How to Calculate Maximum Photoelectron Speed from Wavelength

A metal surface is illuminated with light of wavelength 300 nm. Electrons are emitted, but how fast are the fastest ones moving? Knowing the wavelength alone isn't quite enough. You also need to know the metal's work function: the minimum energy required to release an electron from its surface.

To calculate the **maximum photoelectron speed**, first calculate the maximum kinetic energy using Einstein's photoelectric equation, then convert that energy into speed:

\[
K_{\max}=\frac{hc}{\lambda}-\phi
\]

\[
\boxed{v_{\max}=\sqrt{\frac{2}{m_e}
\left(\frac{hc}{\lambda}-\phi\right)}}
\]

Here, \(h\) is Planck's constant, \(c\) is the speed of light, \(\lambda\) is the light's wavelength, \(\phi\) is the metal's work function, and \(m_e\) is the electron mass.

There is one important condition: the photon energy must be at least equal to the work function. If it is smaller, no photoelectrons are emitted, regardless of how intense the light is.

## 1. From light wavelength to electron speed

Imagine shining light onto a clean metal plate. Some electrons escape from the surface. The fastest escaping electrons are the ones with the greatest kinetic energy, which we call the maximum kinetic energy, \(K_{\max}\).

The key idea is that, in the single-photon model, one photon transfers its energy to one electron. Some of that energy is needed to free the electron from the metal. Any energy left over becomes the electron's kinetic energy.

[IMAGE: An illuminated metal surface showing incoming individual photons, electrons being released from the surface, and a distinction between an electron that escapes with low kinetic energy and the fastest escaping electron. Label photon energy, work function, and maximum kinetic energy.]

### Einstein's photoelectric equation

The energy of a photon is:

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

The symbols are:

- \(E_{\text{photon}}\): energy of one photon, in joules (J).
- \(h=6.626\times10^{-34}\,\text{J s}\): Planck's constant.
- \(f\): light frequency, in hertz (Hz).
- \(c=3.00\times10^8\,\text{m s}^{-1}\): speed of light in a vacuum.
- \(\lambda\): wavelength, in metres (m).

The work function, \(\phi\), is the minimum energy needed to remove an electron from the metal. Once that energy has been supplied, the remaining photon energy becomes kinetic energy.

Conservation of energy gives:

\[
E_{\text{photon}}=\phi+K_{\max}
\]

Rearranging:

\[
\boxed{K_{\max}=\frac{hc}{\lambda}-\phi}
\]

This is the equation you need before calculating electron speed.

The word *maximum* matters. Electrons inside a metal do not all escape with identical kinetic energies. Some lose energy before reaching the surface, while others escape with less energy than the maximum. The equation predicts the upper limit for the emitted electrons.

### A quick way to calculate photon energy

In HSC questions, wavelength is often supplied in nanometres and work function in electronvolts. You can use the convenient approximation:

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

This gives the energy of one photon directly in electronvolts.

For example, a photon with a wavelength of 400 nm has energy:

\[
E_{\text{photon}}=\frac{1240}{400}
=3.10\,\text{eV}
\]

If the metal has a work function of 2.20 eV, the maximum kinetic energy is:

\[
K_{\max}=3.10-2.20
=0.90\,\text{eV}
\]

To convert this kinetic energy into joules, multiply by the elementary charge:

\[
1\,\text{eV}=1.602\times10^{-19}\,\text{J}
\]

Therefore:

\[
K_{\max}=0.90\times1.602\times10^{-19}
=1.442\times10^{-19}\,\text{J}
\]

A metal surface is illuminated with light of wavelength 300 nm. Electrons are emitted, but how fast are the fastest ones moving? Knowing the wavelength alone isn't quite enough. You also need to know the metal's work function: the minimum energy required to release an electron from its surface.

To calculate the **maximum photoelectron speed**, first calculate the maximum kinetic energy using Einstein's photoelectric equation, then convert that energy into speed:

\[
K_{\max}=\frac{hc}{\lambda}-\phi
\]

\[
\boxed{v_{\max}=\sqrt{\frac{2}{m_e}
\left(\frac{hc}{\lambda}-\phi\right)}}
\]

Here, \(h\) is Planck's constant, \(c\) is the speed of light, \(\lambda\) is the light's wavelength, \(\phi\) is the metal's work function, and \(m_e\) is the electron mass.

There is one important condition: the photon energy must be at least equal to the work function. If it is smaller, no photoelectrons are emitted, regardless of how intense the light is.

## 1. From light wavelength to electron speed

Imagine shining light onto a clean metal plate. Some electrons escape from the surface. The fastest escaping electrons are the ones with the greatest kinetic energy, which we call the maximum kinetic energy, \(K_{\max}\).

The key idea is that, in the single-photon model, one photon transfers its energy to one electron. Some of that energy is needed to free the electron from the metal. Any energy left over becomes the electron's kinetic energy.

[IMAGE: An illuminated metal surface showing individual incoming photons and electrons escaping from the surface. Illustrate two emitted electrons with different kinetic energies. Label photon energy, work function, and maximum kinetic energy. Show that the work function must be supplied before an electron can escape.]

### Einstein's photoelectric equation

The energy of a photon is:

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

The symbols are:

- \(E_{\text{photon}}\): energy of one photon, in joules (J).
- \(h=6.626\times10^{-34}\,\text{J s}\): Planck's constant.
- \(f\): light frequency, in hertz (Hz).
- \(c=3.00\times10^8\,\text{m s}^{-1}\): speed of light in a vacuum.
- \(\lambda\): wavelength, in metres (m).

The work function, \(\phi\), is the minimum energy needed to remove an electron from the metal. Once that energy has been supplied, the remaining photon energy becomes kinetic energy.

Conservation of energy gives:

\[
E_{\text{photon}}=\phi+K_{\max}
\]

Rearranging:

\[
\boxed{K_{\max}=\frac{hc}{\lambda}-\phi}
\]

This is the equation you need before calculating electron speed.

The word *maximum* matters. Electrons inside a metal do not all escape with identical kinetic energies. Some lose energy before reaching the surface, while others escape with less energy than the maximum. The equation predicts the upper limit for the emitted electrons.

### A quick way to calculate photon energy

In HSC questions, wavelength is often supplied in nanometres and work function in electronvolts. You can use the convenient approximation:

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

This gives the energy of one photon directly in electronvolts.

For example, a photon with a wavelength of 400 nm has energy:

\[
E_{\text{photon}}=\frac{1240}{400}
=3.10\,\text{eV}
\]

If the metal has a work function of 2.20 eV, the maximum kinetic energy is:

\[
K_{\max}=3.10-2.20
=0.90\,\text{eV}
\]

To convert this kinetic energy into joules, multiply by the elementary charge:

\[
1\,\text{eV}=1.602\times10^{-19}\,\text{J}
\]

Therefore:

\[
K_{\max}=0.90\times1.602\times10^{-19}
=1.442\times10^{-19}\,\text{J}
\]

### What if the wavelength is too long?

Not every photon has enough energy to release an electron.

A metal has a threshold frequency, the minimum frequency needed for photoelectric emission. Equivalently, it has a threshold wavelength, the maximum wavelength that can cause emission.

At the threshold:

\[
hf_0=\phi
\]

Using \(f=c/\lambda\):

\[
\boxed{\lambda_0=\frac{hc}{\phi}}
\]

When the work function is expressed in electronvolts, a convenient version is:

\[
\lambda_0(\text{nm})=\frac{1240}{\phi(\text{eV})}
\]

The conditions are:

| Incident wavelength | Result |
|---|---|
| \(\lambda<\lambda_0\) | Electrons are emitted with positive kinetic energy |
| \(\lambda=\lambda_0\) | Electrons are just able to escape, with zero maximum kinetic energy in the ideal model |
| \(\lambda>\lambda_0\) | No photoelectric emission occurs |

Notice that a shorter wavelength means a higher photon energy. This is because wavelength and frequency are inversely related.

[IMAGE: An energy-level comparison showing three incoming photon energies relative to a metal's work function. One photon has insufficient energy, one exactly reaches the emission threshold, and one exceeds the threshold and leaves an electron with kinetic energy. Label all three cases clearly.]

If you calculate a negative value for \(K_{\max}\), don't take its square root or use its absolute value. It means the photon energy is insufficient to cause emission.

## 2. How did physicists measure maximum photoelectron energy?

Calculating maximum speed is one thing. Measuring it experimentally is another.

Early investigations into the photoelectric effect revealed a problem with the classical wave model of light. If brighter light carries more energy, shouldn't it make emitted electrons faster?

The experiments produced a surprising result.

### Lenard's experiment

In 1887, Heinrich Hertz observed that ultraviolet light affected the behaviour of electrical sparks. Further experiments investigated how light could cause electrons to leave a metal surface.

Around 1902, Philipp Lenard demonstrated that increasing light intensity increased the rate of electron emission, but did not increase the maximum energy of the emitted electrons. He also observed a threshold frequency below which emission did not occur.

The experimental problem was how to measure the energy of the emitted electrons rather than simply count how many arrived at a collector.

A retarding-potential experiment provided the answer.

[IMAGE: An annotated photoelectric experiment apparatus. Show a monochromatic light source shining through a window onto a metal cathode inside an evacuated glass tube. Show emitted electrons travelling towards a wire-mesh collector (anode). Connect the electrodes to an adjustable power supply, an ammeter measuring photocurrent, and a voltmeter measuring the potential difference. Clearly distinguish the cathode, anode, incoming light, electron travel direction, and electrical connections.]

The apparatus works as follows:

1. A light source produces monochromatic light, meaning light of one frequency or wavelength.
2. The light strikes a metal cathode inside an evacuated tube.
3. Electrons are emitted from the cathode and travel towards the collector electrode.
4. An ammeter measures the photocurrent, which tells us how much charge reaches the collector per second.
5. An adjustable power supply changes the potential difference between the electrodes.

Initially, the collector is made positive relative to the cathode. It attracts the negatively charged electrons, allowing them to reach the collector.

Now reverse the potential difference. The collector becomes negative relative to the cathode, so it repels the electrons.

As the magnitude of this reverse voltage increases, fewer electrons have enough kinetic energy to reach the collector. Eventually, even the fastest electrons are stopped, and the photocurrent falls to zero.

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

### Connecting stopping potential to kinetic energy

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

Their initial kinetic energy is completely removed by the retarding electric field.

The energy relationship is:

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

Here:

- \(K_{\max}\) is the maximum kinetic energy in joules.
- \(e=1.602\times10^{-19}\,\text{C}\) is the magnitude of the electron's charge.
- \(V_s\) is the positive magnitude of the stopping potential, in volts.

An especially useful shortcut follows:

If \(K_{\max}\) is expressed in electronvolts, its numerical value is equal to the stopping potential in volts.

For example:

\[
K_{\max}=1.50\,\text{eV}
\]

corresponds to:

\[
V_s=1.50\,\text{V}
\]

This makes stopping potential a direct experimental measurement of maximum photoelectron kinetic energy.

The sign convention is important. If the emitter is at 0 V and the collector is at -1.50 V, the signed potential difference is -1.50 V, but the stopping potential magnitude is positive:

\[
V_s=1.50\,\text{V}
\]

The electron has negative charge, so this negative collector potential increases its electrical potential energy and reduces its kinetic energy.

The experimental arrangement and stopping-potential relationship are described in [OpenStax's explanation of the photoelectric effect](https://openstax.org/books/university-physics-volume-3/pages/6-2-photoelectric-effect).

### What did the experiment actually show?

There are three different quantities to keep separate:

| Quantity | What it tells us | What affects it? |
|---|---|---|
| Photocurrent | Rate at which electrons reach the collector | Light intensity and collection conditions |
| Stopping potential | Energy of the fastest emitted electrons | Light frequency and work function |
| Maximum kinetic energy | Greatest kinetic energy of an emitted electron | Light frequency and work function |

The distinction becomes clear when we compare two light sources with the same wavelength but different intensities.

The brighter source produces more photons per second. More photons can release more electrons per second, increasing the photocurrent.

However, each photon still has the same energy because the wavelength has not changed. The maximum kinetic energy of the emitted electrons therefore stays the same.

The stopping potential also remains unchanged.

[IMAGE: A labelled photocurrent-versus-applied-voltage graph with two curves for the same wavelength at different light intensities. Show the brighter source reaching a higher saturation current, but both curves reaching zero current at the same negative stopping potential. Label the horizontal axis as collector potential relative to emitter, and the vertical axis as photocurrent.]

This observation was central to the development of the photon model.

In 1905, Albert Einstein explained the results by proposing that light transfers energy in individual packets, now called photons. Each photon has energy \(hf\), and an electron can absorb the energy of an individual photon.

Robert Millikan's subsequent measurements, published in 1916, provided a precise experimental test of Einstein's relationship between maximum kinetic energy and light frequency. You can read more in the [American Physical Society's historical account of Millikan's measurements](https://physics.aps.org/story/v3/st23).

The important lesson for calculations is that photocurrent tells you about the number of electrons, while stopping potential tells you about their maximum energy.

For a more detailed explanation of why brightness does not increase maximum photoelectron energy, see [Why Brighter Light Doesn't Increase Photoelectron Energy](https://tutorgum.com/learn/resources/why-brighter-light-doesn-t-give-photoelectrons-more-maximum-kinetic-energy/).

## 3. A reliable method for calculating maximum speed

You can now solve most HSC photoelectron-speed questions using the same sequence.

**Step 1: Check that electrons can be emitted.**

Calculate photon energy and compare it with the work function:

\[
\frac{1240}{\lambda(\text{nm})}\geq\phi(\text{eV})
\]

If this condition is not satisfied, there are no emitted photoelectrons.

**Step 2: Calculate the photon energy.**

Use:

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

This gives energy in eV when wavelength is in nm.

**Step 3: Calculate maximum kinetic energy.**

Subtract the work function:

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

**Step 4: Convert kinetic energy into joules.**

\[
K_{\max}(\text{J})=
K_{\max}(\text{eV})\times1.602\times10^{-19}
\]

**Step 5: Calculate maximum speed.**

Use the classical kinetic energy equation:

\[
K=\frac{1}{2}mv^2
\]

Rearrange:

\[
\boxed{v_{\max}=\sqrt{\frac{2K_{\max}}{m_e}}}
\]

Use the electron mass:

\[
m_e=9.11\times10^{-31}\,\text{kg}
\]

Your final answer should be in metres per second (\(\text{m s}^{-1}\)).

If the question gives a stopping potential instead of a work function, you can skip the photon-energy calculation and use:

\[
K_{\max}=eV_s
\]

Then calculate the speed directly.

### Worked example: Calculate maximum speed from wavelength

A metal has a work function of 2.30 eV. It is illuminated with monochromatic light of wavelength 300 nm.

Calculate the maximum speed of the emitted photoelectrons.

**Step 1**: Calculate the photon energy.

Use the convenient equation:

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

Substitute the wavelength:

\[
E_{\text{photon}}=\frac{1240}{300}
=4.13\,\text{eV}
\]

**Step 2**: Calculate the maximum kinetic energy.

Subtract the work function:

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

\[
K_{\max}=4.13-2.30
\]

\[
K_{\max}=1.83\,\text{eV}
\]

The kinetic energy is positive, so photoelectron emission occurs.

**Step 3**: Convert kinetic energy into joules.

\[
K_{\max}=1.83\times1.602\times10^{-19}
\]

\[
K_{\max}=2.93\times10^{-19}\,\text{J}
\]

**Step 4**: Calculate the maximum speed.

Rearrange the kinetic energy equation:

\[
v_{\max}=\sqrt{\frac{2K_{\max}}{m_e}}
\]

Substitute:

\[
v_{\max}=
\sqrt{\frac{2(2.93\times10^{-19})}
{9.11\times10^{-31}}}
\]

\[
v_{\max}=\sqrt{6.43\times10^{11}}
\]

\[
\boxed{v_{\max}=8.02\times10^5\,\text{m s}^{-1}}
\]

The fastest emitted electrons travel at approximately 802,000 metres per second.

Notice that we used the maximum kinetic energy, not the total photon energy. The work function must be subtracted before calculating speed.

### Worked example: Calculate speed and stopping potential

A photocell has a metal cathode with a work function of 2.40 eV. Monochromatic light of wavelength 310 nm shines on its surface.

Calculate:

a) The maximum kinetic energy of the emitted electrons.

b) Their maximum speed.

c) The stopping potential.

What happens to these quantities if the light intensity is doubled while the wavelength stays constant?

**Step 1**: Calculate the photon energy.

\[
E_{\text{photon}}=\frac{1240}{310}
\]

\[
E_{\text{photon}}=4.00\,\text{eV}
\]

**Step 2**: Calculate maximum kinetic energy.

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

\[
K_{\max}=4.00-2.40
\]

\[
\boxed{K_{\max}=1.60\,\text{eV}}
\]

**Step 3**: Convert to joules.

\[
K_{\max}=1.60\times1.602\times10^{-19}
\]

\[
K_{\max}=2.56\times10^{-19}\,\text{J}
\]

**Step 4**: Calculate maximum speed.

\[
v_{\max}=\sqrt{\frac{2K_{\max}}{m_e}}
\]

\[
v_{\max}=
\sqrt{\frac{2(2.56\times10^{-19})}
{9.11\times10^{-31}}}
\]

\[
\boxed{v_{\max}=7.50\times10^5\,\text{m s}^{-1}}
\]

**Step 5**: Calculate stopping potential.

Using:

\[
K_{\max}=eV_s
\]

The maximum kinetic energy is 1.60 eV, so:

\[
\boxed{V_s=1.60\,\text{V}}
\]

**Step 6**: Consider what happens when intensity doubles.

The photon energy has not changed because the wavelength remains 310 nm.

Each photon still has an energy of 4.00 eV, so the maximum kinetic energy remains 1.60 eV.

The maximum speed remains \(7.50\times10^5\,\text{m s}^{-1}\), and the stopping potential remains 1.60 V.

However, more photons arrive each second. Assuming the same collection conditions, the photocurrent increases.

This is the difference between increasing the number of emitted electrons and increasing the energy of each electron.

## 4. Common mistakes to avoid

Before attempting the questions, check these common errors.

| Mistake | Why it is wrong | What to do instead |
|---|---|---|
| Using \(hf\) as the electron's kinetic energy | Some photon energy is needed to release the electron | Subtract the work function |
| Forgetting to convert eV into joules | The kinetic energy equation requires joules when using electron mass in kg | Multiply by \(1.602\times10^{-19}\) |
| Using light intensity to calculate speed | Intensity affects photon arrival rate, not individual photon energy | Use wavelength or stopping potential |
| Assuming every emitted electron has maximum speed | Photoelectrons can leave with different kinetic energies | Use \(K_{\max}\) for the fastest electron only |
| Taking the square root of a negative kinetic energy | A negative result means emission is impossible under the model | Check the threshold condition first |
| Using the signed reverse voltage as a negative kinetic energy | Stopping potential is a positive magnitude | Use \(K_{\max}=eV_s\), with \(V_s>0\) |

One final detail: the speed calculated here is the electron's maximum speed immediately after emission from the metal. An electric field between the electrodes can subsequently increase or decrease its speed. The stopping-potential method is designed to determine its initial maximum kinetic energy.

The standard HSC single-photon model also assumes that one photon supplies the energy needed to release one electron. It does not account for unusual high-intensity multiphoton effects or detailed energy losses within a real metal surface. Use the work function supplied in the question rather than assuming every sample of a particular metal has exactly the same surface properties.

## Questions and solutions

The following questions progress from a straightforward wavelength calculation to experimental interpretation and a more challenging situation involving two wavelengths.

### Question 1

A metal has a work function of 2.20 eV. It is illuminated by monochromatic light with a wavelength of 350 nm.

Calculate the maximum speed of the emitted photoelectrons.

### Solution 1

The maximum speed is approximately:

\[
\boxed{v_{\max}=6.87\times10^5\,\text{m s}^{-1}}
\]

To find it, first calculate the energy of an incoming photon.

**Step 1**: Calculate photon energy.

\[
E_{\text{photon}}=\frac{1240}{350}
\]

\[
E_{\text{photon}}=3.543\,\text{eV}
\]

**Step 2**: Subtract the work function.

\[
K_{\max}=3.543-2.20
\]

\[
K_{\max}=1.343\,\text{eV}
\]

The positive result confirms that photoelectrons can be emitted.

**Step 3**: Convert to joules.

\[
K_{\max}=1.343\times1.602\times10^{-19}
\]

\[
K_{\max}=2.151\times10^{-19}\,\text{J}
\]

**Step 4**: Calculate maximum speed.

\[
v_{\max}=\sqrt{\frac{2K_{\max}}{m_e}}
\]

\[
v_{\max}=
\sqrt{\frac{2(2.151\times10^{-19})}
{9.11\times10^{-31}}}
\]

\[
\boxed{v_{\max}=6.87\times10^5\,\text{m s}^{-1}}
\]

The key is subtracting the work function before using the kinetic energy equation.

### Question 2

A metal has a work function of 2.75 eV.

It is exposed to two different monochromatic light sources:

- Source A has a wavelength of 400 nm.
- Source B has a wavelength of 470 nm.

Determine:

a) The threshold wavelength for this metal.

b) Whether each source causes photoelectric emission.

c) The maximum speed of electrons emitted by Source A.

Explain why increasing the intensity of Source B cannot make it produce photoelectrons.

### Solution 2

Source A produces photoelectrons with a maximum speed of approximately:

\[
\boxed{v_{\max}=3.51\times10^5\,\text{m s}^{-1}}
\]

Source B produces no photoelectrons, even if its intensity is increased.

**Step 1**: Calculate the threshold wavelength.

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

\[
\lambda_0=\frac{1240}{2.75}
\]

\[
\boxed{\lambda_0=450.9\,\text{nm}}
\]

**Step 2**: Compare each wavelength with the threshold.

For Source A:

\[
400\,\text{nm}<450.9\,\text{nm}
\]

Its wavelength is shorter than the threshold wavelength, so its photons have enough energy to release electrons.

For Source B:

\[
470\,\text{nm}>450.9\,\text{nm}
\]

Its photons do not have enough energy to release electrons.

**Step 3**: Calculate the maximum kinetic energy from Source A.

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

\[
E_{\text{photon}}=3.10\,\text{eV}
\]

Subtract the work function:

\[
K_{\max}=3.10-2.75
\]

\[
K_{\max}=0.35\,\text{eV}
\]

**Step 4**: Convert to joules.

\[
K_{\max}=0.35\times1.602\times10^{-19}
\]

\[
K_{\max}=5.61\times10^{-20}\,\text{J}
\]

**Step 5**: Calculate maximum speed.

\[
v_{\max}=
\sqrt{\frac{2(5.61\times10^{-20})}
{9.11\times10^{-31}}}
\]

\[
\boxed{v_{\max}=3.51\times10^5\,\text{m s}^{-1}}
\]

Source B provides the important contrast. Increasing its intensity increases the number of photons arriving each second, but every photon still has insufficient energy.

The threshold is determined by the energy of an individual photon, not the total energy delivered by the beam.

### Question 3

A photoelectric cell has an unknown metal cathode. Monochromatic light of wavelength 280 nm is directed onto its surface.

The cathode is maintained at 0 V. The collector is gradually made more negative until the photocurrent reaches zero at -1.40 V.

Determine:

a) The maximum kinetic energy of the emitted electrons.

b) Their maximum speed.

c) The work function of the metal.

d) The threshold wavelength of the metal.

[IMAGE: A photoelectric cell experimental setup showing a metal cathode at 0 V on the left and a collector electrode at -1.40 V on the right, inside an evacuated tube. Show monochromatic 280 nm light entering through a window and striking the cathode. Include a variable power supply, voltmeter, and ammeter. Label the electrode potentials and electron travel direction. Do not show calculated kinetic energy or speed.]

### Solution 3

The maximum speed is:

\[
\boxed{v_{\max}=7.02\times10^5\,\text{m s}^{-1}}
\]

The work function is approximately 3.03 eV, and the threshold wavelength is approximately 409 nm.

**Step 1**: Determine the stopping potential magnitude.

The collector is at -1.40 V and the cathode is at 0 V.

The signed potential difference is:

\[
\Delta V=-1.40\,\text{V}
\]

However, the stopping potential is its positive magnitude:

\[
V_s=1.40\,\text{V}
\]

**Step 2**: Calculate maximum kinetic energy.

Using:

\[
K_{\max}=eV_s
\]

The kinetic energy in electronvolts is:

\[
\boxed{K_{\max}=1.40\,\text{eV}}
\]

Convert to joules:

\[
K_{\max}=1.40\times1.602\times10^{-19}
\]

\[
K_{\max}=2.243\times10^{-19}\,\text{J}
\]

**Step 3**: Calculate maximum speed.

\[
v_{\max}=
\sqrt{\frac{2(2.243\times10^{-19})}
{9.11\times10^{-31}}}
\]

\[
\boxed{v_{\max}=7.02\times10^5\,\text{m s}^{-1}}
\]

**Step 4**: Calculate the incoming photon energy.

\[
E_{\text{photon}}=\frac{1240}{280}
\]

\[
E_{\text{photon}}=4.429\,\text{eV}
\]

**Step 5**: Determine the work function.

Rearrange Einstein's equation:

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

\[
\phi=4.429-1.40
\]

\[
\boxed{\phi=3.03\,\text{eV}}
\]

**Step 6**: Calculate the threshold wavelength.

\[
\lambda_0=\frac{1240}{3.029}
\]

\[
\boxed{\lambda_0=409.4\,\text{nm}}
\]

The negative collector potential does not mean the electron has negative kinetic energy. It means the electric field opposes the electron's motion and removes its initial kinetic energy.

The experiment lets us work backwards from a measured stopping potential to determine both the work function and maximum electron speed.

### Question 4

A student investigates a metal surface using three different light conditions. The following illustrative experimental data are recorded.

| Condition | Wavelength | Relative intensity | Saturation current | Stopping potential magnitude |
|---|---:|---:|---:|---:|
| A | 400 nm | \(I_0\) | 0.30 mA | 0.60 V |
| B | 400 nm | \(3I_0\) | 0.90 mA | 0.60 V |
| C | 300 nm | \(I_0\) | Not supplied | 1.63 V |

Assume all three conditions illuminate the same metal surface.

a) Calculate the maximum speed of photoelectrons under conditions A and B.

b) Calculate the maximum speed under condition C.

c) Use condition A to determine the work function of the metal. Check whether condition C is consistent with the same work function.

d) A student argues that condition B must produce electrons travelling three times faster than condition A because its saturation current is three times greater. Explain the error.

e) Explain why condition C produces faster photoelectrons than condition A.

[IMAGE: A photocurrent-versus-collector-potential graph showing three experimental curves. Curves A and B have the same negative stopping potential of -0.60 V, but B has three times the saturation current of A. Curve C reaches zero current at -1.63 V. Use distinct labelled curves and clearly marked stopping-potential intercepts. Do not add calculated electron speeds.]

### Solution 4

Conditions A and B produce identical maximum speeds of approximately \(4.59\times10^5\,\text{m s}^{-1}\). Condition C produces faster electrons, with a maximum speed of approximately \(7.57\times10^5\,\text{m s}^{-1}\).

**Part a: Maximum speed under conditions A and B**

The stopping potential is 0.60 V.

Therefore:

\[
K_{\max}=0.60\,\text{eV}
\]

Convert to joules:

\[
K_{\max}=0.60\times1.602\times10^{-19}
\]

\[
K_{\max}=9.613\times10^{-20}\,\text{J}
\]

Calculate speed:

\[
v_{\max}=
\sqrt{\frac{2(9.613\times10^{-20})}
{9.11\times10^{-31}}}
\]

\[
\boxed{v_{\max}=4.59\times10^5\,\text{m s}^{-1}}
\]

Both conditions have the same wavelength and stopping potential, so their maximum speeds are identical.

**Part b: Maximum speed under condition C**

The stopping potential is 1.63 V.

Therefore:

\[
K_{\max}=1.63\,\text{eV}
\]

Convert to joules:

\[
K_{\max}=1.63\times1.602\times10^{-19}
\]

\[
K_{\max}=2.612\times10^{-19}\,\text{J}
\]

Calculate speed:

\[
v_{\max}=
\sqrt{\frac{2(2.612\times10^{-19})}
{9.11\times10^{-31}}}
\]

\[
\boxed{v_{\max}=7.57\times10^5\,\text{m s}^{-1}}
\]

**Part c: Determine the work function**

Use condition A first.

The photon energy is:

\[
E_{\text{photon}}=\frac{1240}{400}
=3.10\,\text{eV}
\]

The maximum kinetic energy is 0.60 eV.

Therefore:

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

\[
\phi=3.10-0.60
\]

\[
\boxed{\phi=2.50\,\text{eV}}
\]

Now check condition C.

Its photon energy is:

\[
E_{\text{photon}}=\frac{1240}{300}
=4.133\,\text{eV}
\]

Its maximum kinetic energy is 1.63 eV.

Therefore:

\[
\phi=4.133-1.63
\]

\[
\boxed{\phi=2.503\,\text{eV}}
\]

The small difference is due to rounding. Both measurements are consistent with a work function of approximately 2.50 eV.

**Part d: Explain why condition B does not produce faster electrons**

The student's argument is incorrect.

Saturation current measures the rate at which electrons reach the collector. It does not measure their kinetic energy.

Condition B has three times the intensity of condition A. It therefore produces three times the saturation current in the supplied data.

However, both conditions have the same wavelength. Their photons have identical energies, so the maximum kinetic energy remains unchanged.

Consequently, their maximum speeds are identical.

**Part e: Explain why condition C produces faster electrons**

Condition C uses light with a shorter wavelength.

Since:

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

a shorter wavelength means a higher photon energy.

Condition C therefore supplies more energy to each electron. After the work function is paid, more energy remains as kinetic energy.

This produces a higher maximum speed.

The important distinction is that changing intensity affects the number of emitted electrons, while changing wavelength affects their maximum kinetic energy.

### Question 5

A metal surface has a work function of 2.20 eV. It is illuminated by a light source containing two narrow wavelength bands:

- Ultraviolet light with a wavelength of 250 nm.
- Green light with a wavelength of 500 nm.

Both wavelengths are capable of producing photoelectrons.

The green component produces most of the photocurrent. An experiment records the following observations:

- With both wavelengths present, the stopping potential magnitude is 2.76 V.
- A filter is inserted that completely removes the ultraviolet component without changing the green component.
- The photocurrent decreases by only 2%, but the stopping potential magnitude falls to 0.28 V.

Assume the measured stopping potentials are reliable, the ultraviolet component still produces detectable electrons before filtering, and the metal's work function remains unchanged.

a) Calculate the maximum speed of the emitted electrons before the filter is inserted.

b) Calculate the maximum speed after the ultraviolet component is removed.

c) Explain why the stopping potential changes so dramatically even though the photocurrent decreases by only 2%.

d) A student argues that the initial maximum speed should be calculated using the 500 nm wavelength because the green component produces most of the electrons. Explain precisely why this reasoning fails.

[IMAGE: A photoelectric cell illuminated by two simultaneous light sources. Show a relatively weak ultraviolet beam labelled 250 nm and a much stronger green beam labelled 500 nm entering from the left and striking the same metal cathode. Show the cathode and collector inside an evacuated tube, with a variable reverse-voltage circuit and ammeter. Include a removable filter positioned in the ultraviolet beam to illustrate how that component can be blocked. Do not show kinetic-energy arrows, calculated speeds, or stopping-potential results.]

### Solution 5

Before filtering, the maximum speed is approximately:

\[
\boxed{9.86\times10^5\,\text{m s}^{-1}}
\]

After filtering, it falls to:

\[
\boxed{3.14\times10^5\,\text{m s}^{-1}}
\]

The crucial idea is that stopping potential measures the energy of the fastest electrons, not the energy of the most common electrons.

**Part a: Calculate maximum speed before filtering**

The stopping potential magnitude is 2.76 V.

Therefore:

\[
K_{\max}=eV_s
\]

In electronvolts:

\[
K_{\max}=2.76\,\text{eV}
\]

Convert to joules:

\[
K_{\max}=2.76\times1.602\times10^{-19}
\]

\[
K_{\max}=4.422\times10^{-19}\,\text{J}
\]

Calculate speed:

\[
v_{\max}=
\sqrt{\frac{2(4.422\times10^{-19})}
{9.11\times10^{-31}}}
\]

\[
\boxed{v_{\max}=9.86\times10^5\,\text{m s}^{-1}}
\]

**Part b: Calculate maximum speed after filtering**

Once the ultraviolet component is removed, only the 500 nm green light remains.

The stopping potential is now 0.28 V.

Therefore:

\[
K_{\max}=0.28\,\text{eV}
\]

Convert to joules:

\[
K_{\max}=0.28\times1.602\times10^{-19}
\]

\[
K_{\max}=4.486\times10^{-20}\,\text{J}
\]

Calculate speed:

\[
v_{\max}=
\sqrt{\frac{2(4.486\times10^{-20})}
{9.11\times10^{-31}}}
\]

\[
\boxed{v_{\max}=3.14\times10^5\,\text{m s}^{-1}}
\]

**Part c: Explain the large change in stopping potential**

The ultraviolet component contributes very few electrons compared with the green component.

However, its photons have much higher energy.

We can verify this using the supplied work function.

For ultraviolet light:

\[
E_{\text{UV}}=\frac{1240}{250}
=4.96\,\text{eV}
\]

Therefore:

\[
K_{\max,\text{UV}}=4.96-2.20
\]

\[
K_{\max,\text{UV}}=2.76\,\text{eV}
\]

For green light:

\[
E_{\text{green}}=\frac{1240}{500}
=2.48\,\text{eV}
\]

Therefore:

\[
K_{\max,\text{green}}=2.48-2.20
\]

\[
K_{\max,\text{green}}=0.28\,\text{eV}
\]

These values agree exactly with the measured stopping potentials.

The ultraviolet photons produce the most energetic electrons, even though they contribute only a small proportion of the total photocurrent.

The stopping potential must be large enough to stop even these fastest electrons.

Once the ultraviolet component is removed, the fastest remaining electrons come from the green photons. Their maximum kinetic energy is much smaller, so a much smaller stopping potential is required.

The photocurrent falls by only 2% because the green component was already responsible for approximately 98% of the measured photocurrent.

**Part d: Explain why the most common wavelength does not determine maximum speed**

The student's reasoning confuses the most frequently emitted electrons with the fastest emitted electrons.

The green component produces most of the electrons, but this does not mean its electrons have the greatest kinetic energy.

Before filtering, even a small number of ultraviolet photons can produce electrons with greater kinetic energy than the electrons produced by the green photons.

The stopping-potential experiment detects the upper energy limit. It does not measure the average energy or the energy of the majority of electrons.

This means that the initial maximum speed must be calculated using the 250 nm ultraviolet wavelength, not the 500 nm green wavelength.

The distinction is subtle but important: **the maximum speed is determined by the highest-energy photons that successfully produce detectable photoelectrons, not by the wavelength responsible for most of the photocurrent.**

## Where to go next

Once you can calculate maximum photoelectron speed from wavelength, the next useful step is understanding how stopping potential changes with light frequency.

Combining Einstein's equation with the stopping-potential relationship gives:

\[
eV_s=hf-\phi
\]

Rearranging:

\[
V_s=\frac{h}{e}f-\frac{\phi}{e}
\]

This is a straight-line relationship between stopping potential and frequency. Its gradient can be used to determine Planck's constant, while its intercept provides information about the metal's work function.

If you need to calculate the wavelength of light produced by an atomic transition before using the photoelectric equation, the [Rydberg equation: calculating hydrogen wavelengths](https://tutorgum.com/learn/resources/hsc-physics-the-rydberg-equation/) is a useful next step.