Photon properties — E = hf, p = h - λ: Question bank

Before we begin, every symbol used below in one place, so nothing is assumed:

True or false — justify

Doubling a photon's frequency doubles its energy.

True.

E = h f

is a straight proportionality —

E

and

f

rise together in exact lockstep, so twice the frequency is exactly twice the energy.

Doubling a photon's wavelength doubles its momentum.

False.

p = h / λ

means

p \propto 1/ λ

; doubling

λ

halves

p

. Momentum shrinks for longer (redder) waves, it does not grow.

A brighter beam of the same colour has more energetic photons.

False. Same colour = same

f

= same energy per photon. "Brighter" just means more photons per second; each one still carries exactly

h f

Two photons of the same wavelength always carry the same momentum.

True. Momentum depends only on

λ

through

p = h / λ

; identical wavelength forces identical momentum, regardless of the source or brightness.

A photon can be slowed below

c

in vacuum by making it less energetic.

False. In vacuum every photon moves at exactly

c

regardless of energy. Lowering energy lowers

f

(redder light), not speed — speed is fixed by relativity.

Since a photon is massless, its momentum is zero.

False.

p = m v

is only the low-speed limit. The real relation

E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}

with

m_{0} = 0

gives

p = E / c \neq = 0

. Massless ≠ momentumless.

E = h f

and

E = h c / λ

are two independent facts about a photon.

False. They are the same equation, connected by

c = f λ

. Substituting

f = c / λ

turns one into the other; you never use both at once.

For a photon,

E = p c

holds exactly, with no correction term.

True. The rest-mass term

(m_{0} c^{2})^{2}

vanishes because

m_{0} = 0

, leaving

E = p c

with no leftover, unlike a massive particle where the term stays.

A photon's energy in eV and in joules describe two different physical quantities.

False. They are the same energy in two units; you convert by dividing joules by

1.6 \times 1 0^{- 19}

. eV is just a human-friendly ruler for tiny energies.

Spot the error

p = m v

, and

m = 0

for a photon, so a photon exerts no force on a solar sail."

The error is using

p = m v

outside its valid low-speed regime. The correct

p = h / λ

is nonzero, which is exactly why solar sails and comet tails feel light's push (radiation pressure).

"To find photons per second I multiply power by energy per photon:

N = P \times (h c / λ)

The error is multiplying instead of dividing. Power = (energy per photon) × (photons per second), so

N = P \div (h c / λ)

. Multiplying gives units of W·J, which is meaningless here.

"A 500 nm photon has

E = h f = h \times 500 \times 1 0^{- 9}

joules."

The error is plugging wavelength into the frequency slot.

f

is in Hz, not metres. Either use

f = c / λ

first, or switch to the

E = h c / λ

form directly.

"Red light photons and blue light photons have the same energy because both are just light."

The error ignores that energy scales with frequency. Blue has higher

f

than red, so each blue photon carries more energy — "just light" hides a real per-photon difference.

"Since

E = h f

and momentum needs mass, momentum must be

E

times mass."

Two errors: momentum does not need mass, and the invented formula is dimensionally wrong. The right link is

p = E / c

, momentum equals energy divided by the speed of light.

"If I make a laser twice as powerful at the same wavelength, each photon becomes twice as energetic."

The error is confusing total power with per-photon energy. Same

λ

fixes each photon's energy; twice the power means twice as many photons per second, not fatter ones.

Why questions

Why does a single photon's energy depend on frequency and not on amplitude?

A lone photon has no meaningful amplitude — you cannot have "half a photon" or a "louder" one. Frequency is the only knob distinguishing a red lump from a blue lump, so energy must ride on

f

; amplitude survives only as photon count.

Why do we reach for the relativistic relation

E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}

instead of

p = m v

for a photon?

Because

p = m v

is a non-relativistic approximation and dies when

m_{0} = 0

, giving the false answer "zero." The relativistic relation is the general truth and correctly yields

p = E / c

for a massless particle moving at

c

Why does the photoelectric effect care about the light's frequency but not its brightness?

Ejecting an electron needs a single photon of enough energy

h f

to beat the metal's binding. Brightness adds more photons but not more energy each, so below the threshold frequency no amount of brightness helps.

Why can we substitute

f = c / λ

so freely between the energy formulas?

Because all light in vacuum obeys the wave relation

c = f λ

, locking

f

and

λ

into a fixed reciprocal pair. Knowing one gives the other, so the two energy forms are interchangeable, never independent.

Why does light exert pressure on a surface it hits?

Each photon carries momentum

p = h / λ

; absorbing or reflecting it transfers that momentum to the surface. A stream of photons delivering many tiny pushes per second adds up to a steady force — radiation pressure.

Why is Planck's constant

h

so central to both

E

and

p

h

is the universal conversion rate between a wave's frequency/wavelength and the particle-like energy/momentum it carries. It is the single "exchange rate" that turns wave language into quantum-lump language in both formulas.

Edge cases

What is the momentum of a photon as its wavelength grows without bound (very low frequency radio)?

λ \to \infty

p = h / λ \to 0

and

E = h c / λ \to 0

. Such a photon carries vanishingly little energy and push, though it still exists and still moves at

c

Can a photon have exactly zero energy?

No. That would require

f = 0

λ = \infty

, which is no oscillation at all — not a photon. Every real photon has

f > 0

, hence strictly positive

E

and

p

What happens to a photon's energy and momentum if you view it from a receding frame (redshift)?

Both drop together. The frequency shifts down, so

E = h f

falls, and since

p = E / c

the momentum falls in exact proportion — energy and momentum stay locked by

E = p c

in every frame.

Is there any photon speed other than

c

in vacuum?

No. Massless particles are forced to travel at exactly

c

in vacuum in all inertial frames; there is no rest frame for a photon, so "a stationary photon" is a contradiction.

If two photons combine (e.g. in absorption), do their energies simply add?

Their energies and momenta each add as conserved quantities, but the result generally is not another single photon — energy and momentum conservation together forbid one massless lump from carrying the combined amounts unless they travel identically. See Compton Effect and Relativistic Energy-Momentum Relation for the bookkeeping.

What distinguishes running

p = h / λ

"forwards" for a photon versus "backwards" for an electron?

For a photon we start with light's

λ

and get its momentum. For matter, de Broglie Wavelength runs it the other way — given a particle's momentum

p

, assign it a wave of

λ = h / p

. Same constant, opposite direction.

Recall One-line self-test before you close this page

Cover everything: state why brightness changes photon count not photon energy, and why a massless photon still shoves a wall. Answer ::: Energy per photon is fixed by frequency ( $E = h f$ ), so more brightness just means more photons; and momentum comes from $p = E / c$ (not $p = m v$ ), which stays nonzero even when $m_{0} = 0$ , so each photon delivers a real push.

Connections

Photoelectric Effect — the frequency-not-brightness trap made experimental.
Compton Effect — the "massless means no momentum" trap disproved directly.
Relativistic Energy-Momentum Relation — home of the $E = p c$ reasoning.
de Broglie Wavelength — the $p = h / λ$ run backwards for matter.
Radiation Pressure — the edge case of light pushing surfaces.
Wave-Particle Duality — the big picture these traps live inside.