How to Find Planck’s Constant from Stopping-Voltage Data

A bright red beam can produce a larger emission current than a dim blue beam, yet the blue beam can require a much larger reverse voltage to stop its electrons. If brightness meant more energy per electron, that result would be puzzling.

Before going further, predict which beam produces the faster photoelectrons. The blue beam does. Its higher frequency means that each photon carries more energy. The brighter red beam can still eject more electrons per second, but those electrons do not necessarily have greater kinetic energy.

This difference is exactly what allows us to measure Planck’s constant.

The method is straightforward: measure the stopping potential for several known light frequencies, plot stopping potential against frequency, and multiply the gradient by the elementary charge:

\[
\boxed{h=e\times\text{gradient}}
\]

The important part is understanding why this relationship works, and knowing which measurements actually belong in the calculation.

01The experiment: what are we measuring?

The photoelectric effect occurs when light strikes a suitable metal surface and ejects electrons.

To investigate it, we use a photoelectric cell, also called a photocell. Inside the evacuated tube are two electrodes:

• A photosensitive metal cathode, which emits electrons when illuminated.
• A collector or anode, which receives the emitted electrons.

Monochromatic light (light of a single frequency) enters through a window and strikes the cathode. The emitted electrons travel across the gap towards the anode.

An ammeter measures the resulting photocurrent. A variable power supply changes the potential difference between the electrodes, while a voltmeter measures that potential difference.

How do we measure the stopping potential?

Initially, the anode can be made positive relative to the cathode. This attracts the emitted electrons, allowing a photocurrent to flow.

Now reverse the potential difference.

The anode becomes negative relative to the cathode. Because electrons are negatively charged, the anode repels them.

Increase the reverse potential gradually. Fewer electrons have enough kinetic energy to reach the anode, so the photocurrent decreases.

Eventually, the photocurrent reaches zero.

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

It is the potential required to stop even the fastest emitted electrons from reaching the collector.

There is a sign convention worth getting right.

If the cathode is at zero potential and the anode is at a negative potential, then

\[
V_A-V_C=-V_s
\]

Here, \(V_s\) is the positive magnitude of the stopping potential. We use this positive magnitude in the calculations throughout this article.

Three quantities you must keep separate

These three measurements describe different things.

The emission current tells us how many electrons are being collected each second. It does not tell us how much kinetic energy each electron carries.

The stopping potential tells us about the maximum kinetic energy, not the total number of electrons emitted.

That distinction is the foundation of the experiment.

02Why frequency, rather than intensity, determines the stopping potential

The distinction became important when early experiments produced results that a simple classical wave model struggled to explain.

Consider what that model might suggest.

If light transfers energy continuously, increasing its intensity should increase the energy delivered to individual electrons. We might therefore expect brighter light to produce faster electrons.

But experiments did not support that prediction.

The experimental discoveries

In 1887, Heinrich Hertz observed that ultraviolet light affected the sparks produced across a spark gap.

Subsequent investigations established that light could cause electrons to be emitted from metal surfaces.

In 1902, Philipp Lenard investigated how the intensity of incident light affected photoelectric emission. His measurements showed that increasing intensity increased the rate of electron emission, but did not increase the maximum speed of the emitted electrons.

This was a crucial distinction.

The experimental observations suggested that:

• Increasing light intensity increased the number of emitted electrons.
• Increasing light frequency increased the maximum kinetic energy of those electrons.
• Below a certain frequency, electrons were not emitted, regardless of how intense the light became.

Einstein’s explanation

In 1905, Albert Einstein applied the idea of quantised energy to the photoelectric effect.

Instead of treating light energy as continuously distributed, he proposed that light transfers energy in individual packets.

These packets are now called photons.

Each photon has energy

\[
E=hf
\]

where:

• \(E\) is the energy of one photon, in joules.
• \(h\) is Planck’s constant, in joule seconds.
• \(f\) is the frequency of the light, in hertz.

An individual photon transfers its energy to an electron.

Some of that energy is needed to release the electron from the metal. Any remaining energy becomes the electron’s kinetic energy.

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

For the fastest emitted electrons:

\[
\text{Photon energy}
=
\text{Work function}
+
\text{Maximum kinetic energy}
\]

This gives

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

The single-photon model is the key to understanding why changing frequency and changing intensity produce different experimental results.

If you need to revisit the underlying idea of quantised photon energy, see tutorgum’s guide to Planck’s quantisation of light energy.

How Millikan turned this into a measurement of \(h\)

Einstein’s equation made a testable prediction: maximum electron kinetic energy should increase linearly with light frequency.

Robert Millikan investigated this prediction experimentally and published his photoelectric determination of Planck’s constant in 1916.

His apparatus used a vacuum tube containing a metal surface. Monochromatic light illuminated the surface, and an adjustable reverse potential was used to determine the stopping potential.

One important experimental problem was surface contamination.

Even a thin layer of contamination could change the effective work function of the metal. Millikan developed techniques to clean the metal surface inside the evacuated apparatus, rather than exposing it to air between measurements.

This made the measurements more reliable.

By measuring stopping potentials at different light frequencies, Millikan tested Einstein’s equation and obtained a value of approximately

\[
h=6.57\times10^{-34}\ \mathrm{J\,s}
\]

This was close to the accepted value.

There is an interesting historical detail: Millikan confirmed Einstein’s mathematical relationship, but remained sceptical of the photon interpretation for some time. His experimental results were stronger than his confidence in the underlying explanation.

The American Physical Society’s account of Millikan’s 1916 measurement provides further historical context.

03Turning stopping potential into Planck’s constant

We now have the physical explanation. Let’s connect it to the actual calculation.

From Einstein’s photoelectric equation:

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

Rearranging:

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

We also know that the stopping potential provides a way to measure the maximum kinetic energy.

When an electron moves against the retarding electric field, its kinetic energy is converted into electrical potential energy.

The electron has charge magnitude \(e\), so the electrical energy required to stop it is

\[
K_{\max}=eV_s
\]

Combining the two equations gives

\[
eV_s=hf-\phi
\]

Divide everything by \(e\):

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

This is the equation we need.

Why does this equation give us \(h\)?

Compare the photoelectric equation with the general straight-line equation:

\[
y=mx+b
\]

The stopping-potential equation is

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

Therefore:

This means that a graph of stopping potential against frequency should produce a straight line.

Its gradient is

\[
m=\frac{h}{e}
\]

Rearranging gives our required result:

\[
\boxed{h=em}
\]

This is how we find Planck’s constant from stopping-voltage data.

Understanding the gradient and its units

The gradient is calculated using

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

Stopping potential is measured in volts, while frequency is measured in hertz.

Therefore, the gradient has units of

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

We then multiply by the elementary charge:

\[
h=e\times m
\]

The units become

\[
\mathrm{C}\times\mathrm{V\,s}
\]

Since one volt is one joule per coulomb,

\[
\boxed{[h]=\mathrm{J\,s}}
\]

That is exactly the correct unit for Planck’s constant.

For HSC calculations, you will generally use:

\[
e=1.60\times10^{-19}\ \mathrm{C}
\]

\[
c=3.00\times10^8\ \mathrm{m\,s^{-1}}
\]

The accepted value of Planck’s constant is approximately

\[
h=6.63\times10^{-34}\ \mathrm{J\,s}
\]

The exact SI value is \(6.62607015\times10^{-34}\ \mathrm{J\,s}\), as specified by NIST’s fundamental constants.

What if the data is given as wavelength?

Sometimes the experiment provides wavelengths instead of frequencies.

You cannot directly plot stopping potential against wavelength and expect a straight line.

Remember:

\[
c=f\lambda
\]

Therefore,

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

Convert every wavelength into metres, calculate its corresponding frequency, and then plot stopping potential against frequency.

For example, a wavelength of \(500\ \mathrm{nm}\) becomes

\[
\lambda=500\times10^{-9}\ \mathrm{m}
\]

Therefore,

\[
f=\frac{3.00\times10^8}{500\times10^{-9}}
\]

\[
f=6.00\times10^{14}\ \mathrm{Hz}
\]

Use this frequency in your graph and calculations.

A useful alternative: finding \(h\) from two measurements

You do not always need to draw a graph.

Start with the equation:

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

For two measurements using the same metal:

\[
V_{s1}=\frac{h}{e}f_1-\frac{\phi}{e}
\]

\[
V_{s2}=\frac{h}{e}f_2-\frac{\phi}{e}
\]

Subtract the first equation from the second.

The work function disappears because the same metal is being used:

\[
V_{s2}-V_{s1}
=
\frac{h}{e}(f_2-f_1)
\]

Rearranging:

\[
\boxed{
h=e\frac{V_{s2}-V_{s1}}{f_2-f_1}
}
\]

This is simply the gradient method written using two data points.

It is particularly useful when an exam question gives you two measurements rather than a complete graph.

06Evaluating the data: what can go wrong?

Obtaining a numerical answer is only part of experimental analysis. You also need to decide whether the measurements are reliable.

1. Using intensity instead of frequency

Increasing intensity does not increase the maximum kinetic energy of individual photoelectrons.

If the frequency stays constant, the stopping potential should remain constant even when the intensity changes.

A student who uses different light intensities but keeps the frequency unchanged will not obtain the different stopping potentials needed to determine the gradient.

2. Using wavelength directly

The relationship between stopping potential and frequency is linear.

The relationship between stopping potential and wavelength is not.

Always convert wavelength into frequency before plotting the graph.

3. Using only one measurement

A single stopping-potential measurement cannot independently determine both Planck’s constant and the work function.

The equation contains two unknowns:

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

You need measurements at different frequencies to separate the gradient from the intercept.

4. Changing the metal surface

Different metals have different work functions.

Contamination can also change the effective work function of a metal surface.

Use the same clean photocathode throughout the investigation. Otherwise, a changing intercept may be mistaken for a change in the relationship being measured.

5. Using an unreliable stopping-potential reading

The photocurrent becomes very small near the stopping potential.

A meter with limited sensitivity may not identify the true zero-current point accurately.

Repeat measurements, use an appropriate sensitive current meter, and consider the uncertainty in the voltage readings.

A constant voltage offset can also shift the graph vertically. Interestingly, if the offset is the same for every measurement, the gradient can remain unchanged even though the calculated work function is affected.

6. Including frequencies below the threshold

Below the threshold frequency, the metal does not emit photoelectrons in the ideal single-photon model.

There is therefore no meaningful stopping potential to record.

The straight-line equation can be extrapolated mathematically below the threshold, but that does not mean negative stopping potentials are physically measured there.

07Questions and solutions

The following questions move from straightforward stopping-potential calculations towards more challenging experimental reasoning.

Question 1

A metal surface is illuminated with monochromatic light.

The measured stopping-potential magnitude is \(0.68\ \mathrm{V}\).

Calculate:

1. The maximum kinetic energy of the emitted electrons in joules.
2. The maximum kinetic energy in electronvolts.

If the intensity of the light is doubled while its frequency remains constant, what happens to the maximum kinetic energy?

Use \(e=1.60\times10^{-19}\ \mathrm{C}\).

Solution 1

The maximum kinetic energy is \(1.09\times10^{-19}\ \mathrm{J}\), or \(0.68\ \mathrm{eV}\). Doubling the intensity does not change the maximum kinetic energy.

The stopping potential is related to maximum kinetic energy by

\[
K_{\max}=eV_s
\]

Substituting:

\[
K_{\max}=(1.60\times10^{-19})(0.68)
\]

\[
\boxed{K_{\max}=1.088\times10^{-19}\ \mathrm{J}}
\]

To express this in electronvolts, remember that one electronvolt is the energy gained by an electron moving through a potential difference of one volt.

Therefore,

\[
\boxed{K_{\max}=0.68\ \mathrm{eV}}
\]

Increasing the intensity increases the number of incident photons per second. This can increase the emission current.

However, the energy of each photon remains \(hf\). The maximum kinetic energy therefore remains unchanged.

The tempting mistake is to assume that more intense light must produce more energetic electrons. It produces more electrons, not necessarily faster ones.

Question 2

A photocathode is illuminated with monochromatic light of different wavelengths.

The following illustrative data are collected:

Calculate an experimental value for Planck’s constant.

Use:

\[
c=3.00\times10^8\ \mathrm{m\,s^{-1}}
\]

\[
e=1.60\times10^{-19}\ \mathrm{C}
\]

Explain why plotting stopping potential directly against wavelength would not produce a straight line.

Solution 2

The experimental estimate is

\[
\boxed{h=6.59\times10^{-34}\ \mathrm{J\,s}}
\]

The first step is to convert the wavelengths into frequencies.

Using

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

For \(600\ \mathrm{nm}\):

\[
f_1=
\frac{3.00\times10^8}
{600\times10^{-9}}
\]

\[
f_1=5.00\times10^{14}\ \mathrm{Hz}
\]

For \(500\ \mathrm{nm}\):

\[
f_2=6.00\times10^{14}\ \mathrm{Hz}
\]

For \(400\ \mathrm{nm}\):

\[
f_3=7.50\times10^{14}\ \mathrm{Hz}
\]

We can now use the first and last measurements to calculate the gradient.

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

\[
m=
\frac{1.60-0.57}
{(7.50-5.00)\times10^{14}}
\]

\[
m=\frac{1.03}{2.50\times10^{14}}
\]

\[
m=4.12\times10^{-15}\ \mathrm{V\,s}
\]

Multiply by the elementary charge:

\[
h=em
\]

\[
h=(1.60\times10^{-19})(4.12\times10^{-15})
\]

\[
\boxed{h=6.59\times10^{-34}\ \mathrm{J\,s}}
\]

Why can’t we plot against wavelength directly?

Because frequency and wavelength are inversely proportional:

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

Substituting this into the stopping-potential equation gives

\[
V_s=\frac{hc}{e\lambda}-\frac{\phi}{e}
\]

This is not a straight-line relationship in wavelength.

The graph must use frequency on the horizontal axis if we want its gradient to give \(h/e\).

Question 3

Two monochromatic light sources illuminate the same metal photocathode.

Both sources have the same optical power of \(1.00\ \mathrm{mW}\) and illuminate the same area.

Their frequencies are:

\[
f_A=5.00\times10^{14}\ \mathrm{Hz}
\]

\[
f_B=7.00\times10^{14}\ \mathrm{Hz}
\]

Assume:

• Both frequencies are above the threshold frequency.
• The same fraction of incident photons produces collected photoelectrons for each source.
• Each absorbed photon can produce one emitted electron.
• The collection efficiency is unchanged.

Determine:

1. Which source produces the greater maximum kinetic energy?
2. The ratio of the photon arrival rates, \(\dot N_B/\dot N_A\).
3. The difference between the stopping potentials.

Use \(h=6.63\times10^{-34}\ \mathrm{J\,s}\) and \(e=1.60\times10^{-19}\ \mathrm{C}\).

Solution 3

Source B produces electrons with greater maximum kinetic energy and requires a stopping potential approximately \(0.83\ \mathrm{V}\) greater than source A.

However, source B produces only about \(71.4\%\) of the photon arrival rate of source A. Under the stated assumptions, its emission current is therefore also about \(71.4\%\) as large.

Part 1: Compare maximum kinetic energy.

The photon energy is

\[
E=hf
\]

Since source B has the greater frequency, each of its photons carries more energy.

The metal’s work function is the same for both sources, so source B produces the greater maximum kinetic energy.

Part 2: Compare photon arrival rates.

Optical power is energy transferred per second.

Therefore,

\[
P=\dot N hf
\]

Rearranging:

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

Both sources have equal power, so

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

Substitute:

\[
\frac{\dot N_B}{\dot N_A}
=
\frac{5.00\times10^{14}}
{7.00\times10^{14}}
\]

\[
\boxed{\frac{\dot N_B}{\dot N_A}=0.714}
\]

Source B delivers fewer photons per second.

This happens because each of its photons carries more energy, so fewer photons are needed to transfer the same total energy each second.

Part 3: Compare stopping potentials.

For the same metal, the work function cancels when we subtract the photoelectric equations.

Therefore,

\[
\Delta V_s=\frac{h}{e}\Delta f
\]

Substitute:

\[
\Delta V_s=
\frac{6.63\times10^{-34}}
{1.60\times10^{-19}}
(2.00\times10^{14})
\]

\[
\boxed{\Delta V_s=0.829\ \mathrm{V}}
\]

Source B requires a stopping potential approximately \(0.83\ \mathrm{V}\) greater than source A.

The important distinction is that equal optical power does not mean equal photon arrival rates.

Source B has fewer, more energetic photons. Its emission current is smaller under the stated assumptions, but its maximum electron kinetic energy is greater.

Question 4

An experiment produces a straight-line graph of stopping potential against frequency for a particular metal.

The best-fit line has:

• Gradient: \(4.05\times10^{-15}\ \mathrm{V\,s}\)
• Vertical intercept: \(-1.25\ \mathrm{V}\)

Calculate:

1. Planck’s constant.
2. The work function in joules.
3. The threshold frequency.
4. The threshold wavelength.

Use:

\[
e=1.60\times10^{-19}\ \mathrm{C}
\]

\[
c=3.00\times10^8\ \mathrm{m\,s^{-1}}
\]

A student argues that the negative vertical intercept means the stopping potential becomes negative at zero frequency. Explain why this interpretation is incorrect.

Solution 4

The results are:

\[
\boxed{h=6.48\times10^{-34}\ \mathrm{J\,s}}
\]

\[
\boxed{\phi=2.00\times10^{-19}\ \mathrm{J}}
\]

\[
\boxed{f_0=3.09\times10^{14}\ \mathrm{Hz}}
\]

\[
\boxed{\lambda_0=972\ \mathrm{nm}}
\]

Part 1: Calculate Planck’s constant.

The gradient is

\[
m=\frac{h}{e}
\]

Therefore,

\[
h=em
\]

Substituting:

\[
h=(1.60\times10^{-19})(4.05\times10^{-15})
\]

\[
\boxed{h=6.48\times10^{-34}\ \mathrm{J\,s}}
\]

Part 2: Calculate the work function.

The vertical intercept is

\[
b=-\frac{\phi}{e}
\]

Therefore,

\[
\phi=-eb
\]

Substituting:

\[
\phi=(1.60\times10^{-19})(1.25)
\]

\[
\boxed{\phi=2.00\times10^{-19}\ \mathrm{J}}
\]

Part 3: Calculate the threshold frequency.

At the threshold frequency, the photon has exactly enough energy to release an electron.

Therefore,

\[
hf_0=\phi
\]

Rearranging:

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

\[
f_0=
\frac{2.00\times10^{-19}}
{6.48\times10^{-34}}
\]

\[
\boxed{f_0=3.09\times10^{14}\ \mathrm{Hz}}
\]

Part 4: Calculate the threshold wavelength.

Use

\[
c=f_0\lambda_0
\]

Rearranging:

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

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

\[
\lambda_0=9.72\times10^{-7}\ \mathrm{m}
\]

Therefore,

\[
\boxed{\lambda_0=972\ \mathrm{nm}}
\]

Why is the negative intercept not a physical stopping potential?

The student has extrapolated the mathematical straight line beyond the region where photoelectric emission occurs.

Below the threshold frequency, the photons do not have enough energy to release electrons.

Consequently, there is no photocurrent and no measurable stopping potential.

The negative vertical intercept is a mathematical feature of the linear model. It represents the negative work-function term, not a physically measured negative stopping potential.

Question 5

A particularly sensitive photoelectric experiment is performed using one clean metal photocathode.

Two monochromatic light sources are tested separately.

Their measurements are:

The two sources are then switched on simultaneously.

Source B contributes only \(0.10\%\) of the total emitted electrons. The remaining \(99.90\%\) are produced by source A.

Answer the following:

1. Calculate Planck’s constant from the two separate measurements.
2. Determine the work function of the metal.
3. When both sources operate simultaneously, determine the ideal stopping-potential magnitude. Explain why a frequency-weighted average would give an incorrect answer.
4. The photon arrival rate from source B is now reduced by a factor of \(10^6\), but source B is not switched off completely. What happens to the ideal stopping potential?
5. Explain why a real experiment might nevertheless record a different stopping potential after this reduction.

Assume the same metal surface throughout, one-photon photoelectric emission, no secondary emission, and an ideal detector when determining the theoretical stopping potential.

Solution 5

The results are:

\[
\boxed{h=6.64\times10^{-34}\ \mathrm{J\,s}}
\]

\[
\boxed{\phi=3.19\times10^{-19}\ \mathrm{J}}
\]

The ideal stopping potential with both sources operating is

\[
\boxed{V_s=0.91\ \mathrm{V}}
\]

Even though source B contributes only a tiny fraction of the electrons, it determines the stopping potential because it produces the most energetic electrons.

Part 1: Calculate Planck’s constant.

The two sources illuminate the same metal, so their work functions are identical.

Use the difference equation:

\[
h=e\frac{\Delta V_s}{\Delta f}
\]

Substitute:

\[
h=
(1.60\times10^{-19})
\frac{0.91-0.08}
{(7.00-5.00)\times10^{14}}
\]

\[
h=
(1.60\times10^{-19})
\frac{0.83}{2.00\times10^{14}}
\]

\[
\boxed{h=6.64\times10^{-34}\ \mathrm{J\,s}}
\]

Part 2: Determine the work function.

Use the measurements from source A:

\[
hf_A=\phi+eV_{sA}
\]

Rearrange:

\[
\phi=hf_A-eV_{sA}
\]

Substitute:

\[
\phi=
(6.64\times10^{-34})(5.00\times10^{14})
–
(1.60\times10^{-19})(0.08)
\]

\[
\phi=3.32\times10^{-19}
-1.28\times10^{-20}
\]

\[
\boxed{\phi=3.19\times10^{-19}\ \mathrm{J}}
\]

The work function is approximately \(2.00\ \mathrm{eV}\).

Part 3: Determine the stopping potential when both sources operate.

This is the important conceptual part.

Source A produces electrons with a maximum kinetic energy of

\[
K_{\max,A}=eV_{sA}
\]

\[
K_{\max,A}=(1.60\times10^{-19})(0.08)
\]

\[
K_{\max,A}=1.28\times10^{-20}\ \mathrm{J}
\]

Source B produces electrons with a maximum kinetic energy of

\[
K_{\max,B}=eV_{sB}
\]

\[
K_{\max,B}=(1.60\times10^{-19})(0.91)
\]

\[
K_{\max,B}=1.456\times10^{-19}\ \mathrm{J}
\]

Source B produces the more energetic electrons.

When both sources are operating, the most energetic electrons still come from source B.

Therefore, the reverse potential must be large enough to stop these electrons as well as all the less energetic electrons from source A.

Hence,

\[
\boxed{V_s=0.91\ \mathrm{V}}
\]

A frequency-weighted average would be incorrect because stopping potential is not determined by the average energy of all the photons.

It is determined by the maximum kinetic energy of the emitted electrons.

The two photon populations do not combine to create photons with an intermediate frequency.

Part 4: Reduce the photon arrival rate from source B by a factor of \(10^6\).

The ideal stopping potential remains

\[
\boxed{V_s=0.91\ \mathrm{V}}
\]

This is a subtle result.

As long as source B continues to produce photoelectrons, some electrons can still reach the maximum kinetic energy associated with its frequency.

Reducing the photon arrival rate changes how many of these high-energy electrons are produced. It does not change the energy of each photon.

The highest possible electron kinetic energy therefore remains unchanged.

The total emission current also changes very little because source A already produces \(99.90\%\) of the electrons.

After reducing source B’s contribution by a factor of \(10^6\), its contribution becomes negligible. The total current falls by approximately \(0.10\%\), rather than by a factor of one million.

Part 5: Why might a real experiment record a different stopping potential?

A real ammeter has finite sensitivity.

Initially, the high-energy electrons from source B contribute \(0.10\%\) of the total emission current. After the reduction, their contribution becomes extremely small.

The current associated with these electrons might fall below the meter’s detection limit.

An experimentalist could then conclude that the photocurrent has reached zero before the true theoretical stopping potential is reached.

The measured stopping potential could therefore appear closer to the value associated with source A.

This is a limitation of the measurement, not a change in the underlying photoelectric equation.

The distinction is important: the theoretical stopping potential is determined by the highest-energy emitted electrons, while the experimentally recorded value can depend on whether the apparatus is sensitive enough to detect them.

08Where the measurement leads next

Once you can extract Planck’s constant from stopping-voltage data, the same graph gives you information about the metal itself.

The gradient gives \(h/e\), while the vertical intercept gives \(-\phi/e\). From the work function, you can calculate the threshold frequency and threshold wavelength.

You can also compare different metals. Their stopping-potential graphs should have the same gradient because Planck’s constant is universal, but different vertical intercepts because their work functions differ.

That is the next useful step: using the photoelectric graph not just to measure a fundamental constant, but to investigate how different materials interact with light.