CLASSICAL-MECHANICS

How Many Photons Weigh a Kilogram?

IKey Formulas

Formula	Name	Meaning
$E = mc^2$	Mass–energy equivalence	Energy locked in mass
$E = hf$	Planck relation	Energy of one photon of frequency $f$
$m = \dfrac{hf}{c^2}$	Effective photon mass	"Weight" of a single photon
$N = \dfrac{1}{m} = \dfrac{c^2}{hf}$	Photon count	How many photons make 1 kg

Constants: $h = 6.62607015\times10^{-34}\ \text{J}\cdot\text{s}$ , $\ c = 2.99792458\times10^{8}\ \text{m/s}$ , $\ \Delta\nu_{Cs} = 9\,192\,631\,770\ \text{Hz}$

IIInteractive Visualizations

Visualization 1 — Effective mass of one photon

A photon has zero rest mass, yet $m = hf/c^2$ still assigns it an effective mass that climbs in step with frequency.

Frequency log₁₀(f / Hz) 9.5

Visualization 2 — How many photons weigh a kilogram?

Guess the count before you compute it: $N = c^2/(hf)$ turns one kilogram into a single staggering number.

Your guess 10³³

Visualization 3 — Mass and count across the spectrum

From radio waves to X-rays a photon's effective mass spans eleven orders of magnitude — and its per-kilogram count runs the opposite way.

Photon type

💡 Whichever photon you pick, the fixed constants $h$ and $c$ plus a measured frequency always rebuild exactly the same kilogram.

IIIQuiz Questions

Question 1

Starting from $E = mc^2$ and $E = hf$ , which expression gives the effective mass of a single photon of frequency $f$ ?

✅ Correct! Setting $mc^2 = hf$ gives $m = hf/c^2$ .

❌ Not quite. Start from $mc^2 = hf$ and divide both sides by $c^2$ .

Show solution

Solution:

Set the two energy expressions equal: $mc^2 = hf$

Solve for the effective mass: $m = \frac{hf}{c^2}$

This is the bridge between Planck's relation and mass–energy equivalence.

Question 2

True or False: Because we can assign a photon an effective mass $m = hf/c^2$ , this proves that photons actually have a nonzero rest mass.

✅ Correct! Photon rest mass is zero; the effective mass is just an energy-equivalence bridge.

❌ Not quite. Effective mass comes from energy, not rest mass — photons stay massless at rest.

Show solution

Solution:

A photon's rest mass is exactly zero. The quantity $m = hf/c^2$ is an effective mass — it comes from converting the photon's energy into a mass-equivalent through $E = mc^2$ . It is a measurement bridge, not a claim that light weighs something. So the statement is False.

Question 3

A single cesium-frequency photon has an effective mass of about $6.78\times10^{-41}$ kg. Approximately how many such photons add up to one kilogram?

✅ Correct! $N = 1/m \approx 1.48\times10^{40}$ photons.

❌ Not quite. To count photons, take the reciprocal of the single-photon mass: $N = 1/m$ .

Show solution

Solution:

The number of photons is the reciprocal of the single-photon mass: $N = \frac{1}{m} = \frac{1}{6.78\times10^{-41}\ \text{kg}} \approx 1.48\times10^{40}$

So roughly $1.48\times10^{40}$ cesium photons make one kilogram.

Question 4

Since the 2018 redefinition, what defines the kilogram in the SI system?

✅ Correct! The kilogram is now defined through a fixed value of $h$ .

❌ Not quite. The old cylinder was retired in 2018 — the kilogram now rests on the Planck constant.

Show solution

Solution:

On November 18, 2018, the kilogram was redefined by fixing the Planck constant to the exact value $h = 6.62607015\times10^{-34}\ \text{J}\cdot\text{s}.$

Because $1\ \text{J}\cdot\text{s} = 1\ \text{kg}\cdot\text{m}^2\cdot\text{s}^{-1}$ , and the meter (via $c$ ) and the second (via $\Delta\nu_{Cs}$ ) are already fixed, fixing $h$ pins down the kilogram. No physical artifact is needed anywhere — a lab on Mars could reconstruct the kilogram from the same constants.

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