Why Higher Frequency Means More Energy per Photon, Not Faster Light

Two beams of light travel through a vacuum. One is red, and the other is violet. Violet light has a higher frequency, so each violet photon carries more energy. But which beam reaches a distant wall first?

Neither. Both travel at exactly the same speed: approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

This seems strange until you separate three quantities that are easy to mix up: the speed of light, the frequency of light, and the energy carried by each photon. Higher frequency means more energy per photon, not faster light. The photoelectric effect provides some of the clearest experimental evidence for why.

01Frequency, speed, and photon energy are different things

Start with something familiar: waves passing a fixed point.

Imagine standing beside the ocean and counting how many wave crests pass you each second. That count tells you the frequency of the waves.

Now imagine the crests moving towards you. Their speed tells you how quickly each crest travels through the water.

These quantities are related, but they aren’t interchangeable. You could have more crests passing each second simply because the crests are closer together, rather than because they are travelling faster.

The same distinction applies to light.

What frequency actually measures

The frequency of light, represented by \(f\), measures how many complete wave cycles pass a fixed point each second.

Its unit is the hertz (Hz), where:

\[ 1\ \mathrm{Hz}=1\ \mathrm{cycle\,s^{-1}} \]

The wavelength, represented by \(\lambda\), is the distance between equivalent points on consecutive cycles, such as two neighbouring crests.

The speed of light in a vacuum, represented by \(c\), is related to these quantities by:

\[ c=f\lambda \]

Here:

• \(c\) is the speed of light in a vacuum, measured in \(\mathrm{m\,s^{-1}}\).
• \(f\) is frequency, measured in Hz.
• \(\lambda\) is wavelength, measured in metres.

The crucial detail is that \(c\) is constant in a vacuum.

Therefore, if frequency increases, wavelength must decrease.

For example, doubling the frequency halves the wavelength. The light does not move twice as fast. Its wave cycles are simply twice as closely spaced.

Why higher frequency means higher photon energy

Now introduce the photon.

A photon is a discrete packet, or quantum, of electromagnetic radiation. Rather than treating light’s energy as continuously distributed, the quantum model describes it as being transferred in individual packets.

The energy of one photon is:

\[ E=hf \]

where:

• \(E\) is the energy of one photon, measured in joules (J).
• \(h\) is Planck’s constant, \(6.63\times10^{-34}\ \mathrm{J\,s}\).
• \(f\) is the frequency of the light, measured in Hz.

Notice what this equation tells us.

Photon energy is directly proportional to frequency. If frequency doubles, the energy of each photon doubles.

Combining this with \(c=f\lambda\), we can also write:

\[ E=\frac{hc}{\lambda} \]

This gives us an important relationship:

Shorter wavelength means higher frequency and greater energy per photon.

But none of this changes the speed of light in a vacuum.

The analogy with ocean waves has one limitation: light isn’t a mechanical wave made from moving water. Its electromagnetic field propagates through space, and its energy is quantised into photons. The analogy helps separate frequency from speed, while the photon model explains why frequency determines energy.

02Three quantities you must keep separate

Students often make a second mistake after understanding the frequency relationship. They assume that if each photon carries more energy, a beam of higher-frequency light must also deliver more total energy.

Not necessarily.

A beam’s total energy depends on both the energy of each photon and how many photons are delivered.

This last distinction is particularly important.

Photon energy versus photon number

Imagine two light sources, both producing 2.0 mW of optical power.

One produces red light, while the other produces violet light.

The violet photons carry more energy individually. However, because both sources deliver the same total energy each second, the violet source must produce fewer photons per second.

The relationship is:

\[ P=E\dot N \]

Rearranging:

\[ \dot N=\frac{P}{E} \]

where:

• \(P\) is optical power in watts (W).
• \(E\) is energy per photon in joules.
• \(\dot N\) is the photon emission rate in photons per second.

Since \(E=hf\):

\[ \dot N=\frac{P}{hf} \]

At constant power, increasing frequency decreases the number of photons emitted per second.

This is a useful example of why you must identify what is being held constant before making a prediction.

For a more detailed calculation of this relationship, see How to Calculate Light Power from Photon Emission Rate.

03The experiment that revealed the energy of a photon

How did physicists work out that light transfers energy in individual packets rather than simply delivering more energy when its intensity increases?

The photoelectric effect provided the evidence.

It began with an observation that classical wave theory struggled to explain.

Hertz’s observation: ultraviolet light changes a spark

In 1887, Heinrich Hertz observed that ultraviolet light could make an electrical spark jump more easily across a gap between metal electrodes.

The arrangement involved two spark gaps.

One gap produced a spark that emitted ultraviolet radiation. This radiation reached the metal electrodes at a nearby second gap, making it easier for a spark to occur there.

The observation suggested that light could cause electrons to leave a metal surface.

However, the spark-gap arrangement could not directly measure the energy of the emitted electrons. It revealed an effect, but not the detailed relationship between light’s frequency, intensity, and the energy transferred.

Further experiments were needed.

Lenard’s experiment: measuring emitted electrons

Philipp Lenard investigated the photoelectric effect more systematically in the early 1900s.

His experiments used a photosensitive metal inside an evacuated tube.

The arrangement can be understood by following the path of the light and the electrons.

1. Light enters the tube through a window and strikes a metal plate called the cathode.
2. The light transfers energy to electrons in the metal.
3. Some electrons escape from the cathode and travel across the vacuum towards a second electrode called the anode.
4. An ammeter measures the resulting photocurrent.

The evacuated tube prevents air particles from interfering significantly with the electrons travelling between the electrodes.

What did the classical model predict?

Before the photon model, light was commonly described as a continuous electromagnetic wave.

It seemed reasonable to predict that increasing the intensity of the light would transfer more energy to the electrons.

A stronger wave has a greater electromagnetic field amplitude. Surely that should give each electron more energy?

This prediction sounds sensible, but it doesn’t match the experimental results.

Lenard’s investigations revealed two important patterns:

• Increasing the intensity of light at a suitable frequency increased the number of emitted electrons and therefore the photocurrent.
• Increasing the frequency increased the maximum kinetic energy of the emitted electrons, even when the intensity was unchanged.

There was another important observation: below a certain frequency, electrons were not emitted, regardless of how intense the light became in the ordinary photoelectric effect.

The classical wave model could not adequately explain these results.

The revised measurement: finding the stopping potential

A photocurrent measurement tells us how much electric charge reaches the anode each second.

It does not directly tell us how much kinetic energy the fastest electrons possess.

To measure that energy, investigators used a variable voltage supply to oppose the movement of the emitted electrons.

Instead of making the anode positive relative to the cathode, they made it negative.

This created a retarding potential. Electrons needed enough kinetic energy to overcome the electrical potential difference before reaching the anode.

The retarding voltage was increased until the photocurrent fell to zero.

The magnitude of this voltage is called the stopping potential, \(V_s\).

At this point, even the fastest emitted electrons cannot reach the anode.

The experiment therefore provides a way to determine the maximum kinetic energy of the emitted electrons.

Einstein’s explanation: light transfers energy in photons

In 1905, Albert Einstein proposed that light transfers energy in discrete packets.

Each photon carries energy:

\[ E=hf \]

When an electron absorbs a photon, it receives the photon’s energy.

However, the electron cannot necessarily escape from the metal with all that energy. Some energy is required to overcome the attraction holding it within the metal.

This minimum energy is called the work function, represented by \(\phi\).

The remaining energy becomes the electron’s kinetic energy.

Therefore:

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

where:

• \(K_{\max}\) is the maximum kinetic energy of an emitted electron.
• \(h\) is Planck’s constant.
• \(f\) is the frequency of the incident light.
• \(\phi\) is the work function of the metal.

This equation applies when \(hf\geq\phi\).

If the photon energy is below the work function, an electron cannot escape through the ordinary one-photon photoelectric effect.

The stopping potential is related to maximum kinetic energy by:

\[ eV_s=K_{\max} \]

Here, \(e\) is the magnitude of the electron’s charge, \(1.60\times10^{-19}\ \mathrm C\), and \(V_s\) is the positive magnitude of the retarding potential.

If kinetic energy is expressed in electronvolts (eV), its numerical value is equal to the stopping potential in volts.

For example, an electron with a maximum kinetic energy of 1.5 eV requires a stopping potential of 1.5 V.

Millikan’s precision measurements in the 1910s provided experimental confirmation of Einstein’s relationship between photon frequency and electron energy.

The two investigations established an important distinction:

• Photon frequency determines the energy available from each photon.
• Photon arrival rate influences how many electrons can be emitted per second.

A stronger beam does not make its individual photons more energetic if its frequency stays the same.

Does this mean different colours always travel at the same speed?

There is one important qualification.

The statement that all frequencies of light travel at the same speed applies to a vacuum.

When light enters a material such as glass, its propagation speed changes. The speed is given by:

\[ v=\frac{c}{n} \]

where \(v\) is the speed of light in the material and \(n\) is its refractive index.

In a material that exhibits dispersion, the refractive index depends on frequency.

For ordinary glass, violet light generally has a slightly higher refractive index than red light. This means violet light travels more slowly through the glass.

The frequency, however, remains unchanged when light crosses a stationary boundary. Its wavelength changes instead.

This is why a prism separates white light into different colours.

So don’t take the original statement too far.

Higher frequency does not mean faster light in a vacuum. In a dispersive material, different frequencies can have different propagation speeds, but that is caused by the material’s frequency-dependent refractive index, not by photon energy making light accelerate.

04Why wavelength matters in real applications

The relationship between frequency, wavelength, and photon energy helps explain why different parts of the electromagnetic spectrum are useful for different purposes.

Remember:

\[ E=\frac{hc}{\lambda} \]

Shorter wavelengths correspond to higher-energy photons.

The boundaries are approximate and can vary between conventions.

Notice that the applications depend on more than photon energy alone.

For example, microwaves have relatively low energy per photon compared with ultraviolet light. However, a microwave source can deliver enormous numbers of photons, allowing substantial energy transfer to food.

Meanwhile, ultraviolet photons can cause photochemical changes because each individual photon carries considerably more energy.

The useful question is not simply which radiation carries more total energy. It is how much energy each photon carries, how many photons arrive, and how the material responds.

05Worked examples

Let’s apply these relationships to the kinds of calculations you might encounter in HSC Physics.

For calculations, use:

\[ \begin{aligned} h&=6.63\times10^{-34}\ \mathrm{J\,s}\\ c&=3.00\times10^8\ \mathrm{m\,s^{-1}}\\ e&=1.60\times10^{-19}\ \mathrm C \end{aligned} \]

You can also use the convenient approximation:

\[ hc\approx1240\ \mathrm{eV\,nm} \]

Remember that this shortcut requires wavelength in nanometres and gives energy in electronvolts.

06Questions and solutions

The following questions build from basic photon energy calculations to situations where frequency, photon number, stopping potential, and the speed of light must be considered together.

Question 1

A green light source emits photons with a wavelength of 520 nm.

Calculate:

a) The frequency of the light.

b) The energy of one photon in joules.

c) The energy of one photon in electronvolts.

Solution 1

The frequency is \(5.77\times10^{14}\ \mathrm{Hz}\), and the photon energy is \(3.82\times10^{-19}\ \mathrm J\), or approximately \(2.39\ \mathrm{eV}\).

Part a: Calculate frequency.

Use the wave relationship:

\[ c=f\lambda \]

Rearrange:

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

Convert the wavelength:

\[ 520\ \mathrm{nm}=5.20\times10^{-7}\ \mathrm m \]

Substitute:

\[ f=\frac{3.00\times10^8} {5.20\times10^{-7}} \]

\[ f=5.77\times10^{14}\ \mathrm{Hz} \]

Part b: Calculate photon energy in joules.

Use:

\[ E=hf \]

\[ E=(6.63\times10^{-34}) (5.77\times10^{14}) \]

\[ E=3.82\times10^{-19}\ \mathrm J \]

Part c: Convert to electronvolts.

\[ E=\frac{3.82\times10^{-19}} {1.60\times10^{-19}} \]

\[ E=2.39\ \mathrm{eV} \]

The result represents the energy of a single photon. It does not tell us how many photons the source emits each second.

Question 2

A photocell has a metal cathode with a work function of 2.20 eV. Light with a wavelength of 365 nm is directed onto the cathode.

a) Calculate the maximum kinetic energy of the emitted electrons.

b) Calculate the stopping potential.

c) The intensity of the incident light is increased by a factor of four without changing its wavelength. State what happens to the maximum kinetic energy, stopping potential, and photocurrent.

Assume the photocell operates in the ordinary one-photon photoelectric regime, the quantum efficiency is unchanged, and the collection circuit is not limiting the current.

Solution 2

The maximum kinetic energy is approximately 1.21 eV, and the stopping potential is 1.21 V. Increasing the intensity fourfold leaves both quantities unchanged but increases the photocurrent approximately fourfold.

Part a: Calculate the maximum kinetic energy.

First, calculate the photon energy:

\[ E=\frac{1240}{365} \]

\[ E=3.40\ \mathrm{eV} \]

The photon must supply 2.20 eV to overcome the work function.

Therefore:

\[ K_{\max}=E-\phi \]

\[ K_{\max}=3.40-2.20 \]

\[ K_{\max}=1.20\ \mathrm{eV} \]

Using more precise intermediate values gives approximately 1.21 eV.

Part b: Calculate the stopping potential.

Use:

\[ eV_s=K_{\max} \]

Since the maximum kinetic energy is expressed in eV:

\[ V_s\approx1.21\ \mathrm V \]

Part c: Explain the effect of increasing intensity.

Increasing intensity increases the number of photons arriving at the cathode per second.

However, the wavelength remains 365 nm, so the energy of each photon remains unchanged.

Consequently:

• Maximum kinetic energy remains approximately 1.21 eV.
• Stopping potential remains approximately 1.21 V.
• Photocurrent increases approximately fourfold under the stated assumptions.

The common mistake is to assume that more intense light makes each electron more energetic. It actually increases the number of photons available to eject electrons.

Question 3

Two monochromatic light sources produce equal optical powers of 1.5 mW.

Source A emits light with a wavelength of 450 nm, while source B emits light with a wavelength of 650 nm.

Calculate:

a) The energy of one photon from each source.

b) The photon emission rate of each source.

c) The ratio of the number of photons emitted by source B to the number emitted by source A in the same time interval.

d) Explain why the source with the higher-energy photons does not necessarily deliver more total energy per second.

Solution 3

Source A produces higher-energy photons, but source B emits approximately 1.44 times as many photons per second. Both sources deliver the same total energy per second.

Part a: Calculate photon energies.

For source A:

\[ E_A=\frac{hc}{\lambda_A} \]

\[ E_A= \frac{(6.63\times10^{-34})(3.00\times10^8)} {450\times10^{-9}} \]

\[ E_A=4.42\times10^{-19}\ \mathrm J \]

For source B:

\[ E_B= \frac{(6.63\times10^{-34})(3.00\times10^8)} {650\times10^{-9}} \]

\[ E_B=3.06\times10^{-19}\ \mathrm J \]

Source A has the shorter wavelength, so its photons carry more energy.

Part b: Calculate photon emission rates.

The power of each source is:

\[ P=1.5\times10^{-3}\ \mathrm W \]

Using:

\[ \dot N=\frac{P}{E} \]

For source A:

\[ \dot N_A= \frac{1.5\times10^{-3}} {4.42\times10^{-19}} \]

\[ \dot N_A=3.39\times10^{15}\ \mathrm{s^{-1}} \]

For source B:

\[ \dot N_B= \frac{1.5\times10^{-3}} {3.06\times10^{-19}} \]

\[ \dot N_B=4.90\times10^{15}\ \mathrm{s^{-1}} \]

Part c: Calculate the photon number ratio.

\[ \frac{N_B}{N_A} = \frac{\dot N_B}{\dot N_A} \]

\[ =\frac{4.90\times10^{15}} {3.39\times10^{15}} \]

\[ \frac{N_B}{N_A}=1.44 \]

Source B emits approximately 1.44 times as many photons in any given time interval.

Part d: Explain the energy transfer.

Power is the total energy transferred each second.

Both sources have the same power, so they transfer the same total energy per second.

Source A divides that energy into fewer, larger packets. Source B divides it into more, smaller packets.

A higher photon energy does not automatically mean a higher total energy transfer.

Question 4

Two monochromatic light beams travel through a vacuum before entering a glass block at normal incidence.

Their wavelengths in the vacuum are:

• Red light: 650 nm.
• Violet light: 400 nm.

For this particular glass, assume the following illustrative refractive indices:

• Red light: \(n_{\mathrm{red}}=1.50\).
• Violet light: \(n_{\mathrm{violet}}=1.54\).

Calculate:

a) The speed of each colour inside the glass.

b) The wavelength of each colour inside the glass.

c) Explain whether the frequency changes as the light enters the glass.

d) Identify which colour travels faster inside the glass and explain why this does not contradict the relationship between photon energy and frequency.

Solution 4

Red light travels faster inside this glass, at \(2.00\times10^8\ \mathrm{m\,s^{-1}}\), compared with approximately \(1.95\times10^8\ \mathrm{m\,s^{-1}}\) for violet light. The frequencies remain unchanged, but both wavelengths decrease.

Part a: Calculate the speeds inside the glass.

Use:

\[ v=\frac{c}{n} \]

For red light:

\[ v_{\mathrm{red}}= \frac{3.00\times10^8}{1.50} \]

\[ v_{\mathrm{red}}=2.00\times10^8\ \mathrm{m\,s^{-1}} \]

For violet light:

\[ v_{\mathrm{violet}}= \frac{3.00\times10^8}{1.54} \]

\[ v_{\mathrm{violet}}=1.95\times10^8\ \mathrm{m\,s^{-1}} \]

Violet light travels more slowly because its refractive index is higher.

Part b: Calculate the wavelengths inside the glass.

The wave relationship still applies:

\[ v=f\lambda \]

When light crosses a stationary boundary, its frequency stays constant. Therefore, its wavelength changes in proportion to its speed.

We can use:

\[ \lambda_{\mathrm{glass}}= \frac{\lambda_{\mathrm{vacuum}}}{n} \]

For red light:

\[ \lambda_{\mathrm{red}}= \frac{650}{1.50} \]

\[ \lambda_{\mathrm{red}}=433\ \mathrm{nm} \]

For violet light:

\[ \lambda_{\mathrm{violet}}= \frac{400}{1.54} \]

\[ \lambda_{\mathrm{violet}}=260\ \mathrm{nm} \]

Both wavelengths decrease as the light enters the glass.

Part c: Explain the frequency change.

The frequency does not change.

The incoming electromagnetic wave must oscillate at the same frequency as the wave transmitted across the stationary boundary.

Since the propagation speed changes but frequency stays constant, wavelength must change.

Part d: Explain the apparent contradiction.

Violet photons still have more energy than red photons because:

\[ E=hf \]

However, the refractive index of the glass depends on frequency.

Violet light has a higher refractive index in this example, giving it a lower propagation speed inside the material.

There is no contradiction. Photon energy and propagation speed are different physical quantities, and the material affects the propagation speed.

Question 5

This final problem combines photon energy, photon number, the photoelectric effect, and stopping potential.

A laboratory has a vacuum photocell with a replaceable metal cathode.

Two separate, incoherent monochromatic light beams are available:

The photocell can use either of two clean metal cathodes:

Assume:

• The ordinary one-photon photoelectric effect applies.
• Each photon can eject at most one electron.
• Both beams have the same quantum efficiency whenever emission is possible from a given cathode.
• All emitted electrons are collected when there is no retarding potential.
• Contact potentials and other collection losses are negligible.

Answer the following.

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

b) For each cathode, determine which beam can produce photoelectrons and calculate the maximum kinetic energy of the emitted electrons.

c) Cathode M is illuminated by beam A alone and then beam B alone. Compare the photocurrents at zero retarding potential. Calculate the ratio \(I_B/I_A\).

d) Both beams are now directed onto cathode M simultaneously. The collector is made 0.20 V negative relative to the cathode. Determine which beam’s electrons can still reach the collector. Explain whether the exact remaining photocurrent can be calculated from the supplied information.

e) Determine the stopping potential for cathode M when both beams are incident simultaneously.

f) A student argues that beam B must eventually eject electrons from cathode N if its power is increased sufficiently. Evaluate this claim under the stated assumptions.

Solution 5

Cathode M emits electrons under both beams, while cathode N emits electrons only under beam A. At cathode M, beam B produces 1.25 times the zero-bias photocurrent of beam A, but beam A determines the stopping potential of the combined light.

The important distinction is that the beam producing the greatest number of electrons does not necessarily produce the most energetic electrons.

Part a: Calculate photon energies.

For beam A:

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

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

For beam B:

\[ E_B=\frac{1240}{500} \]

\[ E_B=2.48\ \mathrm{eV} \]

Beam A has higher-energy photons because it has the shorter wavelength.

Part b: Determine which beams eject electrons from each cathode.

An electron can escape if:

\[ E_{\mathrm{photon}}\geq\phi \]

For cathode M, the work function is 2.40 eV.

Using beam A:

\[ K_{\max,A}=3.10-2.40 \]

\[ K_{\max,A}=0.70\ \mathrm{eV} \]

Using beam B:

\[ K_{\max,B}=2.48-2.40 \]

\[ K_{\max,B}=0.08\ \mathrm{eV} \]

Both beams can eject electrons from cathode M.

For cathode N, the work function is 2.55 eV.

Using beam A:

\[ K_{\max,A}=3.10-2.55 \]

\[ K_{\max,A}=0.55\ \mathrm{eV} \]

Beam A can eject electrons.

However, beam B has photon energy of only 2.48 eV:

\[ 2.48<2.55 \]

Therefore, beam B cannot eject electrons from cathode N under the stated one-photon model.

Notice that beam B has the same optical power as beam A. Its greater or smaller photon number does not change whether an individual photon has enough energy to overcome the work function.

Part c: Compare the photocurrents at cathode M.

Both beams have the same power.

However, beam B has lower-energy photons, so it must emit more photons per second to deliver the same total power.

Using:

\[ \dot N=\frac{P}{E} \]

The ratio of photon emission rates is:

\[ \frac{\dot N_B}{\dot N_A} = \frac{P/E_B}{P/E_A} \]

The equal powers cancel:

\[ \frac{\dot N_B}{\dot N_A} = \frac{E_A}{E_B} \]

\[ =\frac{3.10}{2.48} \]

\[ =1.25 \]

The two beams have equal quantum efficiency, and all emitted electrons are collected.

Therefore, photocurrent is proportional to photon emission rate:

\[ I=e\eta\dot N \]

Here, \(\eta\) represents the quantum efficiency, or the fraction of incident photons that produce collected electrons.

Since the efficiency is the same for both beams:

\[ \boxed{\frac{I_B}{I_A}=1.25} \]

Beam B produces 25% more photocurrent than beam A at zero retarding potential.

This is the surprising result: beam A produces more energetic electrons, but beam B produces more electrons per second.

Part d: Apply the 0.20 V retarding potential.

The collector is now 0.20 V negative relative to the cathode.

An electron needs at least 0.20 eV of kinetic energy to overcome this retarding potential.

For beam A:

\[ K_{\max,A}=0.70\ \mathrm{eV} \]

Since:

\[ 0.70>0.20 \]

The fastest electrons produced by beam A can still reach the collector.

For beam B:

\[ K_{\max,B}=0.08\ \mathrm{eV} \]

Since:

\[ 0.08<0.20 \]

None of the electrons produced by beam B can reach the collector.

Therefore, the remaining photocurrent comes entirely from beam A.

However, we cannot calculate its exact value.

The maximum kinetic energy tells us the energy of the fastest electrons, not the complete distribution of electron kinetic energies. Some slower electrons produced by beam A may also be stopped by the retarding potential.

We would need additional information about the electron energy distribution to calculate the remaining current.

Part e: Determine the stopping potential for both beams together.

The stopping potential must be large enough to prevent even the most energetic emitted electrons from reaching the collector.

Beam A produces electrons with a maximum kinetic energy of 0.70 eV.

Beam B produces electrons with a maximum kinetic energy of only 0.08 eV.

Therefore, beam A determines the maximum kinetic energy of the combined electron population.

Using:

\[ eV_s=K_{\max} \]

\[ V_s=0.70\ \mathrm V \]

The stopping potential for cathode M is:

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

The fact that beam B produces more electrons does not change this result. Stopping potential depends on the most energetic electrons, not the total number of electrons.

Part f: Evaluate the student’s claim.

The student’s claim is incorrect under the stated one-photon assumptions.

Beam B photons each carry 2.48 eV, while cathode N requires at least 2.55 eV to release an electron.

Increasing the beam’s power increases the number of photons arriving per second. It does not increase the energy carried by each individual photon.

Even if the number of incident photons becomes enormous, each photon still has insufficient energy to overcome the work function.

Therefore, beam B cannot produce photoelectrons from cathode N through the ordinary one-photon photoelectric effect, regardless of how much its power is increased.

At extremely high intensities, multiphoton photoemission can become possible, but that is a different process outside the assumptions of this question.

The key lesson is that three different questions require three different pieces of information:

• How energetic are the emitted electrons? Look at photon frequency and work function.
• How many electrons are emitted per second? Consider photon rate and quantum efficiency.
• What stopping potential is required? Consider the maximum kinetic energy of the emitted electrons.

07What to take into your next HSC question

When a question involves light, don’t automatically assume that higher frequency means greater intensity, greater power, or greater speed.

Instead, identify what the question is actually asking you to calculate.

The next useful step is becoming confident with the relationship between photon energy and photon emission rate. Once you can move between those quantities, questions about light power, photoelectric current, and electromagnetic radiation become much easier to unpack.