How to Find Work Function from Threshold Wavelength

A metal surface is illuminated with light of different wavelengths. At 450 nm, electrons are emitted. At 550 nm, no electrons are emitted, even when the light is much brighter.

What determines the difference? And if you know the longest wavelength that can eject an electron, can you calculate the energy needed to remove it from the metal?

Yes. The **work function** is found directly from the threshold wavelength using:

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

Here, \(\phi\) is the work function, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda_0\) is the threshold wavelength.

If the work function is required in electronvolts and the threshold wavelength is given in nanometres, use the convenient HSC Physics form:

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

For example, a metal with a threshold wavelength of 500 nm has a work function of:

\[
\phi=\frac{1240}{500}
=2.48\text{ eV}
\]

The important part is knowing why this equation works. It comes from the energy carried by a photon and the precise condition at which an electron can just escape from the metal.

## 1. Why the threshold wavelength determines the work function

Imagine an electron inside a metal. It is not completely free to leave. Energy is needed to overcome the attraction holding it within the metal.

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

Think of it as the minimum energy needed to get over a barrier. Once the electron has enough energy to escape, any additional energy can become kinetic energy.

In the single-photon model, one photon transfers its energy to one electron. The photon energy is:

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

where:
- \(E_{\text{photon}}\) is the energy of one photon, in joules (J).
- \(h=6.626\times10^{-34}\text{ J s}\) is Planck's constant.
- \(f\) is the frequency of the light, in hertz (Hz).

Since light frequency and wavelength are related by \(c=f\lambda\), we can also write:

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

The photon must supply the work function before any energy is available as kinetic energy.

Conservation of energy therefore gives Einstein's photoelectric equation:

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

or:

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

Here, \(K_{\max}\) is the maximum kinetic energy of the emitted photoelectrons, measured in joules.

[IMAGE: An accurate cross-sectional illustration of a metal surface showing individual incoming photons and electrons escaping. Show the work-function energy barrier and distinguish an electron that just escapes from one that escapes with additional kinetic energy. Label photon energy, work function, and maximum kinetic energy. Make the energy transfer physically clear without using generic process boxes.]

### What makes a wavelength the threshold wavelength?

Now consider what happens as the wavelength of the incoming light increases.

A longer wavelength means a lower frequency, and therefore less energy per photon.

Eventually, the photons have just enough energy to release electrons. There is no energy left over for electron motion.

This is the threshold condition:

\[
K_{\max}=0
\]

Substituting into Einstein's photoelectric equation:

\[
0=\frac{hc}{\lambda_0}-\phi
\]

Rearranging:

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

That's the entire derivation.

The threshold wavelength, \(\lambda_0\), is the **maximum wavelength that can produce photoelectric emission**. Its corresponding frequency, \(f_0\), is called the threshold frequency.

At the threshold:

\[
\boxed{hf_0=\phi}
\]

The two forms are equivalent because \(f_0=c/\lambda_0\).

[IMAGE: Three side-by-side energy comparisons at a metal surface. Case A: a short-wavelength photon supplies more than the work function, allowing an electron to escape with kinetic energy. Case B: a threshold photon supplies exactly the work function, with zero maximum kinetic energy. Case C: a longer-wavelength photon supplies insufficient energy, so no electron escapes. Clearly label the three cases and show the relative photon energies.]

The three possible outcomes are worth remembering:

| Incident wavelength | Photon energy | Result |
|---|---|---|
| \(\lambda<\lambda_0\) | Greater than the work function | Electrons are emitted with positive kinetic energy |
| \(\lambda=\lambda_0\) | Equal to the work function | Electrons are just able to escape |
| \(\lambda>\lambda_0\) | Less than the work function | No photoelectric emission |

Notice the direction of the inequality. A shorter wavelength means more energetic photons, so it is easier for them to release electrons.

A common mistake is to assume that the threshold wavelength is the shortest wavelength that produces emission. It is actually the longest.

02How do you measure the threshold wavelength?

Knowing the equation is useful. But where does the threshold wavelength come from in an actual experiment?

Imagine you have a metal cathode inside an evacuated glass tube. You shine light onto the cathode and measure the electrical current produced by electrons reaching a second electrode.

This is a photoelectric cell.

The experiment follows a simple sequence:

1. A monochromatic light source produces light of a known wavelength. Monochromatic means the light has one wavelength, or a very narrow range of wavelengths.
2. The light strikes the metal cathode, causing photoelectrons to be emitted if the photons have sufficient energy.
3. The electrons travel towards the positively charged anode.
4. An ammeter measures the resulting photocurrent.
5. The wavelength is gradually increased while the collecting conditions are kept constant.

Why increase the wavelength?

Because increasing wavelength decreases photon energy. Eventually, the photons no longer have enough energy to release electrons.

The longest wavelength that still produces detectable photoelectric emission gives an experimental estimate of the threshold wavelength.

What should the experiment show?

Suppose you start with ultraviolet light and gradually move towards longer wavelengths.

You should initially observe a photocurrent. As the wavelength increases, the current may decrease, and eventually photoelectric emission disappears.

The important observation is that increasing the intensity of light beyond the threshold wavelength cannot restore emission in the standard single-photon model.

A brighter source supplies more photons each second. It does not increase the energy carried by each photon.

A problem with simply measuring zero current

There is an important experimental complication.

A photocurrent of zero does not automatically mean that no electrons have been emitted.

What if electrons leave the cathode but cannot reach the anode?

This can happen when the anode is made negative relative to the cathode. Because electrons are negatively charged, the anode repels them.

A sufficiently strong reverse voltage prevents even the fastest photoelectrons from reaching the collector. The ammeter then reads zero, despite photoelectric emission still occurring.

The solution is to distinguish two different experimental situations:

• To identify the threshold wavelength directly, use a collecting voltage that attracts emitted electrons towards the anode.
• To measure the maximum kinetic energy, apply a reverse voltage and determine the voltage required to stop the fastest electrons.

The second measurement introduces another important quantity: the stopping potential.

03Photocurrent, stopping potential, and maximum kinetic energy

These three quantities are related, but they do different jobs.

Photocurrent: how many electrons reach the collector?

Imagine illuminating the same metal with two light sources of identical wavelength, but one is much brighter.

The brighter source supplies more photons per second. Assuming the same collection conditions, more electrons can reach the anode each second.

The photocurrent increases.

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

This is why photocurrent cannot be used to calculate the work function by itself.

Stopping potential: how much voltage stops the fastest electrons?

Now reverse the voltage across the electrodes.

Suppose the cathode is held at \(0\text{ V}\), and the collector is gradually made more negative.

The collector repels the electrons. As the reverse voltage increases, fewer electrons have enough kinetic energy to reach it.

Eventually, the photocurrent falls to zero. The magnitude of the reverse voltage at this point is called the stopping potential, \(V_s\).

The maximum kinetic energy is related to the stopping potential by:

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

where:

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

Be careful with the sign convention.

If the cathode is at \(0\text{ V}\) and the collector is at \(-0.80\text{ V}\), then:

\[ V_{\text{collector}}-V_{\text{cathode}} =-0.80\text{ V} \]

But the stopping potential is:

\[ V_s=0.80\text{ V} \]

We use its positive magnitude in \(K_{\max}=eV_s\).

There is also a useful shortcut. If maximum kinetic energy is expressed in electronvolts, its numerical value equals the stopping potential in volts.

For example:

\[ K_{\max}=0.80\text{ eV} \]

corresponds to:

\[ V_s=0.80\text{ V} \]

What happens at the threshold?

At the threshold wavelength, the fastest emitted electrons have zero kinetic energy in the ideal model.

Therefore:

\[ K_{\max}=0 \]

Using the stopping-potential relationship:

\[ eV_s=0 \]

So:

\[ \boxed{V_s=0\text{ V}} \]

This gives us another way to identify the threshold.

Instead of looking for the longest wavelength that produces a detectable photocurrent, we can measure stopping potentials at different light frequencies and extrapolate to where the stopping potential becomes zero.

From Einstein’s equation:

\[ eV_s=hf-\phi \]

Rearranging:

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

This is a straight-line equation of the form \(y=mx+b\).

If we plot stopping potential against frequency:

• The gradient is \(h/e\).
• The horizontal intercept is the threshold frequency, \(f_0\).
• At that intercept, \(V_s=0\).

We can then calculate:

\[ \lambda_0=\frac{c}{f_0} \]

and:

\[ \phi=hf_0 \]

This method is particularly useful because the stopping potential can be measured at several wavelengths where emission is clearly occurring. The threshold frequency is then estimated from the graph rather than relying on a photocurrent that is becoming too small to measure.

Notice that the extrapolation is important. Below the threshold frequency, no photoelectrons are emitted in the single-photon model, so there is no stopping potential to measure.

04A reliable method for finding the work function

Before attempting a calculation, identify exactly what the question gives you.

The distinction between an ordinary incident wavelength and a threshold wavelength is especially important.

Suppose a metal emits electrons when illuminated with 400 nm light. You cannot automatically calculate its work function using \(1240/400\).

Why not?

The 400 nm photons may carry more energy than the electrons need to escape. The remaining energy becomes kinetic energy.

Only when the wavelength is the threshold wavelength can you set the photon energy equal to the work function.

05Common mistakes when finding work function

A final experimental detail: the work function depends on the actual metal surface, not simply the name of the element. Surface contamination, oxidation, and other changes to the surface can affect the measured threshold. In HSC calculations, use the work function or threshold wavelength supplied in the question and assume the surface remains unchanged unless told otherwise.

For the next step, see How to Calculate Maximum Photoelectron Speed from Wavelength. It uses the same energy relationship to calculate the speed of the fastest emitted electrons.

05Questions and solutions

The following questions move from a direct threshold calculation to experimental interpretation, stopping-potential measurements, and a more demanding situation involving a non-uniform metal surface.

Question 1

A metal cathode has a threshold wavelength of \(580\text{ nm}\).

Calculate its work function in:

a) electronvolts (eV)

b) joules (J).

Solution 1

The work function is approximately:

\[ \boxed{\phi=2.14\text{ eV}} \]

or:

\[ \boxed{\phi=3.43\times10^{-19}\text{ J}} \]

Step 1: Use the threshold wavelength equation.

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

Step 2: Substitute the given wavelength.

\[ \phi=\frac{1240}{580} \]

\[ \boxed{\phi=2.14\text{ eV}} \]

Step 3: Convert into joules.

\[ \begin{aligned} \phi&=2.1379\times1.602\times10^{-19}\\\ &=3.425\times10^{-19}\text{ J} \end{aligned} \]

\[ \boxed{\phi=3.43\times10^{-19}\text{ J}} \]

The work function represents the minimum energy needed to release an electron. At \(580\text{ nm}\), the photon supplies exactly this energy in the ideal single-photon model.

Question 2

Two different metal cathodes, A and B, have the following threshold wavelengths:

• Metal A: \(440\text{ nm}\)
• Metal B: \(620\text{ nm}\)

Both cathodes are illuminated with monochromatic light of wavelength \(500\text{ nm}\), with equal photon flux.

a) Calculate the work function of each metal.

b) Determine which metal emits photoelectrons.

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

d) Explain whether increasing the intensity of the light incident on metal A would produce photoelectric emission.

Solution 2

Metal B emits photoelectrons, while metal A does not. The work functions are approximately \(2.82\text{ eV}\) and \(2.00\text{ eV}\), respectively.

The maximum kinetic energy of electrons emitted from metal B is:

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

Part a: Calculate both work functions.

For metal A:

\[ \phi_A=\frac{1240}{440} \]

\[ \boxed{\phi_A=2.82\text{ eV}} \]

For metal B:

\[ \phi_B=\frac{1240}{620} \]

\[ \boxed{\phi_B=2.00\text{ eV}} \]

Metal A has the greater work function, meaning more energy is needed to release its electrons.

Part b: Determine whether emission occurs.

First calculate the energy of an incoming photon:

\[ E_{\text{photon}}=\frac{1240}{500} \]

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

Compare this with each work function.

For metal A:

\[ 2.48\text{ eV}<2.82\text{ eV} \]

The photons do not have sufficient energy to release electrons.

For metal B:

\[ 2.48\text{ eV}>2.00\text{ eV} \]

The photons have enough energy to release electrons.

Part c: Calculate maximum kinetic energy.

Only metal B produces photoelectrons.

Using Einstein’s equation:

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

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

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

In joules:

\[ \begin{aligned} K_{\max}&=0.48\times1.602\times10^{-19}\\\ &=7.69\times10^{-20}\text{ J} \end{aligned} \]

Part d: Consider increasing the intensity.

Increasing the intensity of the light incident on metal A will not produce photoelectric emission.

The individual photons still have an energy of \(2.48\text{ eV}\), which is below the work function of \(2.82\text{ eV}\).

Increasing intensity supplies more photons per second, but each photon remains insufficiently energetic.

The distinction is between the number of photons and the energy of each photon.

Question 3

A photoelectric cell has a metal cathode held at \(0\text{ V}\) and a collector electrode held at \(-0.75\text{ V}\).

The cathode is illuminated with monochromatic light of wavelength \(400\text{ nm}\). The photocurrent just falls to zero at this reverse voltage.

Assume that the zero-current reading represents the stopping potential, rather than a failure of the light source or detector.

a) Calculate the maximum kinetic energy of the emitted photoelectrons in eV.

b) Determine the work function of the metal.

c) Calculate the threshold wavelength.

d) Explain why a student who identifies \(400\text{ nm}\) as the threshold wavelength is incorrect.

Solution 3

The work function is:

\[ \boxed{\phi=2.35\text{ eV}} \]

and the threshold wavelength is:

\[ \boxed{\lambda_0\approx528\text{ nm}} \]

The \(400\text{ nm}\) light is not threshold light. The electrons have additional kinetic energy, which is why a reverse voltage is needed to stop them.

Part a: Calculate maximum kinetic energy.

The collector is at \(-0.75\text{ V}\), while the cathode is at \(0\text{ V}\).

The stopping potential is the positive magnitude:

\[ V_s=0.75\text{ V} \]

Using:

\[ K_{\max}=eV_s \]

In electronvolts, this gives:

\[ \boxed{K_{\max}=0.75\text{ eV}} \]

Part b: Calculate the work function.

First calculate the photon energy:

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

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

Using Einstein’s equation:

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

Rearrange:

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

Substitute:

\[ \phi=3.10-0.75 \]

\[ \boxed{\phi=2.35\text{ eV}} \]

In joules:

\[ \phi=2.35\times1.602\times10^{-19} \]

\[ \boxed{\phi=3.76\times10^{-19}\text{ J}} \]

Part c: Calculate the threshold wavelength.

Now use the work function:

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

\[ \lambda_0=\frac{1240}{2.35} \]

\[ \boxed{\lambda_0=527.7\text{ nm}} \]

To three significant figures:

\[ \boxed{\lambda_0=528\text{ nm}} \]

Part d: Explain the misconception.

The \(400\text{ nm}\) light is shorter than the threshold wavelength of approximately \(528\text{ nm}\).

Its photons therefore carry more energy than the minimum required to release electrons.

The photocurrent reaches zero because the negative collector repels the emitted electrons strongly enough to prevent them from arriving.

It does not mean that photoelectric emission has stopped.

This is the key distinction: the stopping potential measures the maximum kinetic energy of emitted electrons, while the threshold wavelength identifies the minimum photon energy required for emission.

Question 4

A student investigates the relationship between light frequency and stopping potential for a metal cathode.

The following illustrative experimental data are recorded:

Assume the data follow an approximately straight-line relationship.

a) Determine the approximate gradient of the graph using the first and last data points.

b) Estimate the threshold frequency by extrapolating the graph to zero stopping potential.

c) Calculate the threshold wavelength.

d) Determine the work function in joules and electronvolts.

Solution 4

The estimated threshold frequency is:

\[ \boxed{f_0\approx5.81\times10^{14}\text{ Hz}} \]

The corresponding work function is:

\[ \boxed{\phi\approx2.40\text{ eV}} \]

and the threshold wavelength is approximately \(517\text{ nm}\).

Part a: Calculate the gradient.

The gradient is:

\[ m=\frac{\Delta V_s}{\Delta f} \]

Using the first and last data points:

\[ m=\frac{0.70-0.08} {(7.5-6.0)\times10^{14}} \]

\[ m=\frac{0.62}{1.5\times10^{14}} \]

\[ \boxed{m=4.13\times10^{-15}\text{ V s}} \]

This is close to the theoretical gradient \(h/e\), as predicted by Einstein’s photoelectric equation.

Part b: Find the threshold frequency.

The threshold frequency is the horizontal intercept, where:

\[ V_s=0 \]

Using the straight-line relationship:

\[ V_s=mf+b \]

We can use the first data point to estimate the intercept:

\[ 0=4.13\times10^{-15}f_0- \left(4.13\times10^{-15}\times6.0\times10^{14}-0.08\right) \]

Alternatively, rearranging from the first point:

\[ f_0=6.0\times10^{14} -\frac{0.08}{4.13\times10^{-15}} \]

\[ \boxed{f_0\approx5.81\times10^{14}\text{ Hz}} \]

Part c: Calculate the threshold wavelength.

Use:

\[ \lambda_0=\frac{c}{f_0} \]

Substitute:

\[ \lambda_0= \frac{3.00\times10^8} {5.81\times10^{14}} \]

\[ \lambda_0=5.16\times10^{-7}\text{ m} \]

Converting into nanometres:

\[ \boxed{\lambda_0\approx516\text{ nm}} \]

Part d: Calculate the work function.

Use the threshold frequency:

\[ \phi=hf_0 \]

Substitute:

\[ \phi=(6.626\times10^{-34}) (5.81\times10^{14}) \]

\[ \boxed{\phi\approx3.85\times10^{-19}\text{ J}} \]

Convert to electronvolts:

\[ \phi=\frac{3.85\times10^{-19}} {1.602\times10^{-19}} \]

\[ \boxed{\phi\approx2.40\text{ eV}} \]

The graph gives the same physical information as a direct threshold-wavelength measurement. The zero-stopping-potential intercept identifies the point where the fastest electrons have no excess kinetic energy.

The important detail is that the threshold frequency is extrapolated from measurements made above the threshold, where photoelectrons are actually emitted.

Question 5

A photoelectric cell has a cathode with two exposed surface regions, P and Q. The regions have different work functions because their surface conditions are different.

Their work functions are:

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

\[ \phi_Q=2.60\text{ eV} \]

Both regions are illuminated uniformly by two monochromatic light sources:

• A red source with wavelength \(600\text{ nm}\).
• A violet source with wavelength \(400\text{ nm}\).

The red source supplies 100 times as many photons per second as the violet source. The red source is observed to produce most of the total photocurrent.

Assume both regions are capable of collecting emitted electrons, the work functions remain unchanged, and there are no contact-potential or field-induced surface effects.

Determine:

a) The threshold wavelength of each surface region.

b) Which region emits electrons when illuminated by the red light, and which regions emit when illuminated by the violet light.

c) The longest wavelength that can produce photoelectric emission anywhere on the entire cathode.

d) The stopping potential when both light sources are present.

e) The stopping potential after the violet source is removed.

f) Explain why the red source can produce most of the photocurrent while the stopping potential is determined by the much weaker violet source.

g) A student measures the overall threshold wavelength of the cathode and uses it to claim that both regions have the same work function. Explain precisely why this conclusion is incorrect.

Solution 5

The longest wavelength that can produce emission anywhere on the cathode is:

\[ \boxed{\lambda_0=620\text{ nm}} \]

However, this only identifies the work function of region P, which has the lower work function.

With both sources present, the stopping potential is:

\[ \boxed{V_s=1.10\text{ V}} \]

After removing the violet source, it falls to:

\[ \boxed{V_s\approx0.067\text{ V}} \]

The central issue is that different parts of the cathode can have different work functions. The threshold wavelength of the whole surface is determined by the region that can release electrons using the lowest-energy photons.

Part a: Calculate the threshold wavelength of each region.

For region P:

\[ \lambda_{0,P}=\frac{1240}{2.00} \]

\[ \boxed{\lambda_{0,P}=620\text{ nm}} \]

For region Q:

\[ \lambda_{0,Q}=\frac{1240}{2.60} \]

\[ \boxed{\lambda_{0,Q}\approx477\text{ nm}} \]

Region P has a longer threshold wavelength because its work function is smaller.

Part b: Determine which regions emit electrons.

First calculate the photon energies.

For the red source:

\[ E_{\text{red}}=\frac{1240}{600} \]

\[ E_{\text{red}}=2.067\text{ eV} \]

For the violet source:

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

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

Compare these energies with the work functions.

For the red source:

• Region P has a work function of \(2.00\text{ eV}\). Since \(2.067>2.00\), electrons are emitted.
• Region Q has a work function of \(2.60\text{ eV}\). Since \(2.067<2.60\), no electrons are emitted.

For the violet source:

• Region P emits electrons because \(3.10>2.00\).
• Region Q also emits electrons because \(3.10>2.60\).

Therefore, the red source produces electrons only from P, while the violet source produces electrons from both P and Q.

Part c: Find the longest wavelength that produces emission anywhere on the cathode.

Region P has the longest threshold wavelength:

\[ \lambda_{0,P}=620\text{ nm} \]

Region Q requires shorter wavelengths, with a threshold of approximately \(477\text{ nm}\).

Therefore, the longest wavelength capable of producing emission anywhere on the cathode is:

\[ \boxed{\lambda_0=620\text{ nm}} \]

This corresponds to the lowest work function on the cathode.

It does not represent an average work function across both regions.

Part d: Calculate the stopping potential with both sources present.

The stopping potential is determined by the maximum kinetic energy of any emitted electron, not by the average energy or the number of electrons produced.

Calculate the maximum kinetic energy for each possible combination.

For red light on region P:

\[ K_{\max}=2.067-2.00 \]

\[ K_{\max}=0.067\text{ eV} \]

For violet light on region P:

\[ K_{\max}=3.10-2.00 \]

\[ K_{\max}=1.10\text{ eV} \]

For violet light on region Q:

\[ K_{\max}=3.10-2.60 \]

\[ K_{\max}=0.50\text{ eV} \]

The greatest kinetic energy is \(1.10\text{ eV}\), produced by violet photons striking region P.

Therefore:

\[ K_{\max}=eV_s \]

Using the numerical equivalence between eV and V:

\[ \boxed{V_s=1.10\text{ V}} \]

Even though the violet source supplies far fewer photons, it produces the fastest electrons.

Part e: Calculate the stopping potential after removing the violet source.

Only the red source remains.

From part b, red light produces electrons only from region P.

Its maximum kinetic energy is:

\[ K_{\max}=2.067-2.00 \]

\[ K_{\max}=0.067\text{ eV} \]

Therefore:

\[ \boxed{V_s\approx0.067\text{ V}} \]

The stopping potential falls from \(1.10\text{ V}\) to approximately \(0.067\text{ V}\).

Part f: Explain the difference between photocurrent and stopping potential.

The red source supplies 100 times as many photons per second as the violet source.

It can therefore produce most of the total photocurrent, even though its photons have less energy.

However, stopping potential measures the energy of the fastest emitted electrons.

The violet photons have greater energy and produce electrons with a maximum kinetic energy of \(1.10\text{ eV}\).

Removing the violet source therefore removes the most energetic electrons. The stopping potential falls sharply, even though the red source continues to produce substantial photocurrent.

This is why photocurrent and stopping potential must not be treated as interchangeable measurements.

Part g: Explain why one threshold wavelength does not establish both work functions.

The student’s conclusion is incorrect because the cathode is not a uniform surface.

The overall threshold wavelength is determined by the region with the lowest work function.

Here, region P has:

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

and a threshold wavelength of \(620\text{ nm}\).

Region Q has:

\[ \phi_Q=2.60\text{ eV} \]

and a threshold wavelength of approximately \(477\text{ nm}\).

The overall cathode can still emit electrons at wavelengths between approximately \(477\text{ nm}\) and \(620\text{ nm}\), because region P can release electrons even when region Q cannot.

A single threshold measurement therefore identifies the easiest-to-escape electrons on the surface. It does not prove that every region has the same work function.

The deeper lesson is that an experimental measurement can be dominated by an extreme rather than an average. The threshold wavelength identifies the lowest local work function, while the stopping potential identifies the highest kinetic energy among the electrons that are emitted.

06Where to go next

Once you can determine work function from threshold wavelength, you can use the same energy relationship to calculate the maximum kinetic energy of photoelectrons and their maximum speed.

The next useful connection is:

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

From there, you can apply:

\[ K_{\max}=\frac{1}{2}m_ev_{\max}^2 \]

to calculate the maximum photoelectron speed.

Understanding which part of the photon energy becomes kinetic energy is the key to moving confidently between threshold wavelength, work function, stopping potential, and electron speed.