In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of $3.15 \times 10^{-18}\ J$ from the radiations of 600 nm, calculate the number of photons received by the detector.

In astronomical observations, signals observed from the distant stars are generally weak. If the photon detect

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3.15 \times 10^{-18}\ J

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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 total energy by the energy per photon gives the count: 9 photons.

The energy of electromagnetic radiation is quantized—it arrives in discrete packets called photons. When a detector collects light from a distant star, it doesn't receive a continuous stream of energy but rather individual photons, each carrying a fixed amount of energy determined solely by the wavelength (or frequency) of the light.

The energy of a single photon is given by the Planck relation:

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

where $h = 6.626 \times 10^{-34}\ \text{J·s}$ is Planck's constant, $c = 3 \times 10^8\ \text{m/s}$ is the speed of light, and $\lambda$ is the wavelength. Once we know the energy per photon, the total number of photons is simply the total energy divided by this single-photon energy.

Step-by-step calculation:

Convert the wavelength to metres.

The given wavelength is $600\ \text{nm}$ :

\lambda = 600 \times 10^{-9}\ \text{m} = 6 \times 10^{-7}\ \text{m}

Calculate the energy of one photon.

Substitute into the photon energy formula:

E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{6 \times 10^{-7}}

Simplify the numerator:

6.626 \times 3 = 19.878 \approx 19.88

E_{\text{photon}} = \frac{19.88 \times 10^{-26}}{6 \times 10^{-7}} = \frac{19.88}{6} \times 10^{-19}

E_{\text{photon}} = 3.313 \times 10^{-19}\ \text{J}

Find the number of photons.

The total energy received is $E_{\text{total}} = 3.15 \times 10^{-18}\ \text{J}$ . The number of photons is: …

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Q.In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15×10−18 J3.15 \times 10^{-18}\ J3.15×10−18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

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Q.In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of $3.15 \times 10^{-18}\ J$ from the radiations of 600 nm, calculate the number of photons received by the detector.

Unlock everything free for 14 days