Why Increasing Intensity Can Mean More Photons Rather Than More Energetic Photons

Two green laser beams hit the same patch of a wall. Beam B has twice the measured intensity of beam A. Does that mean each photon in beam B carries twice as much energy?

No. If both beams have the same wavelength and illuminate the same area, each photon carries exactly the same energy in both beams. Beam B simply delivers twice as many photons per second per square metre.

This is the key distinction: increasing light intensity does not necessarily increase the energy of individual photons. It can increase the number of photons arriving.

01Two different quantities: photon energy and photon number

Imagine a delivery service transporting identical packages. If twice as many packages arrive every minute, the total amount of energy being delivered each minute might double, even though every individual package contains exactly the same thing.

Light behaves similarly in this respect.

A photon is a discrete packet of electromagnetic energy. Its energy depends on the frequency of the electromagnetic radiation, not on how intense the beam is.

For an individual photon:

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

where:

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

Using the relationship between frequency and wavelength, \(c=f\lambda\), we can also write:

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

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.

These equations tell us something important:

• Higher frequency means more energy per photon.
• Shorter wavelength means more energy per photon.
• Lower frequency means less energy per photon.
• Longer wavelength means less energy per photon.

Notice that intensity does not appear in either equation.

For a refresher on the relationship between frequency, wavelength, and photon energy, see Photon energy and atomic transitions: HSC Physics guide.

What does intensity actually measure?

Light intensity describes how much energy passes through a given area each second.

In other words, it measures the rate of energy transfer per unit area.

\[ I_{\text{light}}=\frac{P}{A} \]

where:

• \(I_{\text{light}}\) is light intensity, measured in watts per square metre (\(\text{W m}^{-2}\)).
• \(P\) is the optical power of the beam, measured in watts (W).
• \(A\) is the area perpendicular to the beam, measured in square metres (\(\text{m}^2\)).

Now consider how that energy is delivered.

If every photon has energy \(E_{\text{photon}}\), and \(\Phi\) photons arrive per square metre each second, then:

\[ \boxed{I_{\text{light}}=\Phi E_{\text{photon}}} \]

The quantity \(\Phi\) is called the photon flux.

It is defined as:

\[ \Phi=\frac{N}{At} \]

where:

• \(N\) is the number of photons arriving.
• \(A\) is the illuminated area.
• \(t\) is the time taken.

Photon flux is measured in photons \(\text{m}^{-2}\text{s}^{-1}\).

This equation is the mathematical explanation for the difference between photon energy and photon number.

Light intensity depends on both the energy of each photon and the number of photons arriving per unit area per second.

What happens when intensity increases?

The answer depends on what you keep constant.

The first row is the situation in the opening example.

If you increase the output power of a green laser without changing its wavelength or beam area, more photons arrive each second. Each photon still has the same energy.

There is another useful distinction in the final row. Focusing a beam increases its intensity because the same optical power is concentrated into a smaller area. The total number of photons emitted each second does not necessarily change.

Similarly, moving closer to a lamp can increase the intensity received by a surface. The photons have not gained energy. They are simply distributed across a smaller area.

02Why wavelength still matters

So far, we’ve looked at increasing intensity while keeping wavelength constant.

But what happens when the wavelength changes?

Remember:

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

A shorter wavelength means that each photon carries more energy.

This is why different parts of the electromagnetic spectrum have different interactions with matter.

The wavelength boundaries are approximate and can vary slightly between conventions.

Here’s a useful comparison.

Imagine two beams with exactly the same intensity:

• A red beam with a wavelength of 700 nm.
• A violet beam with a wavelength of 400 nm.

The violet photons have greater energy because their wavelength is shorter.

However, the red beam must contain more photons per second per square metre to deliver the same total energy each second.

This is a point worth remembering:

At equal intensity, a beam containing more energetic photons must have a lower photon flux.

The reverse is also true. If two beams have the same photon flux, the beam with the shorter wavelength has the greater intensity.

03The photoelectric effect: experimental evidence

The distinction between photon energy and photon number isn’t just a mathematical convenience. It helps explain one of the important experiments that challenged classical physics.

Consider a metal surface exposed to light.

When light of a suitable frequency strikes the metal, electrons can escape from its surface. This phenomenon is called the photoelectric effect.

The interesting part is what happens when we increase the intensity of the incoming light.

You might predict that brighter light gives the electrons more energy, causing them to leave the metal faster.

That prediction seems reasonable if you imagine light continuously transferring energy to the electrons.

However, experiments showed something different.

Increasing intensity increases the number of emitted electrons per second, but does not increase their maximum kinetic energy when the frequency is unchanged.

Let’s see how the experiment reveals this.

Setting up the experiment

A standard photoelectric experiment uses a photoelectric cell.

Inside an evacuated glass tube are two electrodes:

• The cathode is a photosensitive metal surface that releases electrons when illuminated.
• The anode collects the electrons travelling away from the cathode.

Monochromatic light enters through a transparent window and strikes the cathode.

The two electrodes are connected to an external circuit containing a variable voltage supply and an ammeter. A voltmeter measures the potential difference across the electrodes.

The evacuated tube is important because it prevents electrons from losing kinetic energy through collisions with air molecules as they travel towards the anode.

First measurement: increase the intensity

Suppose we keep the following conditions constant:

• The frequency of the incoming light.
• The wavelength of the light.
• The metal used for the cathode.
• The illuminated area.
• The collection conditions in the circuit.

We then increase the intensity of the light.

What should happen?

The photon model predicts that more photons will arrive at the metal each second.

If the photons have enough energy to release electrons, a greater number of electrons should be emitted each second.

This means the photocurrent increases.

Photocurrent is the electric current produced by the movement of the emitted electrons through the external circuit.

It measures the amount of electric charge passing a point each second:

\[ I_{\text{photo}}=\frac{Q}{t} \]

where \(Q\) is the charge passing through the circuit in time \(t\).

For a fixed frequency and material, increasing the photon flux generally increases the number of emitted electrons and therefore increases the photocurrent, provided the collection conditions remain suitable.

The observation agrees with the photon model.

But there is a problem.

Photocurrent tells us how much charge is flowing. It does not directly tell us how much kinetic energy each electron has.

We need another measurement.

Refining the experiment: measure the stopping potential

To investigate electron energy, we reverse the potential difference across the photoelectric cell.

This makes the anode negative relative to the cathode, creating an electric field that opposes the movement of the emitted electrons.

We gradually increase the magnitude of this reverse potential.

Eventually, even the fastest emitted electrons cannot reach the anode, and the photocurrent falls to zero.

The magnitude of the reverse potential at this point is called the stopping potential, \(V_s\).

The stopping potential allows us to calculate the maximum kinetic energy of the emitted electrons:

\[ \boxed{K_{\text{max}}=eV_s} \]

where:

• \(K_{\text{max}}\) is the maximum kinetic energy of an emitted electron, measured in joules.
• \(e\) is the magnitude of the electron’s charge, \(1.60\times10^{-19}\ \text{C}\).
• \(V_s\) is the positive magnitude of the stopping potential, measured in volts.

The reason this works is that the electric field removes kinetic energy from the electrons.

At the stopping potential, the fastest electrons have just enough initial kinetic energy to overcome the electrical potential barrier.

What does the experiment show?

Now repeat the experiment using the same frequency of light but a higher intensity.

Two observations emerge:

1. The photocurrent increases.
2. The stopping potential remains unchanged.

This is the important result.

The higher-intensity beam produces more photoelectrons per second, but the maximum kinetic energy of those electrons remains the same.

The experimental results can be summarised as follows.

The distinction between intensity and frequency explains these results.

Einstein’s explanation: energy comes in packets

In 1905, Albert Einstein proposed an explanation of the photoelectric effect using discrete packets of electromagnetic energy.

These packets are now called photons.

Einstein’s explanation was that an individual photon transfers its energy to an electron.

The electron uses some of this energy to escape from the metal surface. Any remaining energy becomes kinetic energy.

The minimum energy required to release an electron from the metal is called the work function, represented by \(\phi\).

Therefore:

\[ \boxed{K_{\text{max}}=hf-\phi} \]

This is Einstein’s photoelectric equation.

It tells us that the maximum kinetic energy depends on the frequency of the incoming light and the work function of the metal.

Notice that intensity is not part of this equation.

Increasing intensity increases the number of photons arriving, but does not change the energy carried by each photon.

If the frequency stays constant, the maximum kinetic energy stays constant too.

This is exactly what the experiment observed. The photoelectric effect provided important evidence for the photon model of electromagnetic radiation. Einstein’s explanation dates to 1905. 【turn104793search0】【turn104793search5】

Why increasing intensity cannot always produce photoelectrons

There is one more consequence of the photon model.

A photon must have enough energy to overcome the work function before it can release an electron.

The minimum frequency required is called the threshold frequency, \(f_0\).

At the threshold:

\[ hf_0=\phi \]

Rearranging:

\[ f_0=\frac{\phi}{h} \]

What happens if the incoming frequency is below this threshold?

Each photon has insufficient energy to release an electron.

Increasing the intensity simply increases the number of photons that arrive. It does not make any individual photon more energetic.

Therefore, in the standard one-photon photoelectric model used in HSC Physics, no photoelectrons are emitted, regardless of the intensity.

This is one of the strongest demonstrations of why photon energy and photon number must be treated separately.

The one-photon model is the appropriate assumption here; specialised multiphoton processes at extremely high intensities are outside this treatment.

04Worked examples

05Questions and solutions

Question 1

A monochromatic orange laser has a wavelength of \(600\ \text{nm}\) and illuminates a surface with an intensity of \(3.0\ \text{W m}^{-2}\).

a) Calculate the energy of one photon.

b) Calculate the photon flux.

c) The intensity is doubled without changing the wavelength or illuminated area. Determine the new photon energy and photon flux.

Solution 1

The photon energy remains \(3.32\times10^{-19}\ \text{J}\), while the photon flux doubles from \(9.05\times10^{18}\) to \(1.81\times10^{19}\ \text{photons m}^{-2}\text{s}^{-1}\).

Part a: Calculate photon energy.

Use:

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

Substitute:

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

\[ \boxed{E_{\text{photon}}=3.32\times10^{-19}\ \text{J}} \]

Part b: Calculate photon flux.

Use:

\[ \Phi=\frac{I_{\text{light}}}{E_{\text{photon}}} \]

\[ \Phi=\frac{3.0}{3.32\times10^{-19}} \]

\[ \boxed{\Phi=9.05\times10^{18} \ \text{photons m}^{-2}\text{s}^{-1}} \]

Part c: Double the intensity.

The new intensity is:

\[ I_2=6.0\ \text{W m}^{-2} \]

The wavelength has not changed, so the photon energy remains:

\[ E_{\text{photon}}=3.32\times10^{-19}\ \text{J} \]

The new photon flux is:

\[ \Phi_2=\frac{6.0}{3.32\times10^{-19}} \]

\[ \boxed{\Phi_2=1.81\times10^{19} \ \text{photons m}^{-2}\text{s}^{-1}} \]

The photon flux doubles because the energy of each photon is unchanged. This is the expected result when intensity increases at constant wavelength.

Question 2

Two monochromatic light beams have identical intensities and illuminate the same metal surface.

The metal has a work function of \(2.20\ \text{eV}\).

• Beam A has a wavelength of \(400\ \text{nm}\).
• Beam B has a wavelength of \(600\ \text{nm}\).

a) Calculate the energy of a photon in each beam.

b) Determine the ratio of the photon flux of beam B to beam A.

c) Identify which beam produces photoelectrons.

d) Calculate the maximum kinetic energy and stopping potential of the emitted electrons.

Explain why the beam with the greater photon flux does not necessarily produce photoelectrons.

Solution 2

Beam A produces photoelectrons with a maximum kinetic energy of \(0.90\ \text{eV}\) and a stopping potential of \(0.90\ \text{V}\). Beam B produces no photoelectrons, even though it has 1.50 times the photon flux of beam A.

Part a: Calculate photon energies.

Use:

\[ E_{\text{photon}}=\frac{1240}{\lambda}\ \text{eV} \]

For beam A:

\[ E_A=\frac{1240}{400} \]

\[ E_A=3.10\ \text{eV} \]

For beam B:

\[ E_B=\frac{1240}{600} \]

\[ E_B=2.07\ \text{eV} \]

Beam A has the more energetic photons because it has the shorter wavelength.

Part b: Compare photon fluxes.

The two beams have equal intensity.

Using:

\[ I_{\text{light}}=\Phi E_{\text{photon}} \]

we can write:

\[ \Phi_A E_A=\Phi_B E_B \]

Rearranging:

\[ \frac{\Phi_B}{\Phi_A}=\frac{E_A}{E_B} \]

Substitute:

\[ \frac{\Phi_B}{\Phi_A} =\frac{3.10}{2.07} \]

\[ \boxed{\frac{\Phi_B}{\Phi_A}=1.50} \]

Beam B has 1.50 times the photon flux of beam A.

Part c: Identify the beam that produces photoelectrons.

The work function is \(2.20\ \text{eV}\).

For beam A:

\[ 3.10>2.20 \]

Therefore, its photons have sufficient energy to release electrons.

For beam B:

\[ 2.07<2.20 \]

Its photons cannot release electrons.

Part d: Calculate maximum kinetic energy and stopping potential.

For beam A:

\[ K_{\text{max}}=E_A-\phi \]

\[ K_{\text{max}}=3.10-2.20 \]

\[ \boxed{K_{\text{max}}=0.90\ \text{eV}} \]

Using the relationship between kinetic energy and stopping potential:

\[ K_{\text{max}}=eV_s \]

When kinetic energy is expressed in electronvolts, the numerical value of the stopping potential in volts is the same:

\[ \boxed{V_s=0.90\ \text{V}} \]

Beam B produces no photoelectrons, so there is no stopping potential associated with emitted electrons.

The tempting prediction is that beam B should produce more electrons because it contains more photons. However, the work function sets a minimum energy requirement for each photon.

Having more photons does not help if every individual photon has insufficient energy to release an electron.

Question 3

An evacuated photoelectric cell is illuminated with monochromatic light of a frequency above the threshold frequency of its metal cathode.

Two experiments produce the following results:

The wavelength, cathode material, and collection conditions are unchanged.

A student concludes:

“The intensity has tripled, so each emitted electron must have three times the kinetic energy.”

Evaluate this statement.

Calculate the maximum kinetic energy of the emitted electrons in joules and electronvolts, and explain what the experimental results demonstrate.

Solution 3

The student’s conclusion is incorrect. The maximum kinetic energy remains \(0.65\ \text{eV}\), or \(1.04\times10^{-19}\ \text{J}\), in both experiments. The photocurrent triples because more electrons are collected each second.

Step 1: Identify what the current measures.

Photocurrent measures the rate of charge flow:

\[ I_{\text{photo}}=\frac{Q}{t} \]

If the photocurrent triples, the rate at which charge passes through the circuit also triples.

Since each electron carries the same charge, the number of electrons collected per second triples.

Step 2: Calculate the maximum kinetic energy.

The stopping potential is unchanged:

\[ V_s=0.65\ \text{V} \]

Use:

\[ K_{\text{max}}=eV_s \]

Substitute:

\[ K_{\text{max}}= (1.60\times10^{-19})(0.65) \]

\[ \boxed{K_{\text{max}}=1.04\times10^{-19}\ \text{J}} \]

In electronvolts:

\[ \boxed{K_{\text{max}}=0.65\ \text{eV}} \]

This result applies to both experiments.

Step 3: Explain the experimental observations.

The frequency of the incoming light has not changed.

Therefore, the energy of each photon remains constant:

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

Increasing intensity at constant frequency increases the number of photons arriving each second.

This produces more photoelectrons and increases the photocurrent.

However, the maximum kinetic energy depends on photon energy and the work function:

\[ K_{\text{max}}=hf-\phi \]

Neither quantity has changed, so the maximum kinetic energy remains constant.

The experiment demonstrates that photocurrent and maximum kinetic energy measure different properties of the interaction:

• Photocurrent provides information about the rate of electron emission.
• Stopping potential provides information about the maximum kinetic energy of the emitted electrons.

The student has confused the number of electrons with the energy carried by each electron.

Question 4

A beam containing two monochromatic components strikes a metal surface.

The metal has a work function of \(2.40\ \text{eV}\), and the illuminated area is \(1.0\ \text{cm}^2\).

The beam contains:

Assume that every above-threshold photon releases one electron, every emitted electron is collected, and photons below the work function cannot release electrons.

Determine:

a) The photon flux of each component.

b) Which component contributes to the photocurrent.

c) The maximum kinetic energy of the emitted electrons and the stopping potential.

d) The photocurrent produced by the violet component.

Solution 4

Only the violet component produces photoelectrons. The maximum kinetic energy is \(0.70\ \text{eV}\), the stopping potential is \(0.70\ \text{V}\), and the ideal photocurrent is approximately \(64.5\ \mu\text{A}\).

Step 1: Calculate the energy of each photon.

For red light:

\[ E_{\text{red}}=\frac{1240}{650} \]

\[ E_{\text{red}}=1.91\ \text{eV} \]

For violet light:

\[ E_{\text{violet}}=\frac{1240}{400} \]

\[ E_{\text{violet}}=3.10\ \text{eV} \]

Step 2: Calculate the photon flux of each component.

We use:

\[ \Phi=\frac{I_{\text{light}}}{E_{\text{photon}}} \]

First, convert the photon energies into joules.

For red light:

\[ E_{\text{red}}=1.91(1.60\times10^{-19}) \]

\[ E_{\text{red}}=3.06\times10^{-19}\ \text{J} \]

Therefore:

\[ \Phi_{\text{red}}= \frac{6.0}{3.06\times10^{-19}} \]

\[ \boxed{\Phi_{\text{red}}= 1.96\times10^{19}\ \text{photons m}^{-2}\text{s}^{-1}} \]

For violet light:

\[ E_{\text{violet}}=3.10(1.60\times10^{-19}) \]

\[ E_{\text{violet}}=4.96\times10^{-19}\ \text{J} \]

Therefore:

\[ \Phi_{\text{violet}}= \frac{2.0}{4.96\times10^{-19}} \]

\[ \boxed{\Phi_{\text{violet}}= 4.03\times10^{18}\ \text{photons m}^{-2}\text{s}^{-1}} \]

Step 3: Determine which photons release electrons.

The work function is:

\[ \phi=2.40\ \text{eV} \]

For red light:

\[ 1.91<2.40 \]

Red photons cannot release electrons.

For violet light:

\[ 3.10>2.40 \]

Violet photons can release electrons.

Therefore, only the violet component contributes to the photocurrent.

Notice that the red component actually has a greater photon flux. However, its photons do not have enough energy individually to overcome the work function.

Step 4: Calculate the maximum kinetic energy.

Only violet photons produce photoelectrons.

\[ K_{\text{max}}=E_{\text{violet}}-\phi \]

\[ K_{\text{max}}=3.10-2.40 \]

\[ \boxed{K_{\text{max}}=0.70\ \text{eV}} \]

Therefore:

\[ \boxed{V_s=0.70\ \text{V}} \]

Step 5: Calculate the photocurrent.

First, convert the illuminated area into square metres:

\[ A=1.0\times10^{-4}\ \text{m}^2 \]

The number of violet photons arriving each second is:

\[ R=\Phi A \]

\[ R=(4.03\times10^{18})(1.0\times10^{-4}) \]

\[ R=4.03\times10^{14}\ \text{photons s}^{-1} \]

Under the ideal assumption of one emitted electron per photon:

\[ I_{\text{photo}}=eR \]

\[ I_{\text{photo}}= (1.60\times10^{-19})(4.03\times10^{14}) \]

\[ I_{\text{photo}}=6.45\times10^{-5}\ \text{A} \]

Therefore:

\[ \boxed{I_{\text{photo}}=64.5\ \mu\text{A}} \]

The red component contributes optical energy to the beam, but not to the photoelectric current.

This question illustrates why the total intensity of a beam containing several wavelengths is not enough to predict its photocurrent. You must identify which photons have sufficient energy to release electrons.

Question 5

A metal surface has a work function of \(3.00\ \text{eV}\).

Two different light beams illuminate the same area of the metal.

Beam A is monochromatic light with a wavelength of \(400\ \text{nm}\) and an intensity of \(S\).

Beam B contains two monochromatic components:

• \(50\%\) of its total optical power is carried by \(250\ \text{nm}\) ultraviolet light.
• \(50\%\) of its total optical power is carried by \(800\ \text{nm}\) infrared light.

Beam B also has a total intensity of \(S\).

Assume:

• Both beams illuminate the same area.
• Every photon with sufficient energy releases exactly one electron.
• All emitted electrons are collected.
• The quantum efficiency is \(100\%\) for above-threshold photons.
• Multiphoton effects are negligible.

a) Calculate the photon energy of each wavelength and determine which components produce photoelectrons.

b) Calculate the maximum kinetic energy of the emitted electrons for each beam.

c) Determine the ratio of the photocurrent produced by beam B to that produced by beam A when their total intensities are equal.

d) Beam B’s total intensity is now increased to \(3.2S\), without changing the proportion of ultraviolet and infrared radiation. Determine whether its photocurrent now equals the original photocurrent from beam A. Does its stopping potential also become equal to that of beam A?

Solution 5

At equal total intensity, beam B produces only \(31.25\%\) of beam A’s photocurrent, but its photoelectrons have a much greater maximum kinetic energy. Increasing beam B’s intensity to \(3.2S\) makes the photocurrents equal, but its stopping potential remains different.

This problem requires us to distinguish three things:

• The total optical intensity.
• The number of photons capable of releasing electrons.
• The energy carried by each photon that actually releases an electron.

Part a: Calculate photon energies and identify the useful components.

Use:

\[ E_{\text{photon}}=\frac{1240}{\lambda}\ \text{eV} \]

For beam A:

\[ E_A=\frac{1240}{400} \]

\[ E_A=3.10\ \text{eV} \]

For the ultraviolet component of beam B:

\[ E_{\text{UV}}=\frac{1240}{250} \]

\[ E_{\text{UV}}=4.96\ \text{eV} \]

For the infrared component:

\[ E_{\text{IR}}=\frac{1240}{800} \]

\[ E_{\text{IR}}=1.55\ \text{eV} \]

The work function is:

\[ \phi=3.00\ \text{eV} \]

Compare each photon energy with the work function:

The infrared component cannot release electrons because its photons do not have enough energy.

Only the ultraviolet component of beam B contributes to its photocurrent.

Part b: Calculate maximum kinetic energies.

For beam A:

\[ K_{\text{max,A}}=3.10-3.00 \]

\[ \boxed{K_{\text{max,A}}=0.10\ \text{eV}} \]

For the ultraviolet component of beam B:

\[ K_{\text{max,B}}=4.96-3.00 \]

\[ \boxed{K_{\text{max,B}}=1.96\ \text{eV}} \]

The infrared component produces no photoelectrons.

Therefore, beam B produces electrons with a greater maximum kinetic energy than beam A.

The corresponding stopping potentials are:

\[ V_{s,A}=0.10\ \text{V} \]

\[ V_{s,B}=1.96\ \text{V} \]

Part c: Compare the photocurrents at equal intensity.

This is the most important part of the question.

Let the total intensity of both beams be \(S\).

For beam A, all the optical power is carried by 400 nm photons.

The photon flux is proportional to:

\[ \Phi_A=\frac{S}{E_A} \]

For beam B, only half the total intensity is carried by ultraviolet photons.

The other half is infrared radiation, which cannot release electrons from this metal.

Therefore, the useful photon flux is:

\[ \Phi_B=\frac{S/2}{E_{\text{UV}}} \]

The ratio is:

\[ \frac{\Phi_B}{\Phi_A} = \frac{S/(2E_{\text{UV}})}{S/E_A} \]

Cancel \(S\):

\[ \frac{\Phi_B}{\Phi_A} = \frac{E_A}{2E_{\text{UV}}} \]

Substitute:

\[ \frac{\Phi_B}{\Phi_A} = \frac{3.10}{2(4.96)} \]

\[ \boxed{\frac{\Phi_B}{\Phi_A}=0.3125} \]

Since every useful photon releases one electron under the stated assumptions, photocurrent is proportional to the useful photon flux.

Therefore:

\[ \boxed{\frac{I_B}{I_A}=0.3125} \]

Beam B produces only \(31.25\%\) of beam A’s photocurrent.

Why?

There are two reasons.

First, half of beam B’s optical power is carried by infrared photons that cannot release electrons.

Second, even its useful ultraviolet photons carry more energy individually than beam A’s photons. Therefore, the same amount of useful optical power corresponds to fewer photons.

The tempting prediction is that beam B should produce more current because it contains higher-energy photons. However, higher energy per photon does not mean more photons. At a fixed energy intensity, it actually means fewer photons.

Part d: Increase beam B’s intensity to \(3.2S\).

The spectral proportions remain unchanged.

Therefore, half of the new intensity is ultraviolet:

\[ I_{\text{UV}}=0.50(3.2S) \]

\[ I_{\text{UV}}=1.6S \]

The useful photon flux is now proportional to:

\[ \Phi'_B=\frac{1.6S}{4.96} \]

Compare this with the original photon flux from beam A:

\[ \frac{\Phi'_B}{\Phi_A} = \frac{1.6S/4.96}{S/3.10} \]

Cancel \(S\):

\[ \frac{\Phi'_B}{\Phi_A} = \frac{1.6(3.10)}{4.96} \]

\[ \boxed{\frac{\Phi'_B}{\Phi_A}=1.00} \]

Therefore, beam B now produces the same photocurrent as the original beam A.

However, the stopping potentials have not changed.

The photon wavelengths are unchanged, so the energy of each photon is unchanged.

Consequently:

\[ V_{s,A}=0.10\ \text{V} \]

\[ V_{s,B}=1.96\ \text{V} \]

The photocurrents are equal, but the maximum kinetic energies of the emitted electrons are very different.

This is the important final distinction: equal photocurrents do not mean equal electron energies, just as equal light intensities do not mean equal photon energies.

A current measurement tells you about the rate of charge transfer. A stopping-potential measurement tells you about the maximum kinetic energy of the emitted electrons.

Once you can separate photon energy from photon flux, the next useful step is connecting both quantities to optical power. The relationship \(P=RE_{\text{photon}}\) allows you to calculate the power carried by a beam from its photon emission rate. See How to Calculate Light Power from Photon Emission Rate.