Why Photoelectric Current Can Be Small Despite Electron Emission

A metal cathode can emit billions of electrons every second and still produce an ammeter reading of just \(0.03\,\mu\mathrm{A}\). It can even produce zero measured current while electrons are continuing to leave its surface.

There is no contradiction. Emission current measures how quickly electrons are released, while the measured photocurrent depends on how many of those electrons actually reach the collector. Neither quantity tells you the maximum kinetic energy of an individual electron.

This is the important distinction in the photoelectric effect.

Before going further, make a prediction. Suppose a metal is illuminated with light of a fixed frequency, and you increase the light intensity fourfold. Do the emitted electrons become four times more energetic?

No. The light supplies more photons each second, but each photon still has exactly the same energy. More electrons can be emitted each second, increasing the current, while the maximum kinetic energy remains unchanged.

To understand why, we need to separate three measurements:

• Emission current: how much charge is released from the metal each second.
• Stopping potential: the reverse voltage needed to prevent even the fastest photoelectrons from reaching the collector.
• Maximum kinetic energy: the greatest kinetic energy carried by an emitted electron.

01What does the ammeter actually measure?

Start with the apparatus.

A photoelectric cell contains two electrodes inside an evacuated glass tube:

• The cathode (C) is the photosensitive metal being investigated.
• The anode (A) is the collector that receives the emitted electrons.

Light enters through a window and strikes the cathode. If the photons have sufficient energy, electrons escape from the metal and travel across the vacuum towards the anode.

An ammeter is connected in series with the external circuit to measure the resulting current. A variable power supply allows the potential difference between the electrodes to be adjusted, and a voltmeter measures that potential difference.

The important detail is that the ammeter does not count every electron that escapes from the cathode.

It measures the charge that actually reaches the collector and passes through the external circuit.

Emitted electrons versus collected electrons

Suppose \(10^{12}\) electrons leave the cathode each second.

If every electron reaches the anode, the external circuit receives \(10^{12}\) electrons each second.

However, if only \(20\%\) reach the anode, the ammeter registers a current corresponding to just \(2\times10^{11}\) electrons each second.

We can describe these two rates separately:

• \(\dot N_{\mathrm{emit}}\): number of electrons emitted per second.
• \(\dot N_{\mathrm{col}}\): number of electrons collected per second.

The magnitude of electric current is charge transferred per second:

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

Since each electron carries a charge with magnitude

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

the current associated with electron emission is

\[
I_{\mathrm{emit}}=e\dot N_{\mathrm{emit}}
\]

The photocurrent actually measured by the ammeter is

\[
I_{\mathrm{ph}}=e\dot N_{\mathrm{col}}
\]

Therefore,

\[
I_{\mathrm{ph}}\leq I_{\mathrm{emit}}
\]

When the anode is sufficiently positive to collect all emitted electrons, the photocurrent reaches its maximum value. This is called the saturation current, \(I_s\).

In the ideal model,

\[
I_s\approx I_{\mathrm{emit}}
\]

Some textbooks use the term emission current for this saturated current. The distinction here is between the charge leaving the cathode and the charge successfully collected by the anode.

Why does such a small current still represent so many electrons?

An electron carries an extremely small amount of charge.

Consider an ammeter reading of

\[
I=0.03\,\mu\mathrm{A}
\]

Converting to amperes:

\[
I=3.0\times10^{-8}\ \mathrm{A}
\]

The number of electrons passing through the circuit each second is

\[
\dot N=\frac{I}{e}
\]

\[
\dot N=
\frac{3.0\times10^{-8}}
{1.60\times10^{-19}}
\]

\[
\dot N=1.875\times10^{11}\ \mathrm{s^{-1}}
\]

That is approximately 188 billion electrons every second.

So a current that looks tiny on an ammeter can represent an enormous number of electrons.

This is the same charge-per-second idea used when studying electric current in metals. The difference is that photoelectrons cross a vacuum gap before entering the external circuit.

02Why does increasing light intensity increase current?

The single-photon model explains why the number of emitted electrons and their kinetic energy are separate quantities.

According to this model, light transfers energy in individual packets called photons.

The energy of one photon is

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

where:

• \(E_{\mathrm{photon}}\) is the energy of one photon, in joules (J).
• \(h\) is Planck’s constant, \(6.63\times10^{-34}\ \mathrm{J\,s}\).
• \(f\) is the frequency of the light, in hertz (Hz).

One photon transfers its energy to one electron in the simplified photoelectric model.

Some of this energy is needed to release the electron from the metal. The minimum energy required for an electron to escape is called the work function, represented by \(\phi\).

Any remaining energy becomes kinetic energy.

For the fastest emitted electrons:

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

Rearranging:

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

Notice that intensity does not appear in this equation.

At a fixed frequency, every photon has the same energy, regardless of how bright the light is.

Increasing the intensity means increasing the number of photons arriving each second, rather than increasing the energy of each photon.

Consequently, more electrons can be released each second, but the maximum kinetic energy does not change.

The role of photon flux

The number of photons arriving each second is called the photon flux.

Not every incident photon necessarily releases an electron. Some photons may be reflected, and some absorbed photons do not produce an electron that escapes.

We can represent this using the quantum efficiency, \(\eta\), which is the fraction of incident photons that produce emitted electrons in the simplified model.

Therefore,

\[
\dot N_{\mathrm{emit}}=\eta\dot N_\gamma
\]

where \(\dot N_\gamma\) is the number of incident photons per second.

The emission current becomes

\[
I_{\mathrm{emit}}=e\eta\dot N_\gamma
\]

At a fixed frequency, increasing the light intensity increases the photon flux. If the quantum efficiency remains constant, the emission current increases proportionally.

This gives us an important comparison:

There is one important complication in the frequency rows.

Intensity measures energy arriving per second per unit area, not the number of photons arriving per second.

At the same optical power, higher-frequency photons carry more energy individually. Therefore, fewer photons arrive each second.

This means that increasing frequency does not automatically increase current. It increases the energy available to each emitted electron, but the number of emitted electrons also depends on photon flux.

03Why can the current become zero while emission continues?

So far, we’ve considered how many electrons the metal releases.

Now consider what happens after those electrons leave the surface.

Not all photoelectrons have identical kinetic energies or travel in identical directions. Some are faster than others, and their initial directions also affect whether they reach the collector.

The voltage between the electrodes controls how easily they can cross the gap.

We will use the sign convention

\[
V=V_A-V_C
\]

where:

• \(V_A\) is the electric potential of the anode.
• \(V_C\) is the electric potential of the cathode.

We will quote the magnitude of the photocurrent as a positive value.

Case 1: The anode is positive

When the anode is positive relative to the cathode, it attracts negatively charged electrons.

More emitted electrons reach the collector as the positive voltage increases.

The photocurrent therefore increases.

Eventually, the anode collects all the emitted electrons that can reach it. Increasing the voltage further produces no additional current.

The current has reached its saturation value, \(I_s\).

Case 2: The anode is at zero potential difference

You might predict that the current must be zero because there is no applied voltage pushing electrons towards the anode.

That prediction is incorrect.

Photoelectrons already possess kinetic energy when they leave the metal. Some have enough energy, and a suitable direction of motion, to reach the anode without an additional accelerating voltage.

Consequently, the photocurrent can be non-zero even when

\[
V_A-V_C=0
\]

It will generally be smaller than the saturation current because not every emitted electron reaches the collector.

Case 3: The anode is negative

Now the anode repels the emitted electrons.

As the anode becomes increasingly negative, electrons face a larger electrical potential-energy barrier when travelling towards it.

Slower electrons are turned back first. Faster electrons can still reach the collector.

Eventually, the reverse voltage becomes large enough to stop even the fastest emitted electrons.

The measured photocurrent becomes zero.

Stopping potential and maximum kinetic energy

The magnitude of the reverse voltage required to reduce the photocurrent to zero is called the stopping potential, \(V_s\).

Remember our sign convention:

\[
V_A-V_C=-V_s
\]

At this voltage, even the fastest photoelectrons cannot reach the anode.

The electrical potential-energy barrier is equal to the maximum initial kinetic energy of the emitted electrons.

Therefore,

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

where:

• \(K_{\max}\) is the maximum kinetic energy, in joules.
• \(e\) is the magnitude of the electron charge, \(1.60\times10^{-19}\ \mathrm{C}\).
• \(V_s\) is the positive magnitude of the stopping potential, in volts.

If kinetic energy is expressed in electronvolts (eV), the numerical value of the stopping potential in volts is the same.

For example,

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

means

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

or

\[
K_{\max}=1.28\times10^{-19}\ \mathrm{J}
\]

The stopping potential tells us about the fastest emitted electrons. It does not tell us how many electrons are being emitted.

This is why a photocell can continue emitting electrons while its measured current becomes very small or zero.

Reading the two important features of a photocurrent graph

A photocurrent-versus-voltage graph contains two separate pieces of information.

The two features are:

A common mistake is to interpret a larger saturation current as evidence that electrons have greater kinetic energy.

It isn’t.

A larger saturation current means more electrons are reaching the anode each second. A larger stopping potential means the fastest emitted electrons have greater kinetic energy.

For more practice identifying these features, see How to Sketch Photocurrent Against Applied Voltage.

04Questions and solutions

The following questions move from straightforward predictions to more demanding experimental reasoning. Pay particular attention to whether a question is asking about the number of electrons, their maximum energy, or the fraction that reaches the collector.

Question 1

A metal cathode is illuminated by monochromatic light above its threshold frequency.

The initial saturation current is \(0.24\,\mu\mathrm{A}\), and the stopping potential is \(0.55\ \mathrm{V}\).

The light intensity is tripled while its frequency, illuminated area, and the metal surface remain unchanged.

(a) Calculate the new saturation current.

(b) Determine the new stopping potential.

(c) Calculate the maximum kinetic energy of the emitted electrons in joules.

Solution 1

The new saturation current is \(0.72\,\mu\mathrm{A}\), while the stopping potential remains \(0.55\ \mathrm{V}\).

(a) New saturation current

At fixed frequency, tripling the intensity triples the photon flux.

Assuming constant quantum efficiency and complete collection at saturation,

\[
I_{s,\mathrm{new}}=3I_{s,\mathrm{initial}}
\]

\[
I_{s,\mathrm{new}}=3(0.24)
\]

\[
\boxed{I_{s,\mathrm{new}}=0.72\,\mu\mathrm{A}}
\]

(b) New stopping potential

The photon frequency has not changed, so the energy of each photon remains constant.

From

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

the maximum kinetic energy is unchanged.

Therefore, the stopping potential is also unchanged:

\[
\boxed{V_s=0.55\ \mathrm{V}}
\]

(c) Maximum kinetic energy

Using

\[
K_{\max}=eV_s
\]

\[
K_{\max}=(1.60\times10^{-19})(0.55)
\]

\[
\boxed{K_{\max}=8.80\times10^{-20}\ \mathrm{J}}
\]

The important distinction is that increasing intensity produces more electrons per second, not more energetic electrons.

Question 2

Two different monochromatic light beams illuminate the same metal cathode in separate experiments.

Both photocurrent measurements are taken at saturation.

(a) Calculate the number of electrons collected per second in each experiment.

(b) Calculate the maximum kinetic energy of the emitted electrons for each beam, in joules.

(c) Identify which beam produces more electrons per second and which produces faster photoelectrons.

(d) Can you conclude that Beam B has ten times the light intensity of Beam A? Explain.

Solution 2

Beam B produces ten times as many collected electrons per second, but Beam A produces electrons with four times the maximum kinetic energy.

(a) Electron emission rates

At saturation, assume all emitted electrons are collected.

Using

\[
\dot N=\frac{I_s}{e}
\]

For Beam A:

\[
I_A=0.016\times10^{-6}\ \mathrm{A}
\]

\[
\dot N_A=
\frac{1.6\times10^{-8}}
{1.60\times10^{-19}}
\]

\[
\boxed{\dot N_A=1.0\times10^{11}\ \mathrm{s^{-1}}}
\]

For Beam B:

\[
I_B=0.160\times10^{-6}\ \mathrm{A}
\]

\[
\dot N_B=
\frac{1.60\times10^{-7}}
{1.60\times10^{-19}}
\]

\[
\boxed{\dot N_B=1.0\times10^{12}\ \mathrm{s^{-1}}}
\]

Beam B therefore produces ten times the electron emission rate.

(b) Maximum kinetic energies

Using

\[
K_{\max}=eV_s
\]

For Beam A:

\[
K_{\max,A}=(1.60\times10^{-19})(1.10)
\]

\[
\boxed{K_{\max,A}=1.76\times10^{-19}\ \mathrm{J}}
\]

For Beam B:

\[
K_{\max,B}=(1.60\times10^{-19})(0.275)
\]

\[
\boxed{K_{\max,B}=4.40\times10^{-20}\ \mathrm{J}}
\]

(c) Comparing the beams

Beam B produces more electrons per second.

However, Beam A produces the fastest electrons because its maximum kinetic energy is greater.

In fact,

\[
\frac{K_{\max,A}}{K_{\max,B}}
=\frac{1.10}{0.275}=4
\]

The fastest electrons from Beam A have four times the kinetic energy of the fastest electrons from Beam B.

(d) Can we determine the intensity ratio?

No. The saturation current tells us the electron emission rate, not the incident optical intensity directly.

The number of emitted electrons also depends on photon frequency and quantum efficiency.

Without knowing the photon flux and quantum efficiency for each beam, we cannot conclude that Beam B has ten times the intensity.

The trap is assuming that a tenfold increase in current must mean a tenfold increase in light intensity.

Question 3

A photocell is illuminated continuously with monochromatic light of fixed frequency.

The following measurements are recorded as the potential difference between the anode and cathode is changed.

Assume the photocurrent has reached saturation by \(+1.50\ \mathrm{V}\).

(a) Determine the stopping potential.

(b) Calculate the maximum kinetic energy of the emitted electrons in joules and electronvolts.

(c) Determine the saturation current.

(d) Calculate the number of electrons collected per second at saturation.

(e) What fraction of the emitted electrons are collected when the applied voltage is zero? Assume that the saturation current represents the total emission current and that other losses are negligible.

Solution 3

The stopping potential is \(1.00\ \mathrm{V}\), the maximum kinetic energy is \(1.60\times10^{-19}\ \mathrm{J}\), and approximately \(39\%\) of the emitted electrons are collected at zero applied voltage.

(a) Stopping potential

The stopping potential occurs when the photocurrent becomes zero.

From the table,

\[
V_A-V_C=-1.00\ \mathrm{V}
\]

Therefore, the positive magnitude of the stopping potential is

\[
\boxed{V_s=1.00\ \mathrm{V}}
\]

(b) Maximum kinetic energy

Using

\[
K_{\max}=eV_s
\]

\[
K_{\max}=(1.60\times10^{-19})(1.00)
\]

\[
\boxed{K_{\max}=1.60\times10^{-19}\ \mathrm{J}}
\]

In electronvolts,

\[
\boxed{K_{\max}=1.00\ \mathrm{eV}}
\]

(c) Saturation current

The current reaches a constant value of \(0.80\,\mu\mathrm{A}\) at positive anode voltages.

Therefore,

\[
\boxed{I_s=0.80\,\mu\mathrm{A}}
\]

(d) Number of electrons collected per second

Convert the saturation current into amperes:

\[
I_s=0.80\times10^{-6}\ \mathrm{A}
\]

Using

\[
\dot N=\frac{I_s}{e}
\]

\[
\dot N=
\frac{0.80\times10^{-6}}
{1.60\times10^{-19}}
\]

\[
\boxed{\dot N=5.0\times10^{12}\ \mathrm{s^{-1}}}
\]

At saturation, approximately five trillion electrons reach the anode each second.

(e) Fraction collected at zero voltage

At zero applied voltage, the photocurrent is \(0.31\,\mu\mathrm{A}\).

The fraction collected is the ratio of the measured current to the saturation current:

\[
\text{Fraction collected}
=\frac{I_{\mathrm{ph}}}{I_s}
\]

\[
=\frac{0.31}{0.80}
\]

\[
=0.3875
\]

Therefore,

\[
\boxed{\text{Fraction collected}\approx39\%}
\]

This means approximately \(61\%\) of the emitted electrons do not reach the anode under these conditions.

Notice that the metal continues emitting electrons even though the measured current is smaller than the saturation current.

Question 4

A student performs two experiments using the same metal cathode and illuminated area.

The results are:

The student concludes:

1. Violet light produces electrons with greater maximum kinetic energy.
2. Violet light produces a greater emission current because its frequency is higher.

Evaluate both conclusions.

Then describe an experimental method that would allow the student to investigate the separate effects of light intensity and frequency.

Solution 4

The first conclusion is supported by the measurements, but the second cannot be established from this comparison.

First conclusion: Violet light produces greater maximum kinetic energy.

This is correct.

The stopping potential is greater for Experiment B:

\[
V_{s,B}=1.10\ \mathrm{V}
\]

compared with

\[
V_{s,A}=0.35\ \mathrm{V}
\]

Using

\[
K_{\max}=eV_s
\]

the maximum kinetic energy is greater in Experiment B.

This is consistent with the higher frequency of violet light.

Each violet photon carries more energy because

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

The additional photon energy can be transferred into the kinetic energy of the emitted electron after the work function has been overcome.

Second conclusion: Violet light produces a greater emission current because its frequency is higher.

This conclusion is not justified.

The violet light is also brighter. Therefore, the student has changed two variables simultaneously:

• Light frequency.
• Light intensity.

The higher saturation current could result from a greater photon flux caused by the higher intensity.

It cannot be attributed to frequency alone.

A better experimental method

The student should perform two separate comparisons.

First, investigate the effect of intensity:

• Keep the frequency constant.
• Keep the metal surface and illuminated area unchanged.
• Change the light intensity.
• Measure the saturation current and stopping potential for each intensity.

The expected result is that saturation current increases with intensity, while stopping potential remains unchanged.

Next, investigate the effect of frequency:

• Keep the metal surface and illuminated area unchanged.
• Change the frequency.
• Control the photon flux so that the same number of photons arrive each second.
• Measure the saturation current and stopping potential.

If the quantum efficiency remains constant, the saturation current should remain approximately unchanged, while stopping potential increases with frequency.

Controlling photon flux is important. Holding optical power constant instead would change the number of photons arriving each second as the frequency changes.

This experiment separates two different physical effects: the rate of electron emission and the energy transferred to each emitted electron.

Question 5

A clean metal cathode is illuminated by two different monochromatic light beams.

Beam A has a frequency of

\[
f_A=5.0\times10^{14}\ \mathrm{Hz}
\]

and a total optical power of \(P\).

When the anode is sufficiently positive to reach saturation, the photocurrent is

\[
I_{s,A}=0.40\,\mu\mathrm{A}
\]

The measured stopping potential is

\[
V_{s,A}=0.40\ \mathrm{V}
\]

Beam B has a frequency of

\[
f_B=7.0\times10^{14}\ \mathrm{Hz}
\]

Its total optical power is also \(P\), and it illuminates the same area of the cathode.

Assume:

• Both frequencies are above the threshold frequency.
• The cathode surface and its work function remain unchanged.
• The quantum efficiency is identical for both beams.
• All emitted electrons are collected when the anode is sufficiently positive.
• Other electrical and collection losses are negligible.

(a) Calculate the work function of the metal in electronvolts.

(b) Calculate the saturation current produced by Beam B.

(c) Calculate the stopping potential for Beam B.

(d) The anode is now set to a potential of \(-1.00\ \mathrm{V}\) relative to the cathode. Assume the emitted electrons have a spread of kinetic energies and directions, with some able to cross the resulting potential barrier and others unable to do so.

Will the measured current be zero, equal to the saturation current, or somewhere between these values?

(e) Can the exact current at this reverse voltage be calculated from the information provided? Explain why a calculation based only on the maximum kinetic energy would be insufficient.

Use

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

and

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

Solution 5

Beam B produces a saturation current of approximately \(0.286\,\mu\mathrm{A}\) and has a stopping potential of approximately \(1.23\ \mathrm{V}\). At a reverse voltage of \(-1.00\ \mathrm{V}\), the measured current is non-zero but below saturation. Its exact value cannot be determined from the supplied information.

The key is to recognise that increasing frequency while holding optical power constant decreases the number of photons arriving each second, even though each photon carries more energy.

(a) Calculate the work function

Start with Einstein’s photoelectric equation:

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

Rearrange:

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

We first need the maximum kinetic energy for Beam A.

Using its stopping potential:

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

\[
K_{\max,A}
=(1.60\times10^{-19})(0.40)
\]

\[
K_{\max,A}=6.40\times10^{-20}\ \mathrm{J}
\]

Now calculate the energy of a Beam A photon:

\[
E_A=hf_A
\]

\[
E_A=(6.63\times10^{-34})
(5.0\times10^{14})
\]

\[
E_A=3.315\times10^{-19}\ \mathrm{J}
\]

Therefore,

\[
\phi=3.315\times10^{-19}
-6.40\times10^{-20}
\]

\[
\phi=2.675\times10^{-19}\ \mathrm{J}
\]

Convert to electronvolts:

\[
\phi=
\frac{2.675\times10^{-19}}
{1.60\times10^{-19}}
\]

\[
\boxed{\phi=1.67\ \mathrm{eV}}
\]

(b) Calculate the saturation current for Beam B

This is where the constant optical power condition matters.

The energy of one photon is

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

The number of photons arriving per second is therefore

\[
\dot N_\gamma=\frac{P}{hf}
\]

Since both beams have the same total optical power,

\[
\frac{\dot N_{\gamma,B}}
{\dot N_{\gamma,A}}
=
\frac{f_A}{f_B}
\]

Substitute:

\[
\frac{\dot N_{\gamma,B}}
{\dot N_{\gamma,A}}
=
\frac{5.0\times10^{14}}
{7.0\times10^{14}}
\]

\[
=\frac{5}{7}
\]

Because the quantum efficiencies are identical, the emitted electron rates also have this ratio.

At saturation, the photocurrent is proportional to the emitted electron rate.

Therefore,

\[
\frac{I_{s,B}}{I_{s,A}}=\frac{5}{7}
\]

\[
I_{s,B}=0.40\times\frac{5}{7}
\]

\[
\boxed{I_{s,B}=0.286\,\mu\mathrm{A}}
\]

Although Beam B has a higher frequency, it produces a smaller saturation current.

This is because the same optical power is now distributed among photons with greater individual energy. Fewer photons arrive each second.

(c) Calculate the stopping potential for Beam B

The work function is unchanged because the metal surface is unchanged.

We already calculated:

\[
\phi=2.675\times10^{-19}\ \mathrm{J}
\]

Now calculate the energy of a Beam B photon:

\[
E_B=hf_B
\]

\[
E_B=(6.63\times10^{-34})
(7.0\times10^{14})
\]

\[
E_B=4.641\times10^{-19}\ \mathrm{J}
\]

Using Einstein’s equation:

\[
K_{\max,B}=hf_B-\phi
\]

\[
K_{\max,B}
=4.641\times10^{-19}
-2.675\times10^{-19}
\]

\[
K_{\max,B}=1.966\times10^{-19}\ \mathrm{J}
\]

Convert this into electronvolts:

\[
K_{\max,B}
=\frac{1.966\times10^{-19}}
{1.60\times10^{-19}}
\]

\[
K_{\max,B}=1.23\ \mathrm{eV}
\]

Using

\[
K_{\max}=eV_s
\]

the stopping potential is

\[
\boxed{V_{s,B}=1.23\ \mathrm{V}}
\]

The maximum kinetic energy has increased from \(0.40\ \mathrm{eV}\) to \(1.23\ \mathrm{eV}\), even though the saturation current has decreased.

(d) Predict the current at a reverse voltage of \(-1.00\ \mathrm{V}\)

The measured current will be non-zero but below the saturation current.

The reverse voltage creates an energy barrier of

\[
E_{\mathrm{barrier}}=e(1.00\ \mathrm{V})
\]

In electronvolts, this is

\[
E_{\mathrm{barrier}}=1.00\ \mathrm{eV}
\]

The maximum kinetic energy of the emitted electrons is

\[
K_{\max,B}=1.23\ \mathrm{eV}
\]

Since

\[
1.23\ \mathrm{eV}>1.00\ \mathrm{eV}
\]

some of the fastest electrons can still reach the anode.

However, the reverse potential prevents other electrons from reaching it.

Therefore,

\[
\boxed{0<I_{\mathrm{ph}}<I_{s,B}}
\]

The metal is still emitting electrons, but only a fraction are collected.

(e) Can we calculate the exact current?

No. We cannot determine the exact photocurrent from the maximum kinetic energy alone.

This is the subtle part of the question.

The stopping potential tells us the energy of the fastest emitted electrons. It does not describe the complete distribution of electron energies and directions.

To calculate the current at an intermediate reverse voltage, we would need to know how many electrons have sufficient kinetic energy in the direction of the collector to overcome the potential barrier.

We would also need to account for their directions of motion and the geometry of the photocell.

Knowing only \(K_{\max}\) tells us the upper limit of the energy distribution. It does not tell us how many electrons sit above the \(1.00\ \mathrm{eV}\) barrier.

The tempting approach is to calculate the fraction of maximum kinetic energy remaining after the reverse voltage is applied, then use that fraction to calculate the current.

That does not work. Current depends on the number of electrons collected, not the fraction of kinetic energy remaining.

This is why a maximum-energy measurement cannot, by itself, predict the photocurrent at every applied voltage.

The most useful habit to take into other photoelectric-effect questions is to identify exactly what the experiment is measuring before choosing an equation:

• If the question asks how many electrons are emitted or collected per second, think about current and photon flux.
• If it asks about the fastest electrons, use maximum kinetic energy.
• If it gives a reverse voltage at which the current becomes zero, identify the stopping potential.
• If it asks for current at an intermediate voltage, consider how many emitted electrons can actually reach the anode.

The next step is learning to interpret the complete photocurrent-versus-voltage graph, particularly how its saturation plateau and stopping-potential intercept reveal two different properties of the same photoelectric experiment.