Photon properties — E = hf, p = h - λ — Physics notes

WHAT is a photon?

WHY does this idea even exist? Classical physics said a wave of any frequency could carry any amount of energy continuously. But experiments (blackbody radiation, photoelectric effect) showed energy is exchanged in fixed chunks proportional to frequency. Planck and Einstein resolved this by saying the energy itself comes packaged.

Deriving $E = hf$ from first principles

HOW we get here (the logic chain):

Planck's quantization (1900): to explain blackbody radiation, Planck assumed oscillators in a wall can only have energies that are integer multiples of a quantum: $E_n = nhf$ . The smallest exchangeable chunk is $hf$ .
Einstein's leap (1905): the light field itself is made of these chunks. One photon = one quantum = energy $hf$ .
Using the wave relation $c = f\lambda$ , substitute $f = c/\lambda$ : $E = hf = h\cdot\frac{c}{\lambda} = \frac{hc}{\lambda}.$

Deriving $p = h/\lambda$ from first principles

A photon has zero rest mass, so we cannot use $p = mv$ . We use the full relativistic energy–momentum relation:

HOW:

Set $m_0 = 0$ (photon has no rest mass): $E^2 = (pc)^2 \implies E = pc \implies p = \frac{E}{c}.$
Now plug in $E = hf$ : $p = \frac{hf}{c}.$
Use $c = f\lambda \Rightarrow f/c = 1/\lambda$ : $\boxed{p = \frac{h}{\lambda}}$

Figure — Photon properties — E = hf, p = h - λ

Worked Examples

Common Mistakes (Steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine light is delivered like coins, not water. Water you could pour in any amount; coins come in fixed values. A blue-light coin is worth more than a red-light coin — the "value" depends on the color (frequency), not how many coins you have. And even though each coin weighs nothing, when a whole stream of them hits a wall they still shove it a tiny bit — that shove is the photon's momentum. More coins (brighter light) = bigger total shove, but each coin's worth and push are fixed by its color.

Flashcards

What is a photon?

The quantum of the EM field — a massless particle moving at

c

with energy

hf

and momentum

h/\lambda

Formula for photon energy in terms of frequency?

E = hf

Formula for photon energy in terms of wavelength?

E = hc/\lambda

Why does photon energy depend on frequency and not amplitude?

A single photon has no "amplitude"; frequency is the only property distinguishing photons, so energy must scale with

f

. Amplitude = number of photons.

Derive

p = h/\lambda

From

E^2=(pc)^2+(m_0c^2)^2

with

m_0=0

get

E=pc

, so

p=E/c=hf/c=h/\lambda

Why can a massless photon have momentum?

Because

p=mv

is only non-relativistic; the true relation

E=pc

gives nonzero

p

even when

m_0=0

Relation linking

E

and

p

for a photon?

E = pc

If wavelength doubles, what happens to

E

and

p

Both halve (each

\propto 1/\lambda

Value of Planck's constant?

h = 6.626\times10^{-34}

J·s.

How do you find photons emitted per second by a laser?

Divide power by energy per photon:

N = P/(hc/\lambda)

Energy of a 500 nm photon in eV (approx)?

~2.48 eV.

Connections

Photoelectric Effect — uses $E=hf$ to explain threshold frequency.
Planck's Law and Blackbody Radiation — origin of $E=hf$ .
de Broglie Wavelength — runs $p=h/\lambda$ backwards for matter waves.
Compton Effect — direct proof photons carry momentum $h/\lambda$ .
Relativistic Energy-Momentum Relation — source of $E=pc$ .
Radiation Pressure — macroscopic consequence of photon momentum.
Wave-Particle Duality — the big picture this note lives inside.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, classical physics kehti thi ki light ek smooth wave hai jo energy continuously deti rehti hai — jaise paani ka dhaara. Lekin experiments ne dikhaya ki light energy chhote-chhote packets mein aati hai, jinhe hum photon kehte hain. Har photon ek lump hai jiski energy uski frequency par depend karti hai: $E = hf$ , ya wavelength ke terms mein $E = hc/\lambda$ . Yaad rakho — brightness ka matlab hai kitne photons, na ki har photon kitna powerful. Isiliye photoelectric effect mein frequency matter karti hai, intensity nahi.

Ab momentum ki baat. Tum sochoge "photon ka mass zero hai, toh $p=mv$ se momentum bhi zero hona chahiye!" Ye galti natural lagti hai, par $p=mv$ sirf slow speeds ke liye hai. Asli formula relativity ka hai: $E^2 = (pc)^2 + (m_0c^2)^2$ . Photon ka rest mass $m_0 = 0$ daalo, toh $E = pc$ mil jaata hai, yaani $p = E/c = h/\lambda$ . Matlab massless hone ke baad bhi photon ke paas momentum hai — isiliye solar sails aur comet tails ko light "push" karti hai.

Sabse important connection: $f$ aur $\lambda$ alag formulas nahi hain, dono $c = f\lambda$ se jude hue hain. Jo diya hai (frequency ya wavelength) uske hisaab se formula choose karo, dono ko ek saath mat use karna. Aur ek pyaara cross-check: $E = pc$ hamesha sach hai, toh agar tumne $E$ aur $p$ alag-alag nikaale, toh $E/p$ ko $c$ ke barabar hona chahiye. Agar match ho gaya, samajh lo answer sahi hai!

Photon properties — E = hf, p = h - λ

WHAT is a photon?

Deriving $E = hf$ from first principles

Deriving $p = h/\lambda$ from first principles

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Modern Physics

Connections