How to Calculate Light Power from Photon Emission Rate

An orange LED emits \(2.0\times10^{18}\) photons every second. Each photon has a wavelength of \(600\ \text{nm}\). What is the power of the light?

The quickest way to solve this is to calculate the energy of one photon, then multiply by the number of photons emitted each second.

The key equation is:

\[
\boxed{P=R\frac{hc}{\lambda}}
\]

where \(P\) is light power in watts, \(R\) is the photon emission rate in photons per second, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength in metres.

For the LED:

\[
P=\frac{(2.0\times10^{18})(6.63\times10^{-34})(3.00\times10^8)}
{600\times10^{-9}}
\]

\[
\boxed{P=0.66\ \text{W}}
\]

The LED emits approximately \(0.66\) joules of light energy every second.

The important part is that two separate quantities are involved: how much energy each photon carries, and how many photons are emitted each second. Keep those quantities separate, and these questions become much easier.

[IMAGE: An annotated comparison of two monochromatic light beams, one blue at 450 nm and one near-infrared at 900 nm. Show equal numbers of photons passing a reference point each second, with individual photon packets represented distinctly. Label the shorter-wavelength photons as higher energy and show that the blue beam therefore carries twice the optical power at the same photon emission rate.]

01Why the calculation works

Think of light as a stream of energy packets. Each photon carries a particular amount of energy.

If you know the energy carried by one photon and how many photons pass each second, you can calculate the total energy transferred each second.

This is exactly what power measures.

There are two separate calculations.

1. Calculate the energy of one photon

The energy of a photon depends on its frequency:

\[ E_{\text{photon}}=hf \]

where:

• \(E_{\text{photon}}\) is the energy of one photon, in joules (J).
• \(h\) is Planck’s constant, \(6.63\times10^{-34}\ \text{J s}\).
• \(f\) is the frequency of the electromagnetic wave, in hertz (Hz).

If the question gives you wavelength rather than frequency, use the wave equation:

\[ c=f\lambda \]

Rearranging:

\[ f=\frac{c}{\lambda} \]

Substituting into the photon energy equation gives:

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

Here, \(c=3.00\times10^8\ \text{m s}^{-1}\) is the speed of light in a vacuum, and \(\lambda\) is the wavelength in metres.

This gives you the energy carried by one photon.

For a useful explanation of why this relationship works, see Photon energy and atomic transitions: HSC Physics guide.

2. Multiply by the photon emission rate

Now consider how many photons are emitted each second.

The photon emission rate is:

\[ R=\frac{N}{t} \]

where:

• \(R\) is the photon emission rate, in photons per second.
• \(N\) is the total number of photons emitted.
• \(t\) is the time taken, in seconds.

Power is energy transferred per unit time:

\[ P=\frac{E_{\text{total}}}{t} \]

If every photon has the same energy, the total energy is simply the number of photons multiplied by the energy of one photon.

Therefore:

\[ P=\frac{NE_{\text{photon}}}{t} \]

Since \(N/t=R\):

\[ P=RE_{\text{photon}} \]

And substituting the photon energy equation:

\[ \boxed{P=\frac{Rhc}{\lambda}} \]

This is the equation to remember when you are given a photon emission rate and wavelength.

Check your units

The units provide a useful check:

\[ P=RE_{\text{photon}} \]

\[ P=\text{s}^{-1}\times\text{J} \]

\[ P=\text{J s}^{-1}=\text{W} \]

Remember to convert nanometres into metres before substituting:

\[ 1\ \text{nm}=1.0\times10^{-9}\ \text{m} \]

For example, \(450\ \text{nm}=4.50\times10^{-7}\ \text{m}\).

02Why wavelength changes the power

Here is a useful prediction to make.

Imagine two light sources emitting exactly \(1.0\times10^{18}\) photons each second.

One emits blue light at \(450\ \text{nm}\). The other emits near-infrared light at \(900\ \text{nm}\).

Which source has greater power?

The blue source does. Its photons have shorter wavelengths, so each photon carries twice as much energy.

Since both sources emit the same number of photons per second, the blue source transfers twice as much energy each second.

Mathematically:

\[ P\propto\frac{1}{\lambda} \]

when the photon emission rate is constant.

There is an important contrasting situation.

Suppose the two sources have the same optical power instead. The infrared source must emit twice as many photons each second to compensate for the lower energy carried by each photon.

So:

This distinction is a common source of mistakes in HSC questions.

Wavelengths and their applications

Different wavelengths are useful for different purposes because photons carry different amounts of energy.

For visible light, a violet or blue photon carries more energy than a red photon.

This does not automatically mean a blue light source has greater power. Its photon emission rate must also be considered.

03How would you measure the photon emission rate?

So far, we have assumed that someone has already given us the photon emission rate.

But how could an experimenter measure it?

Consider a monochromatic laser with a known wavelength. Place the laser in a dark enclosure and direct its beam towards a photon-counting detector.

The detector is connected to an electronic counter, which records the number of detection pulses over a measured time interval.

The measurement process follows this sequence:

1. The laser emits photons at a particular wavelength.
2. The photons travel along the beam towards the detector.
3. Each detected photon produces an electrical pulse.
4. The electronic counter records the pulses.
5. A timer measures the interval over which the pulses are counted.

What should the experiment show?

If the laser emits photons at a constant rate, doubling the measurement time should approximately double the number of detected photons.

For example, suppose an ideal detector records \(2.4\times10^6\) photon pulses in \(0.80\ \text{s}\).

The emission rate is:

\[ R=\frac{N}{t} \]

\[ R=\frac{2.4\times10^6}{0.80} \]

\[ R=3.0\times10^6\ \text{photons s}^{-1} \]

If the laser wavelength is \(500\ \text{nm}\), the energy of each photon is:

\[ E_{\text{photon}}= \frac{(6.63\times10^{-34})(3.00\times10^8)} {500\times10^{-9}} \]

\[ E_{\text{photon}}=3.98\times10^{-19}\ \text{J} \]

Therefore:

\[ P=(3.0\times10^6)(3.98\times10^{-19}) \]

\[ P=1.19\times10^{-12}\ \text{W} \]

The source emits approximately \(1.19\) picowatts of optical power.

These are illustrative measurement values, not experimental results.

Why the first measurement design can be misleading

There is a limitation in this setup.

A detector might not capture every photon emitted by the source. It might only intercept part of the beam, and it might fail to register some photons that reach it.

Consequently, the number of recorded pulses could be lower than the actual number of photons emitted.

A more accurate experiment calibrates two quantities:

• The collection fraction, \(g\), which is the fraction of emitted photons that reach the detector.
• The detection efficiency, \(\eta\), which is the fraction of arriving photons that produce a recorded pulse.

The corrected emission rate is:

\[ R_{\text{source}}=\frac{C}{g\eta t} \]

where \(C\) is the recorded number of pulses.

For a source emitting light in many directions, an integrating sphere or a calibrated collection arrangement can help account for photons travelling away from the detector.

Real detectors also have operating limits. At sufficiently high counting rates, they can miss closely spaced events.

The important experimental distinction is that a measured detection rate is not automatically the same as the source’s total emission rate.

04A reliable method for HSC questions

Before calculating, identify exactly what information the question gives you.

A few important decisions will keep your working accurate.

• If the question gives photons per second, you already have a rate. Do not divide by time again.
• If the question gives a total photon count over a time interval, calculate the rate first.
• If the question gives multiple wavelengths, calculate the power contribution from each wavelength separately.
• If the question gives a wavelength in nanometres, convert it into metres.
• If the question asks for optical power, do not confuse it with the electrical power consumed by the light source.

The last point matters for real devices. An LED converts some electrical energy into light and some into other forms, including thermal energy. The equation here calculates the optical power carried by the emitted photons, not the electrical input power.

05Worked examples

06Questions and solutions

The following questions build from straightforward photon-power calculations to situations involving multiple wavelengths, detector efficiency, and changing emission conditions.

Question 1

A green LED emits \(3.0\times10^{17}\) photons per second. Each photon has a wavelength of \(520\ \text{nm}\).

Calculate:

(a) The energy of one photon.

(b) The optical power emitted by the LED.

Solution 1

The photon energy is \(3.82\times10^{-19}\ \text{J}\), and the optical power is approximately \(0.115\ \text{W}\).

Part (a): Calculate photon energy.

Use:

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

Substitute the values:

\[ E_{\text{photon}}= \frac{(6.63\times10^{-34})(3.00\times10^8)} {520\times10^{-9}} \]

\[ E_{\text{photon}}=3.82\times10^{-19}\ \text{J} \]

Part (b): Calculate power.

Use:

\[ P=RE_{\text{photon}} \]

\[ P=(3.0\times10^{17})(3.82\times10^{-19}) \]

\[ \boxed{P=0.115\ \text{W}} \]

The LED transfers approximately \(0.115\) joules of optical energy each second.

The main trap here is forgetting to convert nanometres into metres. Substituting \(520\) directly into the wavelength equation would give an incorrect result.

Question 2

A specialised light source emits two wavelengths simultaneously.

The emission data are:

Calculate:

(a) The optical power contributed by each wavelength.

(b) The total optical power.

(c) Which wavelength contributes more power, and how many times greater is its contribution?

Solution 2

The total optical power is approximately \(0.398\ \text{W}\). The \(400\ \text{nm}\) photons contribute three times as much power as the \(800\ \text{nm}\) photons.

Part (a): Calculate the energy of each photon.

For the \(400\ \text{nm}\) photons:

\[ E_{400}=\frac{hc}{400\times10^{-9}} \]

\[ E_{400}=4.97\times10^{-19}\ \text{J} \]

For the \(800\ \text{nm}\) photons:

\[ E_{800}=\frac{hc}{800\times10^{-9}} \]

\[ E_{800}=2.49\times10^{-19}\ \text{J} \]

Now calculate the power from each group.

For \(400\ \text{nm}\):

\[ P_{400}=RE \]

\[ P_{400}=(6.0\times10^{17})(4.97\times10^{-19}) \]

\[ P_{400}=0.298\ \text{W} \]

For \(800\ \text{nm}\):

\[ P_{800}=(4.0\times10^{17})(2.49\times10^{-19}) \]

\[ P_{800}=0.0995\ \text{W} \]

Part (b): Add the two power contributions.

\[ P_{\text{total}}=P_{400}+P_{800} \]

\[ P_{\text{total}}=0.298+0.0995 \]

\[ \boxed{P_{\text{total}}=0.398\ \text{W}} \]

Part (c): Compare the contributions.

Calculate the ratio:

\[ \frac{P_{400}}{P_{800}} = \frac{0.298}{0.0995} \]

\[ \boxed{\frac{P_{400}}{P_{800}}=3.0} \]

The \(400\ \text{nm}\) photons contribute three times as much power.

Although their emission rate is only 1.5 times greater, each \(400\ \text{nm}\) photon carries twice the energy.

The tempting mistake is to compare the photon rates alone. Both the number of photons and their individual energies determine the power contribution.

Question 3

Two monochromatic lasers each emit an optical power of \(0.80\ \text{W}\).

• Laser A emits light at \(500\ \text{nm}\).
• Laser B emits light at \(1000\ \text{nm}\).

Both lasers are directed towards separate photon-counting detectors.

Detector A has a detection efficiency of 90%, while detector B has a detection efficiency of 60%.

Calculate:

(a) The photon emission rate of each laser.

(b) The number of detection pulses recorded by each detector per second.

(c) Which detector records more pulses per second, and by what factor?

Solution 3

Laser B has twice the photon emission rate of laser A. Despite its lower detector efficiency, detector B records approximately 1.33 times as many pulses per second.

Part (a): Calculate the photon emission rates.

Rearrange the power equation:

\[ P=\frac{Rhc}{\lambda} \]

Therefore:

\[ R=\frac{P\lambda}{hc} \]

For laser A:

\[ R_A= \frac{(0.80)(500\times10^{-9})} {(6.63\times10^{-34})(3.00\times10^8)} \]

\[ R_A=2.01\times10^{18}\ \text{photons s}^{-1} \]

For laser B:

\[ R_B= \frac{(0.80)(1000\times10^{-9})} {(6.63\times10^{-34})(3.00\times10^8)} \]

\[ R_B=4.02\times10^{18}\ \text{photons s}^{-1} \]

Laser B emits twice as many photons each second.

This makes sense because its photons have half the energy of laser A’s photons. It needs twice as many to deliver the same total energy per second.

Part (b): Calculate the detected photon rates.

The detection efficiency tells us what fraction of arriving photons produce a recorded pulse.

For detector A:

\[ R_{\text{detected,A}}=0.90R_A \]

\[ R_{\text{detected,A}} =0.90(2.01\times10^{18}) \]

\[ R_{\text{detected,A}}=1.81\times10^{18}\ \text{pulses s}^{-1} \]

For detector B:

\[ R_{\text{detected,B}}=0.60R_B \]

\[ R_{\text{detected,B}} =0.60(4.02\times10^{18}) \]

\[ R_{\text{detected,B}}=2.41\times10^{18}\ \text{pulses s}^{-1} \]

Part (c): Compare the detected rates.

\[ \text{Ratio}= \frac{R_{\text{detected,B}}} {R_{\text{detected,A}}} \]

\[ =\frac{2.41\times10^{18}}{1.81\times10^{18}} \]

\[ \boxed{\text{Ratio}=1.33} \]

Detector B records approximately 1.33 times as many pulses per second.

The trap is assuming that equal optical power means equal photon emission rates. It does not.

There is a second trap: comparing detector efficiencies without considering the number of photons arriving. Detector B is less efficient, but it receives enough additional photons to record more pulses overall.

Question 4

A photon-counting detector records \(2.4\times10^{17}\) pulses over a period of \(30\ \text{s}\).

The source emits monochromatic light at \(600\ \text{nm}\).

The experimental arrangement has the following properties:

• Only 25% of the photons emitted by the source reach the detector.
• The detector records 80% of the photons that reach it.

Assume the source emits at a constant rate throughout the measurement.

Calculate:

(a) The actual photon emission rate of the source.

(b) The total optical power emitted by the source.

Solution 4

The actual photon emission rate is \(4.0\times10^{16}\ \text{photons s}^{-1}\), giving an optical power of approximately \(0.0133\ \text{W}\).

Part (a): Calculate the detected photon rate.

The detector records \(2.4\times10^{17}\) pulses over \(30\ \text{s}\).

\[ R_{\text{detected}}=\frac{C}{t} \]

\[ R_{\text{detected}}= \frac{2.4\times10^{17}}{30} \]

\[ R_{\text{detected}}=8.0\times10^{15}\ \text{s}^{-1} \]

However, this is not the actual emission rate.

Only 25% of the photons reach the detector, and only 80% of those photons are registered.

Step 2: Calculate the combined detection fraction.

\[ g\eta=0.25\times0.80 \]

\[ g\eta=0.20 \]

Therefore, the detector records only 20% of the photons emitted by the source.

Step 3: Calculate the actual emission rate.

\[ R_{\text{source}}= \frac{R_{\text{detected}}}{g\eta} \]

\[ R_{\text{source}}= \frac{8.0\times10^{15}}{0.20} \]

\[ \boxed{R_{\text{source}}=4.0\times10^{16}\ \text{photons s}^{-1}} \]

Part (b): Calculate the total optical power.

First calculate the energy of one photon:

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

\[ E_{\text{photon}}= \frac{(6.63\times10^{-34})(3.00\times10^8)} {600\times10^{-9}} \]

\[ E_{\text{photon}}=3.315\times10^{-19}\ \text{J} \]

Now calculate the optical power:

\[ P=RE_{\text{photon}} \]

\[ P=(4.0\times10^{16})(3.315\times10^{-19}) \]

\[ \boxed{P=0.0133\ \text{W}} \]

The source emits approximately \(13.3\ \text{mW}\) of optical power.

The important reasoning is that the detector count must be corrected before it is treated as the source emission rate.

Using the raw pulse count would underestimate the actual power by a factor of five.

Question 5

Two monochromatic light sources, A and B, each operate for two seconds.

Each source changes its wavelength and photon emission rate after the first second. Within each one-second interval, its wavelength and emission rate remain constant.

Their operating conditions are:

Both sources therefore have:

• The same average photon emission rate of \(2.0\times10^{18}\ \text{photons s}^{-1}\).
• The same simple time-averaged wavelength of \(600\ \text{nm}\).

A student argues that the two sources must have identical average optical powers because they have the same average emission rate and average wavelength.

The student calculates:

\[ P=\frac{Rhc}{\lambda} \]

using \(R=2.0\times10^{18}\ \text{photons s}^{-1}\) and \(\lambda=600\ \text{nm}\).

Is the student’s conclusion correct?

Calculate the actual average optical power of both sources and explain why the student’s method fails.

Solution 5

The student’s conclusion is incorrect.

Source A has an average optical power of approximately \(0.870\ \text{W}\), while source B has an average optical power of approximately \(0.622\ \text{W}\).

Although both sources have the same average emission rate and the same time-averaged wavelength, their photons are distributed differently across the two wavelengths.

That difference changes the total energy transferred.

Step 1: Calculate the energy of a photon at each wavelength.

For \(400\ \text{nm}\):

\[ E_{400}=\frac{hc}{400\times10^{-9}} \]

\[ E_{400}=4.97\times10^{-19}\ \text{J} \]

For \(800\ \text{nm}\):

\[ E_{800}=\frac{hc}{800\times10^{-9}} \]

\[ E_{800}=2.49\times10^{-19}\ \text{J} \]

The shorter-wavelength photons carry twice the energy.

Step 2: Calculate the power of source A in each interval.

During the first second, source A emits \(3.0\times10^{18}\) photons per second at \(400\ \text{nm}\).

\[ P_{A1}=R_{A1}E_{400} \]

\[ P_{A1}=(3.0\times10^{18})(4.97\times10^{-19}) \]

\[ P_{A1}=1.49\ \text{W} \]

During the second second, source A emits \(1.0\times10^{18}\) photons per second at \(800\ \text{nm}\).

\[ P_{A2}=R_{A2}E_{800} \]

\[ P_{A2}=(1.0\times10^{18})(2.49\times10^{-19}) \]

\[ P_{A2}=0.249\ \text{W} \]

Step 3: Calculate the average power of source A.

Because each interval lasts one second, the average power is the arithmetic mean of the two interval powers.

\[ P_{A,\text{avg}}= \frac{P_{A1}+P_{A2}}{2} \]

\[ P_{A,\text{avg}}= \frac{1.49+0.249}{2} \]

\[ \boxed{P_{A,\text{avg}}=0.870\ \text{W}} \]

Step 4: Calculate the power of source B in each interval.

During the first second:

\[ P_{B1}=(1.0\times10^{18})(4.97\times10^{-19}) \]

\[ P_{B1}=0.497\ \text{W} \]

During the second second:

\[ P_{B2}=(3.0\times10^{18})(2.49\times10^{-19}) \]

\[ P_{B2}=0.747\ \text{W} \]

Step 5: Calculate the average power of source B.

\[ P_{B,\text{avg}}= \frac{P_{B1}+P_{B2}}{2} \]

\[ P_{B,\text{avg}}= \frac{0.497+0.747}{2} \]

\[ \boxed{P_{B,\text{avg}}=0.622\ \text{W}} \]

Step 6: Identify why the student’s calculation fails.

The student’s method gives:

\[ P_{\text{student}}= \frac{(2.0\times10^{18})(6.63\times10^{-34})(3.00\times10^8)} {600\times10^{-9}} \]

\[ P_{\text{student}}=0.663\ \text{W} \]

This is neither source’s actual average power.

The mistake is assuming that multiplying average values gives the correct average power.

The photon emission rate and wavelength change together. Source A emits more photons during the interval when each photon carries more energy. Source B emits more photons during the interval when each photon carries less energy.

The correct calculation must preserve that relationship.

In general, for a source operating in several successive intervals:

\[ P_{\text{avg}}= \frac{1}{T} \sum_i R_i\Delta t_i \frac{hc}{\lambda_i} \]

Here, \(T\) is the total operating time, \(R_i\) is the emission rate during interval \(i\), \(\Delta t_i\) is the duration of that interval, and \(\lambda_i\) is its wavelength.

This equation accounts for the actual energy emitted during each interval before averaging over the total time.

The deeper lesson is that average photon energy must be weighted by the number of photons emitted. A simple average of wavelengths does not give you the average energy per photon.

That is why two sources with identical average photon rates and identical time-averaged wavelengths can still have different average optical powers.

07Where to go next

You now have the tools to move between photon energy, photon emission rate, and optical power.

The next useful step is understanding how photons transfer energy to atoms and electrons. This lets you connect numerical photon calculations to absorption, emission, and atomic energy-level transitions.

Continue with Photon energy and atomic transitions: HSC Physics guide.