What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Exercises · 2.8

Q. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

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Concept understanding — Photon Energy Calculation

Photon Energy Calculation

The Intuition First

Imagine you're holding a rope tied to a wall. If you flick your wrist once, a single pulse travels down the rope. If you flick faster — more frequently — each pulse carries more energy; the rope vibrates more violently. Light behaves the same way. A photon is the smallest possible "flick" of the electromagnetic field — a single, indivisible packet of light energy.

What determines how much energy that one photon carries? Two things: how fast the wave is oscillating (its frequency), and a universal constant that connects the wave world to the particle world.

The Precise Statement

The energy $E$ of a single photon is directly proportional to its frequency $f$ (or inversely proportional to its wavelength $\lambda$ ). The proportionality constant is Planck's constant $h$ , one of the most fundamental numbers in physics.

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

Where:

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

Why Two Forms?

The first form $E = hf$ is the most direct: higher frequency means higher energy. The second form $E = hc/\lambda$ is often more practical because wavelength is easier to measure than frequency. Since $c = f\lambda$ , you can always swap between them.

What This Tells You

Blue light (short wavelength, high frequency) has more energy per photon than red light (long wavelength, low frequency).
Gamma rays have enormous photon energies; radio waves have tiny photon energies.
The energy is quantised — you cannot have half a photon. Either the full energy $hf$ is absorbed/emitted, or none at all.

A Worked Example

Question: Calculate the energy of a single photon of violet light with wavelength $400 \ \text{nm}$ .

Step 1: Convert wavelength to metres.

400 \ \text{nm} = 400 \times 10^{-9} \ \text{m} = 4.00 \times 10^{-7} \ \text{m}

Step 2: Use $E = hc/\lambda$ .

E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.00 \times 10^{-7}}

Step 3: Compute.

E = \frac{1.9878 \times 10^{-25}}{4.00 \times 10^{-7}} = 4.97 \times 10^{-19} \ \text{J}

Note

This is an incredibly tiny amount of energy — about $5 \times 10^{-19}$ joules. That's why we often use electronvolts (eV) for photon energies in atomic physics. $1 \ \text{eV} = 1.602 \times 10^{-19} \ \text{J}$ , so this photon has about $3.1 \ \text{eV}$ .

Common Mistake to Avoid …

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Use $E = \frac{hc}{\lambda}$ to find the energy of one photon, then divide 1 J by that value to count how many photons are needed. The answer is approximately $\boxed{2.01 \times 10^{16}}$ photons.

The energy of a single photon depends entirely on its wavelength (or equivalently, its frequency). Shorter wavelengths mean higher energy photons; longer wavelengths mean lower energy. Once we know the energy carried by one photon, counting how many are needed to deliver a macroscopic amount of energy like 1 J becomes straightforward division.

The relationship connecting photon energy to wavelength is:

E_{\text{photon}} = \frac{hc}{\lambda}

where $h = 6.626 \times 10^{-34} \, \text{J·s}$ is Planck's constant, $c = 3.00 \times 10^8 \, \text{m/s}$ is the speed of light, and $\lambda$ is the wavelength.

Step-by-step calculation

1. Convert the wavelength to metres

The given wavelength is 4000 pm (picometres). Since $1 \, \text{pm} = 10^{-12} \, \text{m}$ :

\lambda = 4000 \times 10^{-12} \, \text{m} = 4.000 \times 10^{-9} \, \text{m} = 4.0 \, \text{nm}

This is in the extreme ultraviolet range, quite energetic for individual photons.

2. Calculate the energy of one photon

Substitute into the photon energy formula:

E_{\text{photon}} = \frac{(6.626 \times 10^{-34} \, \text{J·s})(3.00 \times 10^8 \, \text{m/s})}{4.000 \times 10^{-9} \, \text{m}}

Numerator:

6.626 \times 3.00 = 19.878 \quad \Rightarrow \quad 19.878 \times 10^{-34+8} = 19.878 \times 10^{-26} \, \text{J·m}

Now divide by $4.000 \times 10^{-9} \, \text{m}$ :

E_{\text{photon}} = \frac{19.878 \times 10^{-26}}{4.000 \times 10^{-9}} = 4.9695 \times 10^{-17} \, \text{J}

Rounding sensibly: …

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Q.What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Step-by-step calculation

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