-9}) = \boxed{6.0\times10^{14}\ \mathrm{Hz}}$ . Green light sits near the middle of the visible band, as expected.

Step 2. Energy in joules from $E = hf$ : $E = (6.63\times10^{-34})(6.0\times10^{14}) = \boxed{3.98\times10^{-19}\ \mathrm{J}}$ . Equivalently $E = hc/\lambda$ gives the same value.

Step 3. Energy in electronvolts: divide by $e$ , or just use the shortcut $E = 1240/\lambda[\mathrm{nm}] = 1240/500 = \boxed{2.48\ \mathrm{eV}}$ . Both routes agree — the electronvolt is the natural unit here because atomic energies live around a few eV.

Step 4. Momentum from $p = h/\lambda$ : $p = (6.63\times10^{-34})/(500\times10^{-9}) = \boxed{1.33\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}}$ . Cross-check with $p = E/c = (3.98\times10^{-19})/(3.00\times10^{8}) = 1.33\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}$ — identical, confirming $p = h/\lambda = E/c$ .

Competitive level — how many photons a laser pointer fires each second (JEE and NEET pattern)

A green laser pointer of wavelength $500\ \mathrm{nm}$ has an output power of $1.0\ \mathrm{mW}$ . How many photons does it emit per second?

Step 1. Each photon carries $E = hc/\lambda = 3.98\times10^{-19}\ \mathrm{J}$ (from the board example above — the energy per photon is a fixed property of the colour, not of the power).

Step 2. "Power" means joules per second; "photons per second" is the total energy delivered each second divided by the energy of one photon: $N = P/E = (1.0\times10^{-3}\ \mathrm{W})/(3.98\times10^{-19}\ \mathrm{J}) = \boxed{2.5\times10^{15}\ \text{photons per second}}$ .

Step 3. Read the size of that number. A milliwatt pointer — feeble by everyday standards — spits out two-and-a-half thousand-trillion photons every second. This is exactly why light looks smooth and continuous: the graininess is real but the grains arrive in unimaginable floods. Quantisation only becomes visible in delicate experiments (single-photon detectors, the photoelectric threshold) where you catch light one packet at a time. Timing note: about 75 seconds. The only real work is one $E = hc/\lambda$ and one division $P/E$ ; do not convert to eV first — keep everything in joules and watts so the units cancel to a pure count.

Competitive level — the energy and momentum of an X-ray photon (JEE-Advanced pattern)

An X-ray used in crystallography has wavelength $0.10\ \mathrm{nm}$ (that is $1.0\ \text{ångström}$ ). Find its photon energy in keV and its momentum, and comment on how it compares with the green photon above.

Step 1. Energy via the shortcut: $E = 1240/\lambda[\mathrm{nm}] = 1240/0.10 = 12{,}400\ \mathrm{eV} = \boxed{12.4\ \mathrm{keV}}$ . That is about five thousand times the $2.48\ \mathrm{eV}$ of the green photon — same formula, vastly bigger punch, because the wavelength is five thousand times shorter.

Step 2. Momentum from $p = h/\lambda = (6.63\times10^{-34})/(0.10\times10^{-9}) = \boxed{6.63\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}}$ . Cross-check $p = E/c = (12{,}400\times1.60\times10^{-19})/(3.00\times10^{8}) = 6.6\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}$ — consistent.

Step 3. This momentum, five thousand times the green photon's, is large enough that when the X-ray bounces off a loosely bound electron it visibly transfers recoil — the wavelength-shift effect Compton measured in 1923, and the subject of the next article. Short wavelength means big momentum, and big momentum is what makes the particle nature of light impossible to ignore. Timing note: under a minute. Recognise "ångström = $0.1\ \mathrm{nm}$ " instantly, use the $1240$ shortcut for energy and $p = h/\lambda$ for momentum, and only convert to $E/c$ if asked to cross-check.

Exam traps

Use frequency for energy, not wavelength — unless you switch formulas. $E = hf$ rises with frequency, but $E = hc/\lambda$ falls with wavelength. Students who write " $E = h\lambda$ " or who think longer wavelength means more energy lose the mark instantly. Long wavelength = low frequency = low energy. Keep the two forms straight: $E = hf = hc/\lambda$ .

The 1240 shortcut only works in eV and nm. Equation (3) is $hc$ dressed in electronvolts and nanometres. Feed it a wavelength in metres or micrometres, or expect an answer in joules, and it is nonsense. If the wavelength is given in ångströms, convert first ( $1\ \text{Å} = 0.1\ \mathrm{nm}$ ).

Photon momentum is $h/\lambda$ , never $mv$ . A photon has zero rest mass, so $p = mv$ gives zero. Its momentum comes from $p = E/c = h/\lambda$ . Writing " $p = m_e v$ " for a photon is a classic slip that reveals you have not grasped what a photon is.

Brightness is not energy per photon. The warning above, restated because it is worth a whole question: intensity sets the number of photons per second; frequency sets the energy of each. A brighter beam of the same colour has identical-energy photons, just more of them.

Keep the blackbody story qualitative. No board at this level asks you to derive Planck's radiation law (it needs calculus we do not use here). You are expected to state why quantisation cures the ultraviolet catastrophe — big lumps $hf$ freeze out the high-frequency modes — not to integrate anything.

Board scope. The photon model, $E = hf = hc/\lambda$ and $p = h/\lambda$ , is examined on CBSE, ICSE, JEE (Main and Advanced), NEET, AP Physics 2 (Unit 15), IB Physics HL (Theme E), and A-Level (CIE 9702). Note it is IB HL, not Standard Level, and that IGCSE 0625 carries no dual-nature content at all — an IGCSE student never needs this. The quantitative Compton momentum transfer (next article) is beyond AP-2, not part of core CBSE/NEET, and is JEE-Advanced / IB-HL / 9702 depth.

Origins · 1 minMax Planck and how this was discovered

Max Planck

1858–1947 · Kiel

Max Planck was a cautious, deeply conservative German theorist — the last man anyone expected to start a revolution. Trained in classical thermodynamics, he spent the 1890s wrestling with the blackbody spectrum, and on 14 December 1900 presented the formula that fit it, built on the assumption that energy is exchanged in discrete quanta $E = hf$ . He regarded the quantum as a provisional trick and tried for years to reconcile it with classical physics; it was Einstein and Bohr who revealed how fundamental it was. Planck's constant $h$ now sits at the heart of all quantum physics and, since 2019, defines the kilogram itself. He won the 1918 Nobel Prize in Physics. His later life was marked by tragedy — a son executed by the Nazis for resisting Hitler — yet he stayed in Germany and worked to preserve its scientific life; the Max Planck Society is named in his honour.

History

1859

— Gustav Kirchhoff poses the blackbody problem, proving the emitted spectrum depends only on temperature.

1896

— Wilhelm Wien gives a distribution law that fits the short-wavelength side of the glow but fails at long wavelengths.

1900

— Max Planck introduces the energy quantum E = hf and reproduces the entire blackbody curve.

1905

— Albert Einstein proposes that light itself is quantised into photons, explaining the photoelectric effect.

1916

— Robert Millikan's precise measurements confirm Einstein's photon relation and pin down the value of h.

1926

— Gilbert N. Lewis coins the name photon for the light quantum.

Going further

Texts. NCERT Physics Class 12, Chapter 11 ("Dual Nature of Radiation and Matter"), the sections on Planck's quantum and the photon · H. C. Verma, Concepts of Physics Vol. 2, "Photoelectric Effect and Wave–Particle Duality" · Halliday, Resnick and Walker, the "Photons and Matter Waves" chapter.

Provenance. Standard results. Recompute: $f(500\ \mathrm{nm}) = 3.00\times10^{8}/500\times10^{-9} = 6.0\times10^{14}\ \mathrm{Hz}$ ; $E(500\ \mathrm{nm}) = hc/\lambda = 3.98\times10^{-19}\ \mathrm{J} = 1240/500 = 2.48\ \mathrm{eV}$ ; $p(500\ \mathrm{nm}) = h/\lambda = E/c = 1.33\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}$ ; $E(400) = 3.10\ \mathrm{eV}$ , $E(656) = 1.89\ \mathrm{eV}$ , $E(700) = 1.77\ \mathrm{eV}$ ; radio ( $1\ \mathrm{m}$ ) $= 1.24\times10^{-6}\ \mathrm{eV}$ , gamma ( $1\ \mathrm{pm}$ ) $= 1.24\ \mathrm{MeV}$ ; laser rate $N = 1.0\times10^{-3}/3.98\times10^{-19} = 2.5\times10^{15}\ \mathrm{s^{-1}}$ ; X-ray ( $0.10\ \mathrm{nm}$ ) $E = 12.4\ \mathrm{keV}$ , $p = 6.63\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}$ ; shortcut constant $hc = 1240\ \mathrm{eV\,nm}$ ; $h = 6.63\times10^{-34}\ \mathrm{J\,s} = 4.14\times10^{-15}\ \mathrm{eV\,s}$ .

Prerequisites. The Photoelectric Effect & Its Laws — the classical failure that this article's photon resolves. Leads to. Einstein's Photoelectric Equation (the very next module, where $E = hf$ becomes $KE_{\max} = hf - \varphi$ ), Dual Nature of Radiation & Compton Scattering (where $p = h/\lambda$ is put to the test), and de Broglie Matter Waves (where the same $p = h/\lambda$ is turned around and applied to electrons). For enrichment, the university seam Blackbody Radiation and the Planck Quantum derives the full radiation law with the calculus we quoted past here (go-deeper only, never an exam dependency).

What actually is a photon — a particle or a wave?

Neither picture is complete on its own, and that is the whole point of "dual nature." A photon is a quantum of the electromagnetic field: it travels at $c$ , has no rest mass, and is always absorbed or emitted as one indivisible packet of energy $hf$ . In experiments like interference and diffraction it behaves like a wave with wavelength $\lambda$ ; in experiments like the photoelectric effect and Compton scattering it behaves like a localised particle carrying energy $hf$ and momentum $h/\lambda$ . It is not that light is "sometimes a wave and sometimes a particle" by choice — it is a single kind of object whose full behaviour needs both descriptions, stitched together by the equations $E = hf$ and $p = h/\lambda$ that link the two.

If a photon has no mass, how can it carry momentum?

Because momentum is not always $mv$ — that formula is a low-speed special case for objects with mass. The general relation from relativity is $E^2 = (pc)^2 + (mc^2)^2$ , and for a massless particle ( $m = 0$ ) this collapses to $E = pc$ , so $p = E/c$ . A photon has energy, therefore it has momentum, even with zero rest mass. You can measure this momentum: light exerts a real pressure on surfaces (radiation pressure), and an X-ray photon visibly recoils an electron in Compton scattering. The everyday intuition "no mass means no momentum" simply does not survive at the quantum and relativistic scale — energy alone is enough to carry momentum $E/c$ .

Why did Planck have to introduce quanta at all?

Because classical physics made an absurd prediction for a glowing object. Treating the radiation as ordinary waves that share energy equally among all vibration modes, the intensity was predicted to climb without limit as the wavelength shortened — infinite energy in the ultraviolet, the "ultraviolet catastrophe." Real objects obviously do not blaze with infinite energy; their spectrum peaks and falls. Planck found that if each oscillator could only take up energy in whole lumps of $hf$ , the high-frequency modes became too "expensive" to excite at ordinary temperatures and were frozen out, bending the curve back down to match experiment exactly. Quantisation was the one assumption that fixed the spectrum, and no continuous-energy theory could do it.

What is the difference between Planck's quantum and Einstein's photon?

Planck (1900) quantised the oscillators in the walls of a blackbody — he said matter can only exchange energy with radiation in lumps of $hf$ , but he still thought of the light travelling through space as a continuous wave. Einstein (1905) made the bolder claim that the radiation itself is quantised: light propagates as a stream of localised packets, each an indivisible $hf$ , not just emitted and absorbed in lumps but existing as lumps. Planck's step explained the blackbody curve; Einstein's step explained the photoelectric effect and gave us the photon as a real object. Planck was uneasy with Einstein's leap for years. Both are needed: Planck introduced the quantum of energy, Einstein turned it into a particle of light.

Where does the number 1240 come from, and can I always trust it?

It is simply $hc$ written in mixed units of electronvolts and nanometres: $hc = 1.99\times10^{-25}\ \mathrm{J\,m}$ , divided by $e$ to get eV and multiplied by $10^{9}$ to get nm, comes to $1240\ \mathrm{eV\,nm}$ (more precisely $1243$ , but $1240$ is the standard exam value). So $E[\mathrm{eV}] = 1240/\lambda[\mathrm{nm}]$ is exact-ish to three figures — but only when the wavelength is in nanometres and you want the answer in electronvolts. Give it a wavelength in metres, micrometres, or ångströms and it fails; convert first. It is the fastest tool in this chapter, and worth memorising, but respect its units.

Does a photon lose energy or slow down when it enters glass?

Its energy is unchanged. Recall from wave optics that the source fixes the frequency and the boundary cannot change it, so $f$ — and therefore the photon's energy $E = hf$ — is the same in glass as in vacuum. What changes inside the glass is the speed ( $v = c/n$ ) and hence the wavelength ( $\lambda = \lambda_0/n$ ). This looks paradoxical next to $p = h/\lambda$ , which would suggest the momentum changes; the resolution is that momentum inside a medium is subtle and beyond this course, but the key exam fact is clean: a photon's frequency and energy are set by its source and survive refraction untouched. A red photon is still a $1.89\ \mathrm{eV}$ red photon under water.

Is E = hf the same as E = hc over lambda?

Yes — they are one equation written two ways, linked by the wave relation $c = f\lambda$ . If you know the frequency, use $E = hf$ ; if you know the wavelength (the more common exam case), substitute $f = c/\lambda$ to get $E = hc/\lambda$ . Both give identical energies. The wavelength form is usually more convenient because light is quoted by wavelength (500 nm, 656 nm), and it is the form the $1240$ shortcut comes from. Just never mix them up into " $E = h\lambda$ ," which is dimensionally wrong and one of the most common careless errors on this topic.