How to Convert Photon Energy (eV) into Wavelength

A photoelectric experiment gives a stopping potential of \(1.20\text{ V}\). The metal’s work function is \(1.80\text{ eV}\). Can you calculate the wavelength of the light that hit the metal?

A common mistake is to put \(1.20\text{ eV}\) straight into the wavelength equation. That gives approximately \(1030\text{ nm}\), but it’s the wrong answer. The stopping potential tells us the maximum kinetic energy of the emitted electrons, not the energy of the incoming photons.

The actual photon energy is \(1.80+1.20=3.00\text{ eV}\). We can now calculate its wavelength:

\[
\lambda=\frac{1240}{3.00}=413\text{ nm}
\]

That’s the conversion you need to know:

\[
\boxed{\lambda(\text{nm})=\frac{1240}{E_\gamma(\text{eV})}}
\]

Here, \(E_\gamma\) is the energy of one photon, measured in electronvolts (eV), and \(\lambda\) is its wavelength in nanometres (nm).

The calculation itself is straightforward. The important part is knowing which energy belongs in the equation, how to handle units, and when you’re actually being asked about the number of photons rather than their individual energies.

01The experiment: what does a stopping voltage actually measure?

Before working through the conversion, let’s look at how a physicist might obtain the photon energy experimentally.

Imagine shining a single-colour light source onto a metal surface inside an evacuated photoelectric tube. The metal surface is called the photocathode.

When light strikes the metal, some electrons can absorb energy from the incoming photons and escape from the surface. These electrons travel across the evacuated gap towards a second metal electrode called the collector.

Why isn’t measuring the current enough?

Suppose you increase the brightness of the light source while keeping its colour unchanged.

What would you expect to happen?

A brighter source delivers more photons each second. This means more electrons can be released, so the photocurrent generally increases.

But there’s a problem. A larger current tells us that more charge is reaching the collector. It doesn’t tell us how much kinetic energy the fastest electrons have.

The original current-only arrangement cannot distinguish between these two things:

• Increasing the number of photons arriving each second.
• Increasing the energy carried by each photon.

To measure electron energy, we need to modify the apparatus.

Introducing the stopping potential

Instead of helping electrons reach the collector, we apply a reverse potential that opposes their motion.

The collector is made negative relative to the photocathode. This creates an electric field that slows down the emitted electrons.

As we increase the reverse potential, fewer electrons have enough kinetic energy to reach the collector. Eventually, even the fastest electrons are turned back, and the photocurrent falls to zero.

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

The maximum kinetic energy of the emitted electrons is:

\[
K_{\max}=eV_s
\]

where:

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

There’s a particularly useful shortcut here.

If kinetic energy is expressed in electronvolts, its numerical value is equal to the stopping potential in volts:

\[
K_{\max}(\text{eV})=V_s(\text{V})
\]

For example, a stopping potential of \(1.20\text{ V}\) corresponds to a maximum electron kinetic energy of \(1.20\text{ eV}\).

Be careful with the sign. The collector is negative relative to the photocathode, but \(V_s\) in these equations is the positive magnitude of the stopping potential.

Connecting the measurement to photon energy

An electron needs a minimum amount of energy to escape from a metal surface. This is called the work function, represented by \(\phi\).

The incoming photon supplies energy that is divided between:

• Overcoming the metal’s work function.
• Giving the emitted electron kinetic energy.

Conservation of energy gives:

\[
E_\gamma=\phi+K_{\max}
\]

Therefore, when the energies are expressed in eV:

\[
\boxed{E_\gamma=\phi+V_s}
\]

This is the important distinction from the opening problem.

The stopping potential measures the maximum kinetic energy of the emitted electrons. To find the original photon energy, we must also account for the work function.

The energy balance assumes the electron receives energy from one photon and escapes with the maximum possible kinetic energy. Some electrons lose energy before escaping, so they have less kinetic energy.

What should the experiment show?

Consider two changes to the light source.

1. Increase its brightness while keeping its frequency constant. The photocurrent increases, but the stopping potential stays the same.
2. Increase its frequency while keeping the metal unchanged. Provided the photons are energetic enough to release electrons, the maximum kinetic energy and stopping potential increase.

This is the experimental distinction between photon number and photon energy.

The stopping-potential method gives us a way to determine photon energy experimentally. Once that energy is known, converting it into wavelength is much simpler.

02Converting electronvolts into wavelength

Step 1: Understand the relationship

Photon energy is related to frequency by:

\[
E_\gamma=hf
\]

where:

• \(E_\gamma\) is the energy of one photon in joules (J).
• \(h\) is Planck’s constant, \(6.626\times10^{-34}\text{ J s}\).
• \(f\) is frequency in hertz (Hz).

Frequency and wavelength are related by:

\[
c=f\lambda
\]

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

Rearranging the wave equation:

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

Substituting this into the photon energy equation:

\[
E_\gamma=h\frac{c}{\lambda}
\]

Therefore:

\[
E_\gamma=\frac{hc}{\lambda}
\]

Rearranging to make wavelength the subject:

\[
\boxed{\lambda=\frac{hc}{E_\gamma}}
\]

This equation gives wavelength in metres when photon energy is supplied in joules.

But HSC questions often give photon energy in electronvolts. We need one more conversion.

Step 2: Convert electronvolts into joules

An electronvolt is a unit of energy.

One electronvolt is the energy transferred to a particle carrying one elementary charge when it moves through a potential difference of one volt.

The conversion is:

\[
1\text{ eV}=1.60\times10^{-19}\text{ J}
\]

Therefore, if a photon has an energy of \(2.50\text{ eV}\), its energy in joules is:

\[
E_\gamma=2.50\times1.60\times10^{-19}
\]

\[
E_\gamma=4.00\times10^{-19}\text{ J}
\]

We can substitute this into the wavelength equation.

Step 3: Combine the constants

Using:

\[
\lambda=\frac{hc}{E_\gamma}
\]

and substituting the electronvolt conversion:

\[
\lambda=
\frac{(6.626\times10^{-34})(3.00\times10^8)}
{E_\gamma(\text{eV})(1.60\times10^{-19})}
\]

Simplifying the constants gives approximately:

\[
\lambda=\frac{1.24\times10^{-6}}{E_\gamma(\text{eV})}
\text{ m}
\]

Since \(1\text{ nm}=10^{-9}\text{ m}\), we can express this much more conveniently as:

\[
\boxed{\lambda(\text{nm})=\frac{1240}{E_\gamma(\text{eV})}}
\]

This is the formula to remember.

You don’t need to convert electronvolts into joules separately when using this shortcut. The conversion has already been included in the constant 1240.

The three-step conversion method

Whenever you’re given photon energy in eV and asked to find wavelength:

1. Check that the energy is for one photon, not an entire beam or pulse.
2. Divide 1240 by the photon energy to obtain wavelength in nanometres.
3. Convert nanometres into metres if the question requests SI units.

For example:

\[
E_\gamma=2.00\text{ eV}
\]

\[
\lambda=\frac{1240}{2.00}
\]

\[
\lambda=620\text{ nm}
\]

If the question wants metres:

\[
\lambda=620\times10^{-9}
\]

\[
\lambda=6.20\times10^{-7}\text{ m}
\]

That’s it.

What wavelengths do different photon energies produce?

The relationship is inversely proportional:

\[
\lambda\propto\frac{1}{E_\gamma}
\]

This means a photon with twice the energy has half the wavelength.

The following values are useful reference points. The spectral boundaries are approximate.

These conversions have practical uses.

For example, a red LED emits photons with relatively low energy compared with a violet or ultraviolet source. Changing the LED’s emitted wavelength changes the energy carried by each photon.

The calculation is also useful when considering solar cells. Silicon has a band-gap energy of approximately \(1.1\text{ eV}\). Converting this energy into wavelength gives a threshold of roughly \(1130\text{ nm}\) in the simplified one-photon model.

This helps identify which incoming photons have enough energy to excite electrons across the band gap. The actual generation and collection of charge carriers involves further processes, explained in How Solar Cells Convert Light into Electrical Energy.

03Photon energy versus photon number

Here’s another distinction that often causes problems in HSC questions.

Photon energy tells us how much energy one photon carries.

Photon number tells us how many photons are present in a beam or pulse.

These are separate quantities.

The total energy of a collection of identical photons is:

\[
E_{\text{total}}=NE_\gamma
\]

where:

• \(E_{\text{total}}\) is the total energy of the light pulse in joules.
• \(N\) is the number of photons.
• \(E_\gamma\) is the energy of one photon in joules.

Rearranging:

\[
\boxed{N=\frac{E_{\text{total}}}{E_\gamma}}
\]

Notice that both energies must be in the same units.

If the total pulse energy is given in joules and individual photon energy is given in eV, convert the photon energy into joules first:

\[
N=
\frac{E_{\text{total}}}
{E_\gamma(\text{eV})\times1.60\times10^{-19}}
\]

The result is a number of photons.

Why brightness doesn’t change wavelength

Imagine two monochromatic light beams with the same wavelength.

One beam is much brighter than the other.

The brighter beam contains more photons arriving each second. However, each photon still has exactly the same energy as a photon in the dimmer beam.

Consequently:

• Increasing intensity increases the number of photons arriving per second.
• Increasing photon energy decreases wavelength.
• Increasing photon number does not change wavelength.

This distinction becomes particularly important in photoelectric questions.

If a metal requires at least \(2.2\text{ eV}\) to release an electron, sending it millions of photons with energies of \(2.0\text{ eV}\) will not normally produce photoelectrons through the one-photon photoelectric effect.

Making the beam brighter increases the number of incoming photons. It doesn’t increase the energy of each photon.

By contrast, photons with energies of \(2.5\text{ eV}\) can release electrons from that metal, even if there are fewer of them.

04What happens to wavelength inside glass?

There’s one more detail worth understanding before tackling the harder questions.

The shortcut:

\[
\lambda=\frac{1240}{E_\gamma}
\]

calculates the wavelength in a vacuum. For most HSC calculations, air is treated as approximately equivalent to a vacuum.

But what if the light enters glass?

Light travels more slowly in glass than in a vacuum. Its frequency remains unchanged when it crosses the boundary, but its wavelength decreases.

The refractive index, \(n\), is defined by:

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

where \(v\) is the speed of light in the material.

The wavelength inside the material is:

\[
\boxed{\lambda_{\text{material}}=\frac{\lambda_0}{n}}
\]

Here, \(\lambda_0\) is the wavelength in a vacuum.

The important point is that the photon energy does not change simply because the light enters glass.

Photon energy depends on frequency:

\[
E_\gamma=hf
\]

Since frequency remains constant, photon energy remains constant.

The shorter wavelength is accompanied by a lower wave speed, not a higher photon energy.

For a material with refractive index \(1.50\), the wavelength inside it is two-thirds of its vacuum wavelength.

This distinction matters whenever a question asks for wavelength inside a material rather than in air or a vacuum.

05Worked examples

06Questions and solutions

The following questions progress from direct wavelength conversions to problems involving photoelectric measurements, photon counts, and changes in the medium through which light travels.

Question 1

A photon has an energy of \(2.75\text{ eV}\).

Calculate:

(a) Its wavelength in nanometres.

(b) Its wavelength in metres.

(c) The approximate region of the electromagnetic spectrum in which it belongs.

Solution 1

The wavelength is approximately \(451\text{ nm}\), or \(4.51\times10^{-7}\text{ m}\). It belongs to the blue region of visible light.

Step 1: Apply the electronvolt-to-wavelength equation.

\[
\lambda=\frac{1240}{E_\gamma}
\]

Step 2: Substitute the photon energy.

\[
\lambda=\frac{1240}{2.75}
\]

\[
\lambda=450.9\text{ nm}
\]

Step 3: Convert into metres.

\[
\lambda=450.9\times10^{-9}
\]

\[
\lambda=4.51\times10^{-7}\text{ m}
\]

Step 4: Identify the region.

A wavelength of approximately \(451\text{ nm}\) is in the blue region of visible light.

The key principle is that photon energy and wavelength are inversely proportional. A photon with greater energy would have a shorter wavelength.

Question 2

Two monochromatic laser pulses each contain a total optical energy of \(0.80\text{ mJ}\).

• Pulse A contains photons with an energy of \(1.60\text{ eV}\) each.
• Pulse B contains photons with an energy of \(3.20\text{ eV}\) each.

For each pulse, calculate:

(a) The wavelength of its photons.

(b) The number of photons in the pulse.

(c) Which pulse contains more photons, and explain why.

Solution 2

Pulse A has a wavelength of \(775\text{ nm}\) and contains approximately \(3.13\times10^{15}\) photons. Pulse B has a wavelength of \(388\text{ nm}\) and contains approximately \(1.56\times10^{15}\) photons.

Pulse A contains twice as many photons because its individual photons carry half as much energy, while both pulses have the same total energy.

Step 1: Calculate the wavelength of pulse A.

\[
\lambda_A=\frac{1240}{1.60}
\]

\[
\lambda_A=775\text{ nm}
\]

Step 2: Convert the energy of one A photon into joules.

\[
E_A=1.60\times1.60\times10^{-19}
\]

\[
E_A=2.56\times10^{-19}\text{ J}
\]

Step 3: Calculate the number of photons in pulse A.

Convert the total pulse energy:

\[
0.80\text{ mJ}=8.00\times10^{-4}\text{ J}
\]

Then:

\[
N_A=\frac{8.00\times10^{-4}}
{2.56\times10^{-19}}
\]

\[
N_A=3.13\times10^{15}
\]

Step 4: Calculate the wavelength of pulse B.

\[
\lambda_B=\frac{1240}{3.20}
\]

\[
\lambda_B=387.5\text{ nm}
\]

Step 5: Convert the energy of one B photon into joules.

\[
E_B=3.20\times1.60\times10^{-19}
\]

\[
E_B=5.12\times10^{-19}\text{ J}
\]

Step 6: Calculate the number of photons in pulse B.

\[
N_B=\frac{8.00\times10^{-4}}
{5.12\times10^{-19}}
\]

\[
N_B=1.56\times10^{15}
\]

Step 7: Compare the results.

\[
\frac{N_A}{N_B}
=
\frac{3.13\times10^{15}}
{1.56\times10^{15}}
\]

\[
\frac{N_A}{N_B}\approx2
\]

Pulse A contains twice as many photons.

The tempting prediction is that equal total energy should mean equal numbers of photons. That would only be true if the photons in both pulses had the same individual energy.

Since pulse B’s photons have twice the energy, it needs only half as many photons to deliver the same total energy.

Question 3

An evacuated photoelectric cell contains a metal photocathode with a work function of \(2.25\text{ eV}\).

A monochromatic light source produces a stopping potential with a magnitude of \(0.85\text{ V}\).

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

(b) Calculate the energy of each incident photon.

(c) Calculate the wavelength of the incident light in metres.

(d) The intensity of the light is tripled while its frequency remains unchanged. Explain what happens to the photon arrival rate, maximum electron kinetic energy, stopping potential, and photocurrent.

Solution 3

The incident photons have an energy of \(3.10\text{ eV}\) and a wavelength of \(4.00\times10^{-7}\text{ m}\). Tripling the intensity increases the photon arrival rate and photocurrent, but does not change the photon energy, wavelength, or stopping potential.

Step 1: Calculate the maximum kinetic energy.

Using:

\[
K_{\max}=V_s
\]

when kinetic energy is expressed in eV:

\[
K_{\max}=0.85\text{ eV}
\]

Step 2: Calculate the incident photon energy.

Apply conservation of energy:

\[
E_\gamma=\phi+K_{\max}
\]

Substitute:

\[
E_\gamma=2.25+0.85
\]

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

Step 3: Calculate the wavelength.

\[
\lambda=\frac{1240}{3.10}
\]

\[
\lambda=400\text{ nm}
\]

Convert into metres:

\[
\lambda=400\times10^{-9}
\]

\[
\lambda=4.00\times10^{-7}\text{ m}
\]

Step 4: Analyse the effect of increasing intensity.

The frequency remains constant, so:

\[
E_\gamma=hf
\]

does not change.

The wavelength also remains constant because:

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

The brighter source supplies more photons per second. Therefore, more electrons can be released per second, increasing the photocurrent in an unsaturated apparatus.

However, the maximum kinetic energy of each emitted electron remains:

\[
K_{\max}=0.85\text{ eV}
\]

The stopping potential also remains:

\[
V_s=0.85\text{ V}
\]

The important distinction is that intensity controls the number of incoming photons, while frequency controls the energy carried by each photon.

Question 4

Two monochromatic light sources, A and B, are directed towards the same metal photocathode in separate experiments.

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

The sources have the following properties:

(a) Calculate the wavelength of photons from each source.

(b) Calculate the number of photons delivered by each source.

(c) Determine which source produces photoelectrons.

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

(e) Explain why the source containing more photons does not necessarily produce photoelectric emission.

Solution 4

Only source B produces photoelectrons. Its photons have an energy of \(2.50\text{ eV}\), which exceeds the metal’s \(2.20\text{ eV}\) work function.

Although source A contains more photons, none of its photons has enough energy to release an electron through the ordinary one-photon photoelectric effect.

Step 1: Calculate the wavelength of source A.

\[
\lambda_A=\frac{1240}{2.00}
\]

\[
\lambda_A=620\text{ nm}
\]

Step 2: Calculate the wavelength of source B.

\[
\lambda_B=\frac{1240}{2.50}
\]

\[
\lambda_B=496\text{ nm}
\]

Source B has the shorter wavelength because its photons have greater energy.

Step 3: Calculate the number of photons from source A.

Convert the photon energy into joules:

\[
E_A=2.00\times1.60\times10^{-19}
\]

\[
E_A=3.20\times10^{-19}\text{ J}
\]

The total pulse energy is:

\[
E_{\text{total}}=4.00\times10^{-4}\text{ J}
\]

Therefore:

\[
N_A=\frac{4.00\times10^{-4}}
{3.20\times10^{-19}}
\]

\[
N_A=1.25\times10^{15}
\]

Step 4: Calculate the number of photons from source B.

\[
E_B=2.50\times1.60\times10^{-19}
\]

\[
E_B=4.00\times10^{-19}\text{ J}
\]

Therefore:

\[
N_B=\frac{4.00\times10^{-4}}
{4.00\times10^{-19}}
\]

\[
N_B=1.00\times10^{15}
\]

Source A delivers 25% more photons than source B.

Step 5: Determine whether photoelectric emission occurs.

The work function is:

\[
\phi=2.20\text{ eV}
\]

For source A:

\[
E_A=2.00\text{ eV}<2.20\text{ eV}
\]

Its photons do not have sufficient energy to release electrons.

For source B:

\[
E_B=2.50\text{ eV}>2.20\text{ eV}
\]

Its photons have enough energy to release electrons.

Step 6: Calculate the maximum kinetic energy and stopping potential for source B.

Using:

\[
K_{\max}=E_\gamma-\phi
\]

\[
K_{\max}=2.50-2.20
\]

\[
K_{\max}=0.30\text{ eV}
\]

Therefore, the stopping potential magnitude is:

\[
V_s=0.30\text{ V}
\]

Step 7: Explain the underlying physics.

This question exposes an important limitation of using brightness to predict photoelectric emission.

Source A supplies more photons, but every photon has insufficient energy to overcome the work function.

Source B supplies fewer photons, but each photon has enough energy to release an electron.

Increasing the intensity of source A would increase the number of incoming photons, but would not overcome the energy threshold in the standard one-photon model.

The energy of an individual photon matters before the number of photons becomes relevant to the emission process.

Question 5

A monochromatic light pulse travels through air and strikes a plane-parallel glass slab at normal incidence.

The following information is provided:

• The energy of each photon in air is \(2.48\text{ eV}\).
• The total energy of the incident pulse is \(0.80\text{ mJ}\).
• The refractive index of the glass is \(1.50\).
• The glass transmits 60% of the incident pulse energy.

Assume there is no reflection, no frequency conversion, and the reduction in transmitted energy is entirely due to absorption. All photons that pass through the glass retain the original frequency.

The transmitted light then strikes a metal photocathode with a work function of \(2.00\text{ eV}\).

(a) Calculate the wavelength of the photons in air.

(b) Calculate the wavelength of the photons inside the glass.

(c) Calculate the number of photons in the incident pulse and the number transmitted through the glass.

(d) Calculate the maximum kinetic energy of photoelectrons released from the metal and the stopping potential required to prevent them reaching the collector.

(e) A student claims:

“The wavelength becomes shorter inside the glass, so each photon gains energy. The glass then removes 40% of the energy from every remaining photon, making the transmitted photons less energetic.”

Evaluate this claim using the relevant physical relationships.

Solution 5

The wavelength is \(500\text{ nm}\) in air and approximately \(333\text{ nm}\) inside the glass. However, each transmitted photon retains its original energy of \(2.48\text{ eV}\).

The transmitted pulse contains approximately \(1.21\times10^{15}\) photons, and the photocathode produces electrons with a maximum kinetic energy of \(0.48\text{ eV}\). The stopping potential magnitude is \(0.48\text{ V}\).

Step 1: Calculate the wavelength in air.

Use the standard electronvolt-to-wavelength relationship:

\[
\lambda_0=\frac{1240}{E_\gamma}
\]

Substitute:

\[
\lambda_0=\frac{1240}{2.48}
\]

\[
\lambda_0=500\text{ nm}
\]

Therefore, the wavelength in air is approximately \(500\text{ nm}\).

Step 2: Calculate the wavelength inside the glass.

The refractive index is:

\[
n=1.50
\]

Use:

\[
\lambda_{\text{glass}}=\frac{\lambda_0}{n}
\]

Substitute:

\[
\lambda_{\text{glass}}=\frac{500}{1.50}
\]

\[
\lambda_{\text{glass}}=333.3\text{ nm}
\]

The wavelength becomes shorter inside the glass.

However, the frequency remains unchanged, so the photon energy remains:

\[
E_\gamma=2.48\text{ eV}
\]

Step 3: Calculate the number of photons in the incident pulse.

First, convert the energy of one photon into joules:

\[
E_\gamma=2.48\times1.60\times10^{-19}
\]

\[
E_\gamma=3.968\times10^{-19}\text{ J}
\]

Convert the total pulse energy:

\[
E_{\text{total}}=0.80\text{ mJ}
\]

\[
E_{\text{total}}=8.00\times10^{-4}\text{ J}
\]

Now calculate the number of incident photons:

\[
N_{\text{incident}}=
\frac{8.00\times10^{-4}}
{3.968\times10^{-19}}
\]

\[
N_{\text{incident}}=2.02\times10^{15}
\]

Step 4: Calculate the number of transmitted photons.

The glass transmits 60% of the incident pulse energy.

Therefore:

\[
E_{\text{transmitted}}
=0.60(0.80)
\]

\[
E_{\text{transmitted}}=0.48\text{ mJ}
\]

Convert to joules:

\[
E_{\text{transmitted}}=4.80\times10^{-4}\text{ J}
\]

The energy of each surviving photon is unchanged, so:

\[
N_{\text{transmitted}}=
\frac{4.80\times10^{-4}}
{3.968\times10^{-19}}
\]

\[
N_{\text{transmitted}}=1.21\times10^{15}
\]

Notice that the number of photons has decreased by 40%.

The glass has absorbed some photons. It has not reduced the energy of every photon that survives.

Step 5: Calculate the maximum kinetic energy of the emitted electrons.

The photocathode has a work function of:

\[
\phi=2.00\text{ eV}
\]

The photon energy is still \(2.48\text{ eV}\) after passing through the glass.

Therefore:

\[
K_{\max}=E_\gamma-\phi
\]

\[
K_{\max}=2.48-2.00
\]

\[
K_{\max}=0.48\text{ eV}
\]

Step 6: Calculate the stopping potential.

The numerical value of maximum kinetic energy in eV equals the stopping potential magnitude in volts.

Therefore:

\[
V_s=0.48\text{ V}
\]

The glass reduces the number of photons reaching the photocathode, so the photocurrent will generally be smaller. However, the stopping potential remains unchanged because the energy of each transmitted photon is unchanged.

Step 7: Evaluate the student’s claim.

The student’s claim is incorrect for two separate reasons.

First, the wavelength decreases inside the glass because light travels more slowly in the material. The frequency remains constant, so the photon energy does not increase.

The mistake is using the vacuum speed of light, \(c\), when applying \(E=hc/\lambda\) inside glass.

Inside the material, the correct wave relationship is:

\[
E_\gamma=h\frac{v}{\lambda_{\text{glass}}}
\]

Since:

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

and:

\[
\lambda_{\text{glass}}=\frac{\lambda_0}{n}
\]

we obtain:

\[
E_\gamma
=h\frac{c/n}{\lambda_0/n}
\]

\[
E_\gamma=\frac{hc}{\lambda_0}
\]

The refractive index cancels. The photon energy is unchanged.

Second, the glass absorbs 40% of the pulse energy by removing photons from the transmitted beam under the stated assumptions. It doesn’t reduce the energy of every surviving photon by 40%.

This is why the transmitted beam has fewer photons but the same photon energy.

The distinction affects the two photoelectric measurements differently:

• Photocurrent decreases because fewer photons reach the metal.
• Stopping potential remains unchanged because each transmitted photon has the same energy.

The result is a useful reminder that wavelength, photon energy, and photon number describe different properties of light. They are connected, but they aren’t interchangeable.

07Where to go next

You can now move between photon energy, wavelength, and photon number without mixing up the quantities.

The next useful connection is understanding why atoms emit or absorb only particular photon energies. An electron transition determines the energy of the photon, and that energy determines its wavelength.

For that connection, continue with Photon energy and atomic transitions: HSC Physics guide.

Keep the central conversion handy:

\[
\boxed{\lambda(\text{nm})=\frac{1240}{E_\gamma(\text{eV})}}
\]

Use the energy of one photon, not the total energy of the beam. If the question gives you a stopping potential, account for the work function first. If it gives you a pulse energy, use it to calculate the photon number separately.