Why Increasing Light Frequency Does Not Always Increase Photoelectric Current

A metal photocathode is illuminated with light. You increase the light frequency, making the photons more energetic, but keep the beam’s intensity (power per unit area) constant.

Do you expect the photocurrent to increase?

Not necessarily. In fact, under the usual single-photon model, if the emission efficiency stays constant and both frequencies are above the threshold, increasing frequency at constant intensity decreases the saturation current.

The reason is simple: each photon carries more energy, so fewer photons arrive each second when the total light power stays constant.

However, the maximum kinetic energy of the emitted electrons increases, as does the magnitude of the stopping potential.

These are three different quantities, and keeping them separate is the key to answering photoelectric-effect questions correctly.

01The important distinction: energy per photon versus photons per second

Imagine two light beams striking the same metal surface.

• Beam A contains lower-frequency red light.
• Beam B contains higher-frequency violet light.

Both beams deliver the same amount of energy to the surface every second.

Predict: Which beam contains more photons arriving each second?

Beam A does. Each red photon carries less energy, so more photons are needed to deliver the same total energy.

Beam B contains fewer photons, but each carries more energy.

This gives us an important distinction:

• Frequency determines the energy carried by each photon.
• Light intensity determines the energy arriving per second per unit area.
• Photon flux determines how many photons arrive per second per unit area.
• Photocurrent depends on how many electrons are emitted and collected each second.

The distinction becomes clearer when we use the equations.

Why higher frequency means fewer photons at constant intensity

The energy of an individual photon is:

\[ E_\gamma=hf \]

where:

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

Now consider light intensity.

Here, we use intensity to mean irradiance: the power arriving per unit area, measured in \(\text{W m}^{-2}\).

If the irradiance is \(S\), the photon flux is:

\[ \boxed{\Phi_\gamma=\frac{S}{hf}} \]

Photon flux is the number of photons arriving per square metre per second.

Notice that frequency is in the denominator.

If the irradiance stays constant and the frequency doubles:

\[ \Phi_{\gamma,2}=\frac{1}{2}\Phi_{\gamma,1} \]

The photon flux halves.

This is the central reason that increasing frequency does not automatically increase photocurrent.

Think of a conveyor delivering a fixed amount of energy every second. If each package contains more energy, fewer packages need to pass along the conveyor.

The analogy is useful for understanding the energy budget, although real photons interact probabilistically with the metal. Not every incident photon necessarily releases an electron.

02How the photoelectric experiment revealed the difference

The distinction between the number of emitted electrons and their energy came from an important experimental problem.

Early experiments showed that ultraviolet light could cause electrons to leave metal surfaces. But observing a metal lose charge did not tell physicists how many electrons were escaping each second or how much kinetic energy those electrons carried.

From observing charge loss to measuring electron energy

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

In 1888, Wilhelm Hallwachs demonstrated that ultraviolet light could cause a negatively charged zinc plate to lose its charge.

These observations established that light could help remove charge from a metal surface. However, an electroscope could not independently measure the emission rate and the kinetic energy of the escaping electrons.

A more controlled experiment was needed.

In 1902, Philipp Lenard investigated the properties of the electrons emitted when light struck a metal surface.

His experimental arrangement used an evacuated tube containing two electrodes.

The experiment worked as follows.

Step 1: Illuminate the cathode.

Light struck the metal cathode, causing photoelectrons to escape from its surface.

Step 2: Collect the electrons.

The electrons travelled through the evacuated space towards the anode.

The evacuated tube reduced collisions with air molecules, allowing the emitted electrons to reach the collector without losing significant kinetic energy through collisions.

Step 3: Measure the current.

An ammeter in the external circuit measured the photocurrent.

A larger current indicated that more charge was reaching the collector each second.

Step 4: Apply a reverse voltage.

Lenard could make the collector negative relative to the cathode.

The negative anode repelled the emitted electrons. Only electrons with sufficient kinetic energy could reach it.

As the reverse voltage increased, fewer electrons reached the collector.

Eventually, even the fastest electrons could no longer reach the anode. The current fell to zero.

This gave a way to determine the maximum kinetic energy of the emitted electrons.

What did Lenard discover?

Consider what a simple classical wave model might have predicted.

If light transferred energy continuously to electrons, increasing the intensity should increase the energy absorbed by individual electrons. More intense light should therefore produce more energetic photoelectrons.

But Lenard’s observations did not support this prediction.

He found that:

• Increasing light intensity increased the number of emitted electrons.
• Increasing light intensity did not increase their maximum velocity.
• Increasing light frequency increased the velocity of the emitted electrons.
• A minimum frequency was required for photoelectric emission.

The important observation was that the rate of emission and the energy of the emitted electrons responded differently to changes in the light.

Einstein proposed an explanation in 1905 using light quanta, now called photons. Each photon transfers its energy to an individual electron. The electron uses some of that energy to escape from the metal, and the remainder becomes kinetic energy.

This explanation connected the experimental observations to the single-photon model.

The historical sequence is discussed in the Nobel Prize’s educational account of the photoelectric effect and the University of Virginia’s explanation of Lenard’s measurements.

03Three measurements you must keep separate

The experiment gives us three quantities that are related but describe different physical properties.

1. Emission current: how many electrons leave the metal?

Electric current is the rate of charge flow.

If \(\dot N_e\) electrons are emitted each second, the emission current is:

\[ \boxed{I_{\mathrm{em}}=e\dot N_e} \]

where:

• \(I_{\mathrm{em}}\) is the emission current, measured in amperes (A).
• \(\dot N_e\) is the number of electrons emitted per second.
• \(e\) is the magnitude of the electron’s charge, \(1.60\times10^{-19}\ \text{C}\).

A larger emission current means more electrons are leaving the metal each second.

However, the current measured by an external ammeter depends on how many of those electrons actually reach the collector.

When the anode is sufficiently positive, all emitted electrons can be collected. The current reaches a maximum value called the saturation current.

Under these conditions:

\[ \boxed{I_s=I_{\mathrm{em}}} \]

where \(I_s\) is the saturation current.

This is the most useful current to compare when investigating how frequency affects the emission rate.

2. Maximum kinetic energy: how energetic are the fastest electrons?

Before an electron can escape the metal, it must overcome the metal’s work function.

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

A photon supplies energy according to:

\[ E_\gamma=hf \]

Some of this energy is used to release the electron. Any remaining energy becomes kinetic energy.

For the fastest emitted electrons:

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

where \(K_{\max}\) is the maximum kinetic energy, measured in joules.

This is Einstein’s photoelectric equation.

Notice that intensity does not appear in the equation.

For the same metal, increasing frequency increases the energy of each photon. Therefore, the maximum kinetic energy of the emitted electrons increases.

If the frequency changes from \(f_1\) to \(f_2\):

\[ \begin{aligned} K_{\max,1}&=hf_1-\phi\\ K_{\max,2}&=hf_2-\phi \end{aligned} \]

Subtracting gives:

\[ \boxed{\Delta K_{\max}=h\Delta f} \]

The increase in maximum kinetic energy depends directly on the change in frequency.

3. Stopping potential: how much voltage is needed to stop the fastest electrons?

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

Picture an electron travelling towards a negatively charged anode.

The negative anode creates an electrical potential-energy barrier. Electrons with lower kinetic energy are turned back first.

As the reverse voltage increases, the barrier becomes harder to overcome.

Eventually, even the fastest photoelectrons cannot reach the anode.

At this point, the photocurrent becomes zero.

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

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

where:

• \(V_s\) is the positive magnitude of the stopping potential, measured in volts (V).
• \(e\) is the magnitude of the electron’s charge.
• \(K_{\max}\) is measured in joules when using SI units.

Combining this with Einstein’s equation:

\[ \boxed{eV_s=hf-\phi} \]

Therefore, increasing frequency increases the stopping potential magnitude.

Be careful with the sign convention.

If the applied voltage is defined as

\[ V=V_A-V_C \]

where \(V_A\) is the anode potential and \(V_C\) is the cathode potential, the stopping point occurs at:

\[ V=-V_s \]

The stopping potential magnitude \(V_s\) is positive, even though the anode voltage at the stopping point is negative relative to the cathode.

Also, remember that stopping potential measures the maximum kinetic energy, not the average kinetic energy of all emitted electrons.

04How frequency changes the current and stopping potential

We can now combine the relationships.

The photon flux is:

\[ \Phi_\gamma=\frac{S}{hf} \]

Suppose the metal has a constant quantum efficiency, \(\eta\).

Quantum efficiency is the proportion of incident photons that result in emitted electrons in the simple single-photon model.

The number of emitted electrons per second is:

\[ \dot N_e=\eta A\Phi_\gamma \]

where \(A\) is the illuminated area of the metal.

Substituting into the current equation:

\[ I_{\mathrm{em}}=e\dot N_e \]

gives:

\[ I_s=e\eta A\Phi_\gamma \]

Therefore:

\[ \boxed{I_s=\frac{\eta eAS}{hf}} \]

This equation is particularly useful.

For constant irradiance, illuminated area, and quantum efficiency:

\[ \boxed{I_s\propto\frac{1}{f}} \]

Meanwhile:

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

and:

\[ V_s=\frac{hf-\phi}{e} \]

So increasing frequency can decrease the saturation current while increasing the maximum kinetic energy and stopping potential.

Compare the different situations

*Assuming unchanged quantum efficiency and complete electron collection.

There are two particularly important comparisons here.

Case 1: Increasing frequency while keeping irradiance constant.

The energy arriving each second stays the same, but each photon carries more energy.

There are fewer photons arriving each second, so fewer electrons are emitted if the emission efficiency remains constant.

The saturation current decreases.

However, the fastest electrons have greater kinetic energy, so the stopping potential magnitude increases.

Case 2: Increasing frequency while keeping photon flux constant.

Now the same number of photons arrives each second.

If the emission efficiency remains constant, the same number of electrons is emitted each second.

The saturation current stays constant.

However, each photon carries more energy, so the maximum kinetic energy and stopping potential increase.

Notice that maintaining the same photon flux while increasing frequency requires increasing the irradiance. These are two different experimental conditions.

What does the photocurrent-voltage graph show?

The distinction becomes particularly useful when interpreting a photocurrent-voltage graph.

The horizontal axis represents the applied potential difference, and the vertical axis represents photocurrent.

The graph contains two important features.

• The negative-voltage intercept tells you the stopping potential magnitude, which determines maximum kinetic energy.
• The positive-voltage plateau tells you the saturation current, which measures the rate of electron emission.

For the higher-frequency beam, under the constant-irradiance and constant-efficiency assumptions:

• The stopping potential magnitude is larger.
• The saturation current is smaller.

This is why you should never use the height of the current-voltage graph to determine the maximum kinetic energy.

The two features measure different things.

Why might a real experiment give a different current?

There is one more qualification worth understanding.

The relationship \(I_s\propto 1/f\) assumes that quantum efficiency remains constant.

In a real photoelectric experiment, the probability that an incident photon releases an electron can depend on frequency and the properties of the metal surface.

Consequently, increasing frequency can increase, decrease, or leave the saturation current approximately unchanged, depending on how the quantum efficiency changes.

There is also a difference between emission current and the current actually measured at a particular anode voltage.

If the anode is not sufficiently positive to collect every emitted electron, some electrons will return to the cathode. More energetic electrons may be able to reach the collector even when slower electrons cannot.

Therefore, a higher-frequency beam can sometimes produce a larger measured current at a particular voltage, even if its total emission rate is lower.

This does not contradict the photon model. It means the experiment is measuring both electron emission and electron collection.

For standard calculations, pay close attention to whether the question gives a saturation current or a current measured at a particular applied voltage.

05A reliable method for HSC questions

When a question changes light frequency, intensity, or wavelength, use this sequence.

Step 1: Identify what is being held constant.

Is it:

• Light intensity (power per unit area)?
• Total light power?
• Photon flux?
• Something else?

Do not assume these are interchangeable.

Step 2: Check the threshold frequency.

Calculate:

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

If \(f<f_0\), no photoelectrons are emitted in the standard single-photon model, regardless of the intensity.

Step 3: Determine the photon flux.

If irradiance is constant:

\[ \Phi_\gamma=\frac{S}{hf} \]

If photon flux is constant, the number of incident photons per second does not change.

Step 4: Determine the emission rate.

If quantum efficiency is constant:

\[ I_s=e\eta A\Phi_\gamma \]

Remember that saturation current measures the rate of electron emission when all electrons are collected.

Step 5: Calculate maximum kinetic energy separately.

Use:

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

Then determine the stopping potential magnitude:

\[ V_s=\frac{K_{\max}}{e} \]

Never assume that a larger stopping potential means a larger photocurrent.

06Worked examples

07Questions and solutions

Question 1

A metal surface is illuminated by monochromatic light with a photon flux of \(4.0\times10^{18}\ \text{photons m}^{-2}\text{s}^{-1}\).

The illuminated area is \(1.5\ \text{cm}^2\), and the quantum efficiency is 0.10.

Assuming all emitted electrons are collected, calculate the saturation current.

If the frequency increases while photon flux, illuminated area, and quantum efficiency remain constant, determine what happens to the saturation current.

Solution 1

The saturation current is \(9.6\ \mu\text{A}\), and it remains unchanged when frequency increases under the stated conditions.

Step 1: Convert the illuminated area into square metres.

\[ A=1.5\times10^{-4}\ \text{m}^2 \]

Step 2: Calculate the number of incident photons per second.

\[ \dot N_\gamma=\Phi_\gamma A \]

\[ \dot N_\gamma =(4.0\times10^{18})(1.5\times10^{-4}) \]

\[ \dot N_\gamma=6.0\times10^{14}\ \text{photons s}^{-1} \]

Step 3: Calculate the number of emitted electrons per second.

The quantum efficiency is 0.10, so:

\[ \dot N_e=0.10(6.0\times10^{14}) \]

\[ \dot N_e=6.0\times10^{13}\ \text{electrons s}^{-1} \]

Step 4: Calculate the saturation current.

\[ I_s=e\dot N_e \]

\[ I_s=(1.60\times10^{-19}) (6.0\times10^{13}) \]

\[ \boxed{I_s=9.6\times10^{-6}\ \text{A}} \]

Therefore:

\[ \boxed{I_s=9.6\ \mu\text{A}} \]

Increasing frequency does not change the current here because the photon flux is held constant.

The maximum kinetic energy and stopping potential would increase, provided both frequencies remain above threshold.

Question 2

A photocell has a work function of \(1.80\ \text{eV}\).

It is illuminated by two monochromatic beams with frequencies:

\[ f_1=5.0\times10^{14}\ \text{Hz} \]

\[ f_2=7.0\times10^{14}\ \text{Hz} \]

Both beams have an irradiance of \(1.5\ \text{W m}^{-2}\).

The illuminated area is \(0.80\ \text{cm}^2\), and the quantum efficiency is 0.25 for both frequencies.

Assume all emitted electrons are collected.

Calculate:

(a) The threshold frequency and whether emission occurs for both beams.

(b) The photon flux for each beam.

(c) The saturation current for each beam.

(d) The maximum kinetic energy and stopping potential magnitude for each beam.

Solution 2

Both frequencies produce photoelectric emission. The saturation current decreases from approximately \(14.5\ \mu\text{A}\) to \(10.3\ \mu\text{A}\), while the stopping potential magnitude increases from \(0.27\ \text{V}\) to \(1.10\ \text{V}\).

Step 1: Calculate the threshold frequency.

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

Using \(h=4.14\times10^{-15}\ \text{eV s}\):

\[ f_0=\frac{1.80}{4.14\times10^{-15}} \]

\[ \boxed{f_0=4.35\times10^{14}\ \text{Hz}} \]

Both incident frequencies are above this value, so both beams produce photoelectrons.

Step 2: Calculate the photon fluxes.

Use:

\[ \Phi_\gamma=\frac{S}{hf} \]

For the first beam:

\[ \Phi_{\gamma,1} = \frac{1.5} {(6.63\times10^{-34})(5.0\times10^{14})} \]

\[ \boxed{\Phi_{\gamma,1} =4.52\times10^{18}\ \text{photons m}^{-2}\text{s}^{-1}} \]

For the second beam:

\[ \Phi_{\gamma,2} = \frac{1.5} {(6.63\times10^{-34})(7.0\times10^{14})} \]

\[ \boxed{\Phi_{\gamma,2} =3.23\times10^{18}\ \text{photons m}^{-2}\text{s}^{-1}} \]

The higher-frequency beam has a lower photon flux.

Step 3: Calculate the saturation currents.

First convert the area:

\[ A=0.80\times10^{-4}\ \text{m}^2 \]

For the first beam:

\[ \dot N_{e,1}=\eta A\Phi_{\gamma,1} \]

\[ \dot N_{e,1} =0.25(8.0\times10^{-5})(4.52\times10^{18}) \]

\[ \dot N_{e,1}=9.04\times10^{13}\ \text{electrons s}^{-1} \]

Therefore:

\[ I_{s,1}=e\dot N_{e,1} \]

\[ I_{s,1} =(1.60\times10^{-19})(9.04\times10^{13}) \]

\[ \boxed{I_{s,1}=14.5\ \mu\text{A}} \]

For the second beam:

\[ \dot N_{e,2} =0.25(8.0\times10^{-5})(3.23\times10^{18}) \]

\[ \dot N_{e,2}=6.46\times10^{13}\ \text{electrons s}^{-1} \]

Therefore:

\[ I_{s,2} =(1.60\times10^{-19})(6.46\times10^{13}) \]

\[ \boxed{I_{s,2}=10.3\ \mu\text{A}} \]

Step 4: Calculate maximum kinetic energy.

For the first beam:

\[ K_{\max,1}=hf_1-\phi \]

\[ K_{\max,1}=2.07-1.80 \]

\[ \boxed{K_{\max,1}=0.27\ \text{eV}} \]

For the second beam:

\[ K_{\max,2}=hf_2-\phi \]

\[ K_{\max,2}=2.898-1.80 \]

\[ \boxed{K_{\max,2}=1.098\ \text{eV}} \]

Step 5: Determine the stopping potential magnitudes.

The numerical value of maximum kinetic energy in eV equals the stopping potential magnitude in volts.

Therefore:

\[ \boxed{V_{s,1}=0.27\ \text{V}} \]

\[ \boxed{V_{s,2}=1.10\ \text{V}} \]

The higher-frequency beam produces fewer photoelectrons per second, but the fastest electrons have greater kinetic energy.

This is exactly why photocurrent and stopping potential must be analysed separately.

Question 3

An experiment is performed using the same metal photocathode at two different frequencies.

The illuminated area, irradiance, and surface conditions are unchanged. The anode voltage is sufficiently positive to collect all emitted electrons.

The following illustrative experimental data are recorded:

Determine:

(a) Whether the current measurements are consistent with constant quantum efficiency.

(b) The work function of the metal using each beam.

(c) Whether the stopping-potential measurements support Einstein’s photoelectric equation.

Solution 3

Yes. The data are consistent with constant quantum efficiency and Einstein’s photoelectric equation, within rounding. Both measurements give a work function of approximately \(1.98\ \text{eV}\).

Step 1: Compare the saturation currents.

Since irradiance is constant:

\[ \Phi_\gamma\propto\frac{1}{f} \]

If quantum efficiency is unchanged, the saturation current follows the same relationship.

Therefore, the predicted current ratio is:

\[ \frac{I_{s,B}}{I_{s,A}} = \frac{f_A}{f_B} \]

\[ \frac{I_{s,B}}{I_{s,A}} = \frac{6.0}{8.0} \]

\[ \frac{I_{s,B}}{I_{s,A}}=0.75 \]

Now compare this with the measured current ratio:

\[ \frac{3.6}{4.8}=0.75 \]

The measured result matches the prediction.

The current data are consistent with constant quantum efficiency.

Step 2: Calculate the work function using Beam A.

The stopping potential gives the maximum kinetic energy:

\[ K_{\max}=eV_s \]

In electronvolts:

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

The photon energy is:

\[ E_{\gamma,A}=hf_A \]

\[ E_{\gamma,A} =(4.14\times10^{-15})(6.0\times10^{14}) \]

\[ E_{\gamma,A}=2.484\ \text{eV} \]

Using Einstein’s equation:

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

\[ \phi_A=2.484-0.50 \]

\[ \boxed{\phi_A=1.984\ \text{eV}} \]

Step 3: Calculate the work function using Beam B.

The photon energy is:

\[ E_{\gamma,B} =(4.14\times10^{-15})(8.0\times10^{14}) \]

\[ E_{\gamma,B}=3.312\ \text{eV} \]

The maximum kinetic energy is:

\[ K_{\max,B}=1.33\ \text{eV} \]

Therefore:

\[ \phi_B=3.312-1.33 \]

\[ \boxed{\phi_B=1.982\ \text{eV}} \]

The two values agree closely.

Step 4: Check the change in stopping potential.

The measured change is:

\[ \Delta V_s=1.33-0.50 \]

\[ \Delta V_s=0.83\ \text{V} \]

Einstein’s equation predicts:

\[ \Delta K_{\max}=h\Delta f \]

In electronvolts:

\[ \Delta K_{\max} =(4.14\times10^{-15})(2.0\times10^{14}) \]

\[ \Delta K_{\max}=0.828\ \text{eV} \]

This corresponds to an increase in stopping potential magnitude of \(0.828\ \text{V}\).

The measured increase of \(0.83\ \text{V}\) agrees with the prediction.

Interpretation: The experiment demonstrates both effects at once. Increasing frequency reduces the number of electrons emitted per second at constant irradiance, while increasing the maximum kinetic energy of those electrons.

Question 4

A metal has a work function of \(2.30\ \text{eV}\).

It is illuminated by two monochromatic beams with frequencies:

\[ f_1=5.0\times10^{14}\ \text{Hz} \]

\[ f_2=6.0\times10^{14}\ \text{Hz} \]

Both beams have an irradiance of \(2.0\ \text{W m}^{-2}\).

The illuminated area is \(1.0\ \text{cm}^2\), and the quantum efficiency is 0.10 whenever photoelectric emission occurs.

Assume complete electron collection.

(a) Calculate the threshold frequency.

(b) Determine whether photoelectric emission occurs for each beam.

(c) Calculate the saturation current for each beam.

(d) Calculate the stopping potential magnitude for the second beam.

Explain why the saturation current increases even though the photon flux decreases.

Solution 4

The saturation current increases from zero to approximately \(8.05\ \mu\text{A}\). This happens because the first beam is below the threshold frequency, while the second beam is above it.

Step 1: Calculate the threshold frequency.

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

\[ f_0= \frac{2.30}{4.14\times10^{-15}} \]

\[ \boxed{f_0=5.56\times10^{14}\ \text{Hz}} \]

Step 2: Determine whether emission occurs.

For the first beam:

\[ f_1=5.0\times10^{14}\ \text{Hz} \]

Since:

\[ f_1<f_0 \]

there is insufficient energy per photon to overcome the work function.

Therefore:

\[ \boxed{I_{s,1}=0} \]

For the second beam:

\[ f_2=6.0\times10^{14}\ \text{Hz} \]

Since:

\[ f_2>f_0 \]

photoelectric emission occurs.

Step 3: Calculate the photon flux of the second beam.

\[ \Phi_{\gamma,2}=\frac{S}{hf_2} \]

\[ \Phi_{\gamma,2} = \frac{2.0} {(6.63\times10^{-34})(6.0\times10^{14})} \]

\[ \boxed{\Phi_{\gamma,2} =5.03\times10^{18}\ \text{photons m}^{-2}\text{s}^{-1}} \]

Convert the illuminated area:

\[ A=1.0\times10^{-4}\ \text{m}^2 \]

The incident photon rate is:

\[ \dot N_\gamma=\Phi_\gamma A \]

\[ \dot N_\gamma =(5.03\times10^{18})(1.0\times10^{-4}) \]

\[ \dot N_\gamma=5.03\times10^{14}\ \text{photons s}^{-1} \]

The emission rate is:

\[ \dot N_e=0.10(5.03\times10^{14}) \]

\[ \dot N_e=5.03\times10^{13}\ \text{electrons s}^{-1} \]

Therefore:

\[ I_{s,2}=e\dot N_e \]

\[ I_{s,2} =(1.60\times10^{-19})(5.03\times10^{13}) \]

\[ \boxed{I_{s,2}=8.05\ \mu\text{A}} \]

Step 4: Calculate the stopping potential magnitude for the second beam.

The photon energy is:

\[ E_{\gamma,2}=hf_2 \]

\[ E_{\gamma,2} =(4.14\times10^{-15})(6.0\times10^{14}) \]

\[ E_{\gamma,2}=2.484\ \text{eV} \]

Therefore:

\[ K_{\max}=2.484-2.30 \]

\[ K_{\max}=0.184\ \text{eV} \]

Hence:

\[ \boxed{V_{s,2}=0.184\ \text{V}} \]

Step 5: Explain the change in current.

Interestingly, the photon flux of the first beam is actually higher:

\[ \Phi_{\gamma,1} = \frac{2.0}{(6.63\times10^{-34})(5.0\times10^{14})} \]

\[ \Phi_{\gamma,1} =6.03\times10^{18}\ \text{photons m}^{-2}\text{s}^{-1} \]

However, none of those photons have enough energy to release electrons.

The second beam has a lower photon flux, but its photons have sufficient energy to overcome the work function.

This means the current increases from zero to a positive value.

The important lesson: The inverse relationship between frequency and saturation current applies to above-threshold emission under constant-efficiency assumptions. It does not mean that increasing frequency must always decrease current. Crossing the threshold can switch photoelectric emission on.

Question 5

A photocell is tested using two monochromatic light beams.

Both beams have the same irradiance, illuminated area, and quantum efficiency. Both frequencies are above the metal’s threshold frequency.

The frequencies are:

\[ f_A=6.0\times10^{14}\ \text{Hz} \]

\[ f_B=1.2\times10^{15}\ \text{Hz} \]

The anode is held at the same positive potential during both experiments. However, this voltage is not sufficient to collect every emitted electron.

Measurements show that:

• 30% of the electrons emitted under Beam A reach the anode.
• 80% of the electrons emitted under Beam B reach the anode.

Treat these collection fractions as measured values.

A student argues:

“Beam B has twice the frequency, so its photon flux must be half as large. Therefore, the measured photocurrent must also be half as large.”

Evaluate this argument.

Calculate:

(a) The ratio of photon fluxes.

(b) The ratio of total emission currents.

(c) The ratio of the measured photocurrents at the applied voltage.

(d) The increase in stopping potential magnitude for Beam B.

(e) The ratio of saturation currents if the anode voltage is increased sufficiently to collect all emitted electrons.

Solution 5

The student’s conclusion is incorrect because it confuses the emission current with the measured photocurrent.

Beam B produces half the emission current, but its measured photocurrent is \(4/3\) times that of Beam A at the specified voltage.

Step 1: Calculate the ratio of photon fluxes.

Both beams have the same irradiance.

Therefore:

\[ \Phi_\gamma\propto\frac{1}{f} \]

The ratio is:

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

\[ = \frac{6.0\times10^{14}} {1.2\times10^{15}} \]

\[ \boxed{\frac{\Phi_{\gamma,B}}{\Phi_{\gamma,A}}=0.50} \]

Beam B has half the photon flux of Beam A.

Step 2: Calculate the ratio of emission currents.

The quantum efficiency and illuminated area are unchanged.

Therefore, the electron emission rate is proportional to photon flux.

Hence:

\[ \frac{I_{\mathrm{em},B}}{I_{\mathrm{em},A}} = \frac{\Phi_{\gamma,B}}{\Phi_{\gamma,A}} \]

\[ \boxed{\frac{I_{\mathrm{em},B}}{I_{\mathrm{em},A}}=0.50} \]

Beam B produces half as many emitted electrons per second.

Up to this point, the student’s reasoning is correct.

The mistake occurs when they assume every emitted electron reaches the anode.

Step 3: Calculate the ratio of measured photocurrents.

The measured current depends on both the emission rate and the fraction of electrons collected.

For Beam A:

\[ I_A=0.30I_{\mathrm{em},A} \]

For Beam B:

\[ I_B=0.80I_{\mathrm{em},B} \]

Therefore:

\[ \frac{I_B}{I_A} = \frac{0.80I_{\mathrm{em},B}} {0.30I_{\mathrm{em},A}} \]

Substitute the emission-current ratio:

\[ \frac{I_B}{I_A} = \frac{0.80(0.50)}{0.30} \]

\[ \boxed{\frac{I_B}{I_A}=1.33} \]

The measured photocurrent for Beam B is approximately 33% greater.

This is possible because the more energetic electrons in Beam B have a greater collection fraction under the specified experimental conditions.

The emission rate has fallen, but the fraction of electrons reaching the anode has increased enough to produce a larger measured current.

Step 4: Calculate the increase in stopping potential magnitude.

The maximum kinetic energy changes according to:

\[ \Delta K_{\max}=h(f_B-f_A) \]

Using \(h=4.14\times10^{-15}\ \text{eV s}\):

\[ \Delta K_{\max} = (4.14\times10^{-15}) (6.0\times10^{14}) \]

\[ \Delta K_{\max}=2.484\ \text{eV} \]

Since the stopping potential magnitude is numerically equal to the maximum kinetic energy in electronvolts:

\[ \boxed{\Delta V_s=2.48\ \text{V}} \]

Beam B therefore has a stopping potential magnitude 2.48 V greater than Beam A.

Step 5: Determine the ratio of saturation currents.

At saturation, every emitted electron is collected.

The collection fraction becomes 1 for both beams.

Therefore:

\[ I_{s,A}=I_{\mathrm{em},A} \]

\[ I_{s,B}=I_{\mathrm{em},B} \]

Using the emission-current ratio:

\[ \boxed{\frac{I_{s,B}}{I_{s,A}}=0.50} \]

The saturation current for Beam B is half that of Beam A, even though its measured current at the original applied voltage was greater.

Interpretation: The student’s mistake was treating the measured photocurrent as a direct measurement of the total emission rate. That is only true when all emitted electrons are collected.

This is why saturation current, emission current, stopping potential, and maximum kinetic energy must be distinguished.

08Where to go next

When interpreting a photoelectric experiment, keep three questions separate:

• How many electrons leave the metal each second? This determines the emission current.
• How much kinetic energy can the fastest electron have? This determines the maximum kinetic energy.
• What reverse voltage stops the fastest electrons? This determines the stopping potential.

Frequency changes the energy available to an individual photon. Photon flux and quantum efficiency determine the emission rate, while the applied voltage determines how many emitted electrons reach the collector.

The natural next step is learning to extract both the saturation current and stopping potential from a photocurrent-versus-applied-voltage graph. See How to Sketch Photocurrent Against Applied Voltage.