How to Interpret a Negative Intercept on a Photoelectric Graph

A graph of maximum photoelectron kinetic energy against light frequency has a vertical intercept of \(-1.8\ \mathrm{eV}\). Does that mean the electrons have negative kinetic energy?

No. The negative vertical intercept represents the negative of the metal’s work function. The work function itself is positive: \(1.8\ \mathrm{eV}\) in this example.

The important detail is that the straight line has been extrapolated beyond the range where photoelectric emission actually occurs. Below a certain frequency, electrons are not emitted at all, so negative kinetic energy is never physically produced.

Let’s work through why this happens, how the intercept is measured, and how to avoid confusing it with photocurrent or stopping potential.

Before continuing, predict this: if the work function increases, should the negative intercept move closer to zero or further below it? It moves further below zero because more energy is needed to release an electron.

01Why a negative intercept appears

Start with the energy carried by one photon.

A photon of frequency \(f\) has energy:

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

Here:

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

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

However, the electron cannot simply take all that energy as kinetic energy. It first needs enough energy to escape the metal.

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

The remaining energy becomes the electron’s kinetic energy.

Therefore:

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

Rearranging:

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

Here, \(K_{\max}\) is the maximum kinetic energy of the emitted electrons.

The work function is always a positive energy requirement. The minus sign appears because this energy must be subtracted from the photon’s energy.

What happens at different frequencies?

There are three important situations.

1. Frequency below the threshold

Suppose a photon carries \(1.2\ \mathrm{eV}\), but the metal requires \(1.8\ \mathrm{eV}\) to release an electron.

The photon does not have enough energy.

No photoelectron is emitted in the standard single-photon model.

You cannot assign the electron a kinetic energy of \(-0.6\ \mathrm{eV}\). There is no emitted electron to have that kinetic energy.

2. Frequency exactly at the threshold

Now suppose the photon carries exactly \(1.8\ \mathrm{eV}\).

All the energy is used to release the electron. No energy remains for kinetic energy.

\[
K_{\max}=0
\]

This is the threshold condition:

\[
hf_0=\phi
\]

Rearranging:

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

The symbol \(f_0\) represents the threshold frequency: the minimum frequency that allows photoelectric emission.

3. Frequency above the threshold

Suppose the photon carries \(2.5\ \mathrm{eV}\).

The metal still requires \(1.8\ \mathrm{eV}\), leaving:

\[
K_{\max}=2.5-1.8
\]

\[
K_{\max}=0.7\ \mathrm{eV}
\]

The electron escapes with a maximum kinetic energy of \(0.7\ \mathrm{eV}\).

So why is the vertical intercept negative?

Return to:

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

Compare this with the general straight-line equation:

\[
y=mx+b
\]

The maximum kinetic energy is the vertical variable, and frequency is the horizontal variable.

The equation therefore gives:

\[
\text{Gradient}=h
\]

\[
\boxed{y\text{-intercept}=-\phi}
\]

The vertical intercept is negative because the work function is subtracted from the photon energy.

To find the vertical intercept mathematically, set \(f=0\):

\[
K_{\max}=h(0)-\phi
\]

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

This is an extrapolation, not a real photoelectric measurement.

At zero frequency, there is no photon energy available to release an electron. The actual experiment produces no photoelectrons below the threshold frequency.

The negative section of the line simply represents what the straight-line equation would produce if extended outside its physical range.

Remember the distinction:

• Work function: positive energy required to release an electron.
• Vertical intercept: negative of the work function.
• Horizontal intercept: positive threshold frequency.

02Don’t confuse photocurrent, stopping potential, and maximum kinetic energy

These three quantities are related, but they describe different parts of the experiment.

Let’s separate them properly.

1. Photocurrent: how many electrons are collected?

Photocurrent is the electric current produced when emitted electrons reach the collector.

Current measures the rate of charge flow:

\[
I=\frac{\Delta Q}{\Delta t}
\]

Here, \(I\) is current in amperes (A), \(\Delta Q\) is the charge transferred in coulombs (C), and \(\Delta t\) is the time interval in seconds (s).

Imagine illuminating the same metal with light of a fixed frequency, above its threshold.

Now increase the intensity.

What changes?

More photons arrive per second. This generally produces more emitted electrons per second, increasing the photocurrent.

However, each photon still has the same energy because its frequency has not changed.

Therefore, increasing intensity does not increase the maximum kinetic energy of the electrons in the single-photon model.

2. Stopping potential: how much voltage stops the electrons?

Photoelectric experiments use a variable voltage to oppose the motion of emitted electrons.

The stopping potential is the magnitude of the reverse voltage required to prevent even the fastest photoelectrons from reaching the collector.

We define \(V_s\) here as a positive magnitude, measured in volts (V).

The relationship is:

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

Here, \(e\) is the magnitude of the electron’s charge:

\[
e=1.602\times10^{-19}\ \mathrm{C}
\]

The actual electron charge is negative, \(-e\). We use the positive magnitude in this equation because \(V_s\) is defined as a positive stopping-voltage magnitude.

The important connection is that stopping potential measures the maximum kinetic energy indirectly.

A larger stopping potential means the fastest electrons have more kinetic energy.

3. Maximum kinetic energy: how much energy does the fastest electron have?

Maximum kinetic energy is determined by the photon’s energy and the work function:

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

It depends on frequency, not intensity, in the standard single-photon model.

This gives us a useful comparison:

The last row is important. Below threshold, you cannot measure a stopping potential for photoelectrons that are not being emitted.

03How the experiment produces a negative intercept

The negative intercept is not measured directly. It is calculated from measurements made using a photoelectric apparatus.

Understanding the apparatus makes the graph much easier to interpret.

The experimental setup

Imagine an evacuated glass tube containing two metal electrodes.

• The cathode is the metal surface being illuminated.
• The anode is the collector that receives emitted electrons.
• A monochromatic light source shines through a window onto the cathode.
• An ammeter measures the photocurrent in the circuit.
• A variable power supply changes the potential difference between the electrodes.
• A voltmeter measures the potential difference.

The tube is evacuated so that electrons can travel between the electrodes without collisions with air molecules.

What are experimenters trying to find?

The key question is:

Does increasing light intensity make the emitted electrons more energetic, or does it simply produce more electrons?

A measurement of photocurrent alone cannot answer this. It tells us how much charge reaches the collector, not the maximum kinetic energy of individual electrons.

The experimental method therefore needs an additional measurement.

Measuring the stopping potential

First, illuminate the cathode with monochromatic light.

The emitted electrons travel towards the collector, producing a photocurrent.

Next, reverse the potential difference so that the collector becomes negative relative to the cathode.

Because electrons are negatively charged, the electric force opposes their motion towards the collector.

Gradually increase the magnitude of this reverse voltage.

The photocurrent decreases because fewer electrons have enough kinetic energy to reach the collector.

Eventually, the current falls to zero.

At this point, even the fastest photoelectrons cannot reach the collector.

The magnitude of the applied reverse voltage is the stopping potential, \(V_s\).

We can now calculate the maximum kinetic energy:

\[
K_{\max}=eV_s
\]

Repeat this procedure for several different light frequencies, keeping the light monochromatic for each measurement.

Calculate the maximum kinetic energy for each frequency, then plot \(K_{\max}\) against \(f\).

The negative vertical intercept appears when the measured straight-line relationship is extrapolated back to zero frequency.

What did the experiment reveal?

A simple classical wave model suggested that increasing light intensity should increase the energy transferred to electrons.

However, the observations showed something different:

• Increasing intensity increased photocurrent.
• Increasing intensity did not increase stopping potential at a fixed frequency.
• Increasing frequency increased stopping potential.
• Below a threshold frequency, photoelectric emission did not occur.

In 1905, Einstein proposed an explanation using quantised packets of light energy, now called photons.

Each photon carries energy \(hf\), and an electron must receive enough energy from one photon to overcome the work function.

This explains why frequency determines maximum kinetic energy, while intensity primarily affects the number of photons arriving per second.

The experimental arrangement and the photon explanation are discussed in OpenStax’s treatment of the photoelectric effect.

04Read the graph’s axes before interpreting the intercept

There is one more detail worth getting right.

Not every photoelectric graph has the same vertical axis. The numerical value of the negative intercept depends on what is being plotted.

Case 1: Maximum kinetic energy against frequency

The equation is:

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

Therefore:

\[
\boxed{b=-\phi}
\]

If the vertical intercept is \(-1.5\ \mathrm{eV}\), the work function is:

\[
\phi=1.5\ \mathrm{eV}
\]

The negative sign belongs to the intercept, not the work function.

Case 2: Stopping potential against frequency

Start with:

\[
K_{\max}=eV_s
\]

Substitute into the photoelectric equation:

\[
eV_s=hf-\phi
\]

Divide by \(e\):

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

Therefore:

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

The vertical intercept is still negative, but it is now measured in volts rather than joules or electronvolts.

For example, a vertical intercept of \(-1.5\ \mathrm{V}\) corresponds to a work function of \(1.5\ \mathrm{eV}\).

To convert that work function into joules:

\[
\phi=1.5(1.602\times10^{-19})
\]

\[
\phi=2.40\times10^{-19}\ \mathrm{J}
\]

The numerical values in volts and electronvolts match for a single electron, but the units are different.

A sign-convention trap

Some experiments plot the actual signed collector potential rather than the positive magnitude of the stopping potential.

Let the signed potential be:

\[
V=V_{\text{anode}}-V_{\text{cathode}}
\]

Under reverse bias, \(V\) is negative.

Since \(V=-V_s\), the equation becomes:

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

Notice what has happened:

• The gradient is negative.
• The vertical intercept is positive.

This does not contradict the usual negative-intercept graph. The voltage has simply been defined differently.

Always check whether the graph uses the positive magnitude of stopping potential or the signed electrode potential.

For a more detailed treatment of graph gradients, threshold frequencies, and comparisons between metals, see Photoelectric Effect Graphs: Gradients and Intercepts Explained.

05Worked examples

06Questions and solutions

The following questions move from direct graph interpretation to situations where several photoelectric concepts interact.

Question 1

A metal has a maximum-kinetic-energy-versus-frequency graph with a vertical intercept of \(-2.07\ \mathrm{eV}\).

Given:

\[
h=4.14\times10^{-15}\ \mathrm{eV\,s}
\]

Calculate:

(a) The work function in eV.

(b) The work function in joules.

(c) The threshold frequency.

(d) Explain why the graph must not be interpreted as predicting negative kinetic energy below the threshold frequency.

Solution 1

The work function is \(2.07\ \mathrm{eV}\), equivalent to \(3.32\times10^{-19}\ \mathrm{J}\), and the threshold frequency is \(5.00\times10^{14}\ \mathrm{Hz}\).

(a) Work function in eV

The vertical intercept is the negative of the work function.

\[
\phi=-b
\]

\[
\boxed{\phi=2.07\ \mathrm{eV}}
\]

(b) Work function in joules

Use the conversion:

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

\[
\begin{aligned}
\phi&=2.07(1.602\times10^{-19})\\
&=3.32\times10^{-19}\ \mathrm{J}
\end{aligned}
\]

(c) Threshold frequency

At threshold:

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

\[
\begin{aligned}
f_0&=\frac{2.07}{4.14\times10^{-15}}\\
&=5.00\times10^{14}\ \mathrm{Hz}
\end{aligned}
\]

(d) Explain the negative section

The negative section is a mathematical extrapolation of the straight-line relationship.

Below \(5.00\times10^{14}\ \mathrm{Hz}\), photons do not have enough energy to overcome the work function. No photoelectrons are emitted, so there is no physical negative kinetic energy.

The tempting mistake is to treat every point on the extended line as an experimental result. Only the region at or above the threshold represents possible photoelectric emission in the ideal model.

Question 2

A photoelectric experiment plots the signed collector potential \(V_A-V_C\) against light frequency.

The following two points lie on the straight-line graph:

The collector potential is negative because the experiment uses a retarding voltage.

Given:

\[
h=4.14\times10^{-15}\ \mathrm{eV\,s}
\]

Calculate:

(a) The gradient.

(b) The vertical intercept.

(c) The work function in eV and joules.

(d) The threshold frequency.

(e) Explain why the vertical intercept is positive, even though the usual stopping-potential-magnitude graph has a negative intercept.

Solution 2

The gradient is approximately \(-4.10\times10^{-15}\ \mathrm{V\,s}\), the vertical intercept is \(+1.22\ \mathrm{V}\), and the work function is \(1.22\ \mathrm{eV}\).

(a) Calculate the gradient

\[
\begin{aligned}
m&=\frac{-1.24-(-0.83)}
{(6.0-5.0)\times10^{14}}\\
&=\frac{-0.41}{1.0\times10^{14}}\\
&=-4.10\times10^{-15}\ \mathrm{V\,s}
\end{aligned}
\]

(b) Calculate the vertical intercept

Use:

\[
V=mf+b
\]

Substitute the first point:

\[
-0.83=(-4.10\times10^{-15})(5.0\times10^{14})+b
\]

\[
-0.83=-2.05+b
\]

Therefore:

\[
\boxed{b=+1.22\ \mathrm{V}}
\]

(c) Calculate the work function

For the signed collector potential:

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

Therefore:

\[
\phi=eb
\]

In electronvolts:

\[
\boxed{\phi=1.22\ \mathrm{eV}}
\]

In joules:

\[
\begin{aligned}
\phi&=1.602\times10^{-19}\times1.22\\
&=1.95\times10^{-19}\ \mathrm{J}
\end{aligned}
\]

(d) Calculate the threshold frequency

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

\[
\begin{aligned}
f_0&=\frac{1.22}{4.14\times10^{-15}}\\
&=2.95\times10^{14}\ \mathrm{Hz}
\end{aligned}
\]

(e) Explain the positive intercept

The graph uses the signed collector potential, which is negative when the collector repels electrons.

The usual stopping-potential graph instead uses the positive magnitude of the reverse voltage.

Changing from \(V_s\) to \(-V_s\) reverses the sign of the gradient and intercept.

The physical work function has not changed. Only the voltage convention has changed.

This is why checking the axis label is essential before interpreting an intercept.

Question 3

A metal is illuminated with monochromatic light of a fixed frequency above its threshold frequency.

Two photocurrent-versus-potential curves are measured:

• At low intensity, the saturation current is \(0.30\ \mathrm{\mu A}\).
• At high intensity, the saturation current is \(0.90\ \mathrm{\mu A}\).

Both curves reach zero photocurrent when the signed collector potential is \(-0.72\ \mathrm{V}\).

Calculate:

(a) The maximum kinetic energy of the emitted electrons in eV.

(b) The maximum kinetic energy in joules.

(c) Explain why the saturation current increases but the stopping potential remains unchanged.

(d) Predict what happens to the stopping potential if the light frequency is reduced while remaining above threshold.

Solution 3

The maximum kinetic energy is \(0.72\ \mathrm{eV}\), or \(1.15\times10^{-19}\ \mathrm{J}\). Increasing intensity increases the number of emitted electrons, not their maximum kinetic energy.

(a) Maximum kinetic energy in eV

The stopping potential has a magnitude of \(0.72\ \mathrm{V}\).

Use:

\[
K_{\max}=eV_s
\]

For a single electron, the numerical value of the kinetic energy in eV equals the stopping potential in volts.

\[
\boxed{K_{\max}=0.72\ \mathrm{eV}}
\]

(b) Maximum kinetic energy in joules

\[
\begin{aligned}
K_{\max}&=1.602\times10^{-19}\times0.72\\
&=1.15\times10^{-19}\ \mathrm{J}
\end{aligned}
\]

(c) Explain the change in saturation current

The saturation current increases from \(0.30\ \mathrm{\mu A}\) to \(0.90\ \mathrm{\mu A}\).

This is a factor of three.

At a fixed frequency, increasing intensity increases the number of photons arriving per second. This produces more photoelectrons per second, increasing the saturation current.

However, the energy of each photon remains \(hf\).

Therefore, the maximum kinetic energy does not increase, and the stopping potential remains \(0.72\ \mathrm{V}\).

A common misconception is that a larger current means faster electrons. Current measures charge flow, not the kinetic energy of each electron.

(d) Explain the effect of reducing frequency

Reducing frequency reduces the energy of each photon.

Since:

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

the maximum kinetic energy decreases.

Consequently, a smaller reverse voltage is needed to stop the fastest electrons.

The stopping-potential magnitude decreases, provided the frequency remains above threshold.

If the frequency falls below threshold, photoelectric emission stops.

Question 4

Two different metal surfaces, A and B, are investigated using monochromatic light.

Their maximum-kinetic-energy graphs are described by:

Metal A:

\[
K_{\max,A}=4.14\times10^{-15}f-1.24
\]

Metal B:

\[
K_{\max,B}=4.14\times10^{-15}f-2.07
\]

The kinetic energies are measured in eV and frequency is measured in Hz.

Calculate or explain:

(a) The work function of each metal.

(b) The threshold frequency of each metal.

(c) Which metal emits electrons when illuminated with light of frequency \(4.0\times10^{14}\ \mathrm{Hz}\).

(d) The maximum kinetic energy and stopping potential for both metals when illuminated with light of frequency \(5.5\times10^{14}\ \mathrm{Hz}\).

(e) Explain why the gradients are identical but the vertical intercepts are different.

(f) Can you determine which metal produces the larger photocurrent from these equations alone? Explain.

Solution 4

Metal A has a work function of \(1.24\ \mathrm{eV}\), while metal B has a work function of \(2.07\ \mathrm{eV}\). Metal A has the lower threshold frequency and produces higher-energy photoelectrons at the same frequency when both metals emit.

(a) Calculate the work functions

The negative vertical intercept is the negative of the work function.

For metal A:

\[
\boxed{\phi_A=1.24\ \mathrm{eV}}
\]

For metal B:

\[
\boxed{\phi_B=2.07\ \mathrm{eV}}
\]

Metal B requires more energy to release an electron.

(b) Calculate the threshold frequencies

Use:

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

For metal A:

\[
\begin{aligned}
f_{0,A}&=\frac{1.24}{4.14\times10^{-15}}\\
&=3.00\times10^{14}\ \mathrm{Hz}
\end{aligned}
\]

For metal B:

\[
\begin{aligned}
f_{0,B}&=\frac{2.07}{4.14\times10^{-15}}\\
&=5.00\times10^{14}\ \mathrm{Hz}
\end{aligned}
\]

Metal B has a higher threshold frequency because its work function is larger.

(c) Determine which metal emits at \(4.0\times10^{14}\ \mathrm{Hz}\)

Compare the incident frequency with each threshold.

For metal A:

\[
4.0\times10^{14}>3.0\times10^{14}
\]

Emission occurs.

Its maximum kinetic energy is:

\[
\begin{aligned}
K_{\max,A}&=(4.14\times10^{-15})(4.0\times10^{14})-1.24\\
&=1.656-1.24\\
&=0.416\ \mathrm{eV}
\end{aligned}
\]

For metal B:

\[
4.0\times10^{14}<5.0\times10^{14}
\]

No photoelectric emission occurs.

Metal A emits electrons, but metal B does not.

(d) Calculate maximum kinetic energy at \(5.5\times10^{14}\ \mathrm{Hz}\)

First calculate the photon energy:

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

\[
E_{\text{photon}}=(4.14\times10^{-15})(5.5\times10^{14})
\]

\[
E_{\text{photon}}=2.277\ \mathrm{eV}
\]

For metal A:

\[
K_{\max,A}=2.277-1.24
\]

\[
\boxed{K_{\max,A}=1.037\ \mathrm{eV}}
\]

For metal B:

\[
K_{\max,B}=2.277-2.07
\]

\[
\boxed{K_{\max,B}=0.207\ \mathrm{eV}}
\]

Both metals emit electrons because the incident frequency is above both thresholds.

Now calculate the stopping potentials.

Using \(K_{\max}=eV_s\), the numerical values in eV and volts are equal:

\[
\boxed{V_{s,A}=1.037\ \mathrm{V}}
\]

\[
\boxed{V_{s,B}=0.207\ \mathrm{V}}
\]

Metal A produces electrons with greater maximum kinetic energy, so it requires a larger stopping potential.

(e) Explain the identical gradients and different intercepts

Both gradients are equal because the gradient represents Planck’s constant.

Planck’s constant is universal. It does not depend on which metal is being illuminated.

The vertical intercept depends on the work function:

\[
b=-\phi
\]

Metal B has a larger work function, so its vertical intercept is more negative.

(f) Can we determine which metal produces the larger photocurrent?

No. The equations describe maximum kinetic energy, not the number of electrons emitted per second.

Photocurrent also depends on factors such as photon flux, photoemission efficiency, and electron collection.

Even if the two metals receive light of the same frequency and intensity, their photoemission efficiencies may differ.

The tempting mistake is to assume that the metal with higher-energy electrons must produce a larger current. These are different physical quantities.

Question 5

A metal surface has a work function of \(2.00\ \mathrm{eV}\).

It is illuminated simultaneously by two monochromatic light beams:

• Beam A contains photons with energy \(3.20\ \mathrm{eV}\).
• Beam B contains photons with energy \(2.50\ \mathrm{eV}\).

Both beams have non-zero photon rates.

Assume the standard single-photon model, a stable metal surface, and an ideal detector capable of detecting any non-zero photocurrent.

A researcher increases the photon rate of beam B by a factor of 100 while leaving beam A unchanged.

Answer the following:

(a) Calculate the maximum kinetic energy of electrons produced by each beam.

(b) Determine the stopping potential when both beams illuminate the metal simultaneously.

(c) Explain what happens to the photocurrent and stopping potential when beam B’s photon rate increases by a factor of 100.

(d) If beam A is completely blocked, calculate the new stopping potential.

(e) A student suggests plotting the measured stopping potential against the intensity-weighted average frequency of the two beams. Explain why this does not produce a valid point on the standard single-frequency photoelectric graph.

(f) Now suppose beam A is weakened until only one photon arrives per minute, while beam B remains intense. Does the ideal stopping potential change? Explain the difference between the theoretical result and what might happen in a real experiment.

Solution 5

With both beams present, the stopping-potential magnitude is \(1.20\ \mathrm{V}\). Increasing the intensity of beam B increases its contribution to photocurrent but does not change the stopping potential.

The crucial point is that stopping potential is determined by the fastest emitted electrons, not the average energy of all the incident photons.

(a) Calculate the maximum kinetic energy from each beam

Use the single-photon equation:

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

For beam A:

\[
K_{\max,A}=3.20-2.00
\]

\[
\boxed{K_{\max,A}=1.20\ \mathrm{eV}}
\]

For beam B:

\[
K_{\max,B}=2.50-2.00
\]

\[
\boxed{K_{\max,B}=0.50\ \mathrm{eV}}
\]

Both beams have enough energy to produce photoelectrons.

However, beam A produces electrons with greater maximum kinetic energy.

(b) Determine the stopping potential

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

Beam A produces the fastest electrons:

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

Therefore:

\[
K_{\max}=eV_s
\]

\[
\boxed{V_s=1.20\ \mathrm{V}}
\]

Beam B does not determine the stopping potential because its electrons have a lower maximum kinetic energy.

(c) Explain the effect of increasing beam B’s photon rate

Increasing beam B’s photon rate produces more photoelectrons from that beam.

Its contribution to photocurrent increases.

However, the energy of each beam B photon remains \(2.50\ \mathrm{eV}\). Its maximum electron kinetic energy remains \(0.50\ \mathrm{eV}\).

Beam A is also still present, producing electrons with a maximum kinetic energy of \(1.20\ \mathrm{eV}\).

Consequently:

• The total photocurrent increases.
• The maximum kinetic energy remains \(1.20\ \mathrm{eV}\).
• The stopping potential remains \(1.20\ \mathrm{V}\).

The exact factor by which total photocurrent increases cannot be calculated without knowing the photoemission efficiencies of the two beams.

This is the important distinction: increasing the number of lower-energy electrons does not remove the higher-energy electrons already being produced.

(d) Calculate the stopping potential when beam A is blocked

Only beam B remains.

Its maximum kinetic energy is:

\[
K_{\max,B}=0.50\ \mathrm{eV}
\]

Therefore:

\[
\boxed{V_s=0.50\ \mathrm{V}}
\]

The stopping potential decreases because the most energetic remaining photoelectrons now come from beam B.

(e) Explain why the average frequency cannot be used

The standard photoelectric graph describes the relationship between maximum kinetic energy and the frequency of monochromatic light.

Here, two different frequencies are present.

Each photon transfers its own energy to an individual electron. The electrons produced by beam A and beam B therefore have different maximum kinetic energies.

The stopping potential is determined by the highest kinetic energy present, not the average kinetic energy.

An intensity-weighted average frequency would describe neither beam accurately. It would not represent the photon energy responsible for the fastest electrons.

Therefore, the mixed-light measurement cannot be treated as an ordinary single-frequency data point.

This is a particularly important distinction: the average energy of a collection of photons does not determine the maximum kinetic energy of the emitted electrons.

(f) Explain what happens when beam A becomes extremely weak

The ideal stopping potential remains:

\[
\boxed{V_s=1.20\ \mathrm{V}}
\]

Even if beam A produces only one photon per minute, that photon can still produce an electron with up to \(1.20\ \mathrm{eV}\) of kinetic energy.

Under the ideal assumptions, any non-zero rate of these high-energy electrons is enough to determine the maximum kinetic energy.

However, a real experiment has limitations.

If the high-energy electrons are extremely rare, the ammeter may not detect them reliably. Measurement noise, limited sensitivity, and the time allowed for measurement can make the apparent stopping potential difficult to determine.

That does not mean the actual maximum kinetic energy has changed. It means the experiment may not have detected the rare electrons responsible for it.

This final distinction is useful whenever a question involves several photon frequencies: the stopping potential is set by the fastest electrons that can be produced and detected, not by the average photon energy.

The central idea is simple: a negative vertical intercept is an energy requirement written with a minus sign, not a negative physical kinetic energy. Once you can distinguish the work function, photocurrent, and stopping potential, you can read these graphs much more confidently.

The next useful step is converting the maximum kinetic energy into the speed of the fastest emitted electrons. See How to Calculate Maximum Photoelectron Speed from Wavelength.