A 25 watt bulb emits monochromatic yellow light of wavelength of $0.57\ \mu m$. Calculate the rate of emission of quanta per second.

A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μ m. Calculate the rate of emission of qu

Exercises · 2.11

0.57\ \mu m

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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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Each photon carries energy $E = \frac{hc}{\lambda}$ ; dividing the bulb's power by the energy per photon gives the photon emission rate. The bulb emits $7.16 \times 10^{19}$ photons per second.

The power rating of a bulb tells us the total energy it radiates per second. When that radiation is monochromatic—meaning all photons have the same wavelength—we can find exactly how many photons are needed to account for that power output. The bridge between the macroscopic world (watts) and the quantum world (individual photons) is Planck's relation connecting a photon's energy to its wavelength.

A photon of wavelength $\lambda$ carries energy

E = \frac{hc}{\lambda}

where $h = 6.626 \times 10^{-34}\ \text{J·s}$ is Planck's constant and $c = 3 \times 10^8\ \text{m/s}$ is the speed of light. If the bulb emits $n$ such photons every second, the total power is simply $P = nE$ . Rearranging gives us the emission rate.

Step-by-step calculation:

Convert the wavelength to metres.

The given wavelength is $0.57\ \mu\text{m} = 0.57 \times 10^{-6}\ \text{m} = 5.7 \times 10^{-7}\ \text{m}$ .
Calculate the energy of one photon.

Substitute into the photon energy formula:

E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{5.7 \times 10^{-7}}

The numerator is:

6.626 \times 3 = 19.878 \quad \Rightarrow \quad 19.878 \times 10^{-34+8} = 19.878 \times 10^{-26}\ \text{J}

Dividing by $5.7 \times 10^{-7}$ :

E = \frac{19.878}{5.7} \times 10^{-26+7} = 3.487 \times 10^{-19}\ \text{J}

Find the number of photons emitted per second. Power is energy per unit time, so if $n$ photons are emitted each second: …

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Q.A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 μm0.57\ \mu m0.57 μm. Calculate the rate of emission of quanta per second.

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Q.A 25 watt bulb emits monochromatic yellow light of wavelength of $0.57\ \mu m$ . Calculate the rate of emission of quanta per second.

Unlock everything free for 14 days