Estimating the following two numbers should be interesting. The first number will tell you why radio engineers

Additional Exercises · 11.25

Q. Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never 'count photons', even in barely detectable light.

(a) The number of photons emitted per second by a Medium wave transmitter of

10\ \text{kW}

power, emitting radiowaves of wavelength

500\ \text{m}

(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (

\sim 10^{-10}\ \text{W m}^{-2}

). Take the area of the pupil to be about

0.4\ \text{cm}^2

, and the average frequency of white light to be about

6 \times 10^{14}\ \text{Hz}

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Concept understanding — Photon Energy

Photon Energy: The Energy Carried by Light

Light does not carry its energy in a continuous stream. It comes in discrete packets — like individual drops instead of a steady hose. Each packet is called a photon: the smallest possible unit of light of a given frequency. You cannot have half a photon; it is all or nothing.

The Intuition: Colour Determines Energy

Red light and blue light are both light, but they behave differently — blue light causes sunburn and drives chemical reactions more readily than red. The reason is that the colour (frequency) of light directly fixes how much energy each photon carries, and blue photons carry more energy than red photons.

The Planck–Einstein Relation

The energy $E$ of a single photon is directly proportional to its frequency $f$ , and inversely proportional to its wavelength $\lambda$ :

E = h f = \frac{h c}{\lambda}

Where:

$E$ = energy of one photon (joules, J)
$f$ = frequency (hertz, Hz)
$\lambda$ = wavelength (metres, m)
$c$ = speed of light in vacuum ( $3.00 \times 10^8$ m/s)
$h$ = Planck's constant ( $6.626 \times 10^{-34}$ J·s)

Important

Planck's constant $h$ is extremely tiny, so an individual photon carries a very small amount of energy — which is why we never feel single photons striking our skin.

What This Tells You

Higher frequency = higher energy. Gamma rays have extremely energetic photons; radio waves have very low-energy photons.
Shorter wavelength = higher energy. Ultraviolet photons are more energetic than visible-light photons; infrared photons are less.
Energy is quantised. Light energy comes in fixed packets — you can have 1, 2, or 1000 photons, but never 0.5. This was Max Planck's revolutionary idea of 1900.

This is why UV light causes sunburn (UV photons carry enough energy to damage DNA, while visible photons do not), why X-ray photons are energetic enough to pass through soft tissue but get absorbed by bone, and why a photon needs enough energy to actually break a bond or drive a photochemical/biological reaction, rather than merely wavelength-dependent intensity.

A Concrete Example

Note

Question: Which has more energy — a photon of red light ( $\lambda = 700$ nm) or a photon of blue light ( $\lambda = 450$ nm)?

Since $E = \dfrac{hc}{\lambda}$ , energy is inversely proportional to wavelength, so the shorter-wavelength blue photon carries more energy:

Red: $E_{\text{red}} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{700 \times 10^{-9}} \approx 2.84 \times 10^{-19}$ J …

Why this formula?

Why Photon Energy is $E = hf$ — The Reasoning Behind the Formula

The formula $E = hf$ is not something you memorise and plug numbers into. It comes from a deep shift in how physicists understood light.

The problem that forced the formula

By the late 1800s, classical physics said light was a wave — continuous, carrying energy in proportion to its intensity (brightness). But blackbody radiation and the photoelectric effect showed something bizarre: when you shine light on a metal, electrons are ejected only if the light's frequency is above a certain threshold, no matter how bright the light is. Below that frequency, even the brightest lamp won't kick out a single electron. Classically this made no sense — a wave's energy depends on amplitude, not frequency.

Einstein's radical idea (1905)

Einstein proposed that light comes in discrete packets — photons — each carrying a fixed energy that depends only on its frequency, with Planck's constant $h$ as the proportionality:

A single photon's energy is $E = hf$ , where $f$ is the frequency of the light.

This directly explains the photoelectric effect: an electron needs a minimum energy (the work function $\phi$ ) to escape. If a photon's energy $hf < \phi$ , no electron is ejected — regardless of how many photons you send, because each photon interacts with one electron individually.

Important

$E = hf$ is not derived from deeper principles — it is a postulate justified by the experiments it explains, and it unified optics with quantum mechanics.

The equivalent forms

Since $c = f\lambda$ for light in vacuum, the same idea in terms of wavelength is:

E = \frac{hc}{\lambda}

Use this when you're given $\lambda$ instead of $f$ . A photon also carries momentum (despite having no mass): …

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In both parts, find the single-photon energy from the given wavelength/frequency, then divide the total power reaching/leaving the surface by it. The radio transmitter emits an enormous ~2.5×10^31 photons/s (so individual photon quantisation is utterly negligible for radio engineers), while the faintest visible light our eye can detect corresponds to only ~10^4 photons/s (still too many per second for the eye to "count" individually, but the smallness of the number compared to (a) is the point).

Medium-wave transmitter. Energy per photon at $\lambda=500\ \text{m}$ : $E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{500} = 3.978\times10^{-28}\ \text{J}$ Photon emission rate for 10 kW = $10^4$ W of power: $N = \frac{P}{E} = \frac{10^{4}}{3.978\times10^{-28}} \approx 2.51\times10^{31}\ \text{photons/s}$ This is such a vast number that the discreteness of individual radio photons is completely undetectable — radio waves behave essentially as a continuous classical wave for all practical (engineering) purposes.
Faintest perceivable light entering the eye. Power entering the pupil: $P = I \times A = (10^{-10}\ \text{W/m}^2)(0.4\times10^{-4}\ \text{m}^2) = 4\times10^{-15}\ \text{W}$ Energy per photon at $f=6\times10^{14}\ \text{Hz}$ : …

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Q.Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never 'count photons', even in barely detectable light.

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