Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 10²

Exercises · 2.46

5.6 \times 10^{24}

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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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The power of a laser is the total energy delivered per second. For a pulsed laser, we find the energy of a single photon using $E = hc/\lambda$ , multiply by the number of photons to get total energy, then divide by the pulse duration (assumed 1 second here, as no time is given — the problem implies a per-second rate). The result is approximately $3.3 \times 10^6$ W.

The key idea here is that a laser's power isn't directly about the wavelength — it's about how much energy the laser dumps per unit time. Each photon carries a tiny, fixed amount of energy determined by its wavelength. If you know how many photons are emitted, you know the total energy. If you also know the time over which they're emitted, you get power.

But there's a subtlety: the problem gives you the number of photons ( $5.6 \times 10^{24}$ ) but does not give a time interval. In standard exam problems of this type, that number is assumed to be the number of photons emitted per second — otherwise you cannot calculate power. We'll proceed on that assumption.

Find the energy of a single photon. The energy of one photon is given by the Planck-Einstein relation:

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

where $h = 6.626 \times 10^{-34} \ \text{J·s}$ , $c = 3.0 \times 10^8 \ \text{m/s}$ , and $\lambda = 337.1 \ \text{nm} = 337.1 \times 10^{-9} \ \text{m}$ .

Plug in:

E_{\text{photon}} = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{337.1 \times 10^{-9}}

First compute numerator: $6.626 \times 3.0 = 19.878$ , and $10^{-34} \times 10^8 = 10^{-26}$ , so numerator = $19.878 \times 10^{-26} \ \text{J·m}$ .

Divide by $337.1 \times 10^{-9}$ :

E_{\text{photon}} = \frac{19.878 \times 10^{-26}}{337.1 \times 10^{-9}} = \frac{19.878}{337.1} \times 10^{-17}

$\frac{19.878}{337.1} \approx 0.05898$ , so:

E_{\text{photon}} \approx 5.898 \times 10^{-19} \ \text{J}

Tip

A quick sanity check: visible light photons have energies around $10^{-19}$ J. 337 nm is near-UV, so slightly higher energy — this checks out.

Find the total energy emitted per second. If $5.6 \times 10^{24}$ photons are emitted each second, the total energy per second (i.e., power) is: …

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Q.Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6×10245.6 \times 10^{24}5.6×1024, calculate the power of this laser.

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Q.Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is $5.6 \times 10^{24}$ , calculate the power of this laser.

Unlock everything free for 14 days