My name is Alexander FufaeV and here I write about:

# Photoelectric Effect: Understand the Revolutionary Experiment Like Einstein

## Important Formula

What do the formula symbols mean?

## Photon energy

Unit
The energy of a photon is the sum of the kinetic energy (of an electron released with the help of the photon) and the work function of the illuminated material. From the photon energy, the frequency of the light can be determined: $$f~=~ \frac{ W_{\text p} }{ h }$$

## Frequency

Unit
Frequency of the used light source. If you multiply light frequency with Planck's constant $$h$$, you obtain the energy $$W_{\text p}$$ of a photon.

## Work function

Unit
Work function is the energy that must be spent to knock an electron out of a solid (e.g. a metal plate). It is usually given in units of "eV" (electron volt) and can be calculated with the help of the cutoff frequency $$f_0$$: $$W = h\, f_0$$

## Cutoff frequency

Unit
Cutoff frequency is the minimum frequency of light necessary to knock out electrons. Multiply it by the Planck's constant $$h$$ to obtain the work function: $$W = h \, f_0$$.

## Velocity

Unit
Maximum speed of an electron knocked out by a photon. If the material is illuminated by photons with a larger photon energy $$W_{\text p}$$ ( at constant work function $$W$$), then also the electron speed becomes larger.

## Electron mass

Unit
Rest mass of the electron. It is a physical constant with the value: $$m_{\text e} = 9.109 \cdot 10^{-31} \, \mathrm{kg}$$

## Planck's Constant

Unit
Planck's constant $$h$$ is a physical constant from quantum mechanics and has the following exact value: $$h ~=~ 6.626 \, 070 \, 15 ~\cdot~ 10^{-34} \, \mathrm{Js}$$

The photoelectric effect is a phenomenon in which light ejects electrons from matter. Albert Einstein succeeded in the revolutionary explanation of the photoelectric effect in 1905. His explanation paved the way for quantum physics and opened up many modern technical applications, such as solar cells and light sensors.

The goal of this lesson is to understand exactly how the photoelectric effect is explained and why it is represented by the following equations:

## Experiment: Setup for the photoelectric effect

To demonstrate the photoelectric effect, you basically need four things:

1. Monochromatic light source - with this you create single-color light (e.g. green light). A vapor lamp (with sodium or mercury vapor) is fine for this purpose.

2. Plate capacitor - consists of two metal plates (called electrodes) between which a voltage $$U$$ can be applied. For the photoelectric effect, it should also be possible to change the polarity of the voltage. That means: It should be possible to charge one electrode positively (+) and the other negatively (-) and vice versa.

3. Ammeter - is a current meter that allows you to measure the electric current $$I_{\text p}$$ between the two electrodes.

4. Voltmeter - is a voltage meter with which you can read off the set voltage $$U$$ between the two electrodes.

## Light particles for bombarding the electrode

The revolutionary aspect in the explanation of the photoelectric effect is the assumption that the light of the used light source does not propagate wave-like, but particle-like. We assume that the light consists of quite many light particles, which we call photons.

As soon as we illuminate one of the electrodes using the light source, many photons hit the electrode. This can trigger the photoelectric effect, because these light particles have a certain energy, which is then transferred to the electrode. We refer to the energy that a single photon has as photon energy $$W_{\text p}$$. In monochromatic light, all photons have equal energy. This amount of energy that a photon has determines whether it can knock out an electron or not.

The light quantum hypothesis established by Einstein states that we can obtain the energy $$W_{\text p}$$ of a single photon by multiplying the light frequency $$f$$ with the Planck's constant $$h$$:

The Planck's constant $$h = 6.626 \cdot 10^{-34} \, \mathrm{Js}$$ is a physical constant which appears in the equations whenever nature shows quantum effects. In our case, light shows a quantum effect by occurring in 'small portions'.

Since the Planck's constant $$h$$ does not change, you can read off from the equation 2 that the light frequency $$f$$ alone determines how large the energy $$W_{\text p}$$ of a photon is.

Sometimes instead of the light frequency $$f$$, the light wavelength $$\lambda$$ ("Lamda") is known. But this is not a problem, because the frequency is related to the wavelength by the speed of light $$c$$ as follows: $$f = c / \lambda$$. If instead of the frequency $$f$$ the wavelength $$\lambda$$ is given, then you can express the photon energy 2 with the help of the wavelength as follows:

The speed of light is also like the Planck's constant a physical constant and has the following value:

From the equation 3 you can see that photons with a smaller (shorter) wavelength carry a larger energy! Red light, for example, has a longer wavelength than blue light. The photons of the red light have a smaller energy than the photons of the blue light.

Table : Examples of light wavelengths, associated frequencies, energies, and colors.
Wavelength $$\lambda$$ Frequency $$f$$ Photon energy $$W_{\text p}$$
780 nm 3.8 × 1014 Hz 2.5 × 10-19 J
546 nm 5.5 × 1014 Hz 3.6 × 10-19 J
435 nm 6.9 × 1014 Hz 4.6 × 10-19 J
400 nm 7.5 × 1014 Hz 5 × 10-19 J
365 nm (UV light) 8.2 × 1014 Hz 5.4 × 10-19 J

Many photons generated by the light source fly towards the electrode surface at the speed of light and are absorbed by it. If a photon has a sufficiently large energy, it can release an electron. Here we have to answer two important questions:

• How do I know that electrons are being ejected? I can't see them with the naked eye...

• What does 'sufficiently large energy' mean here?

Keep reading and you won't have these questions anymore.

## Accelerating and stopping voltage between the electrodes

In order to be able to answer these questions at all, we must first set up the capacitor's polarity appropriately.

If a voltage $$U$$ is applied between the two electrodes, the electrodes can be polarized in two ways:

1. The illuminated electrode is negatively charged and the opposite electrode is positively charged. In this case, an electron released from the illuminated electrode would be attracted by the opposite positive electrode. With this polarity, the voltage is called accelerating voltage $$U_{\text B}$$ (see illustration 4).

2. The illuminated electrode is positively charged and the opposite electrode is negatively charged. In this case, an electron released from the illuminated electrode would be repelled by the opposite negative electrode. With this polarity, the voltage is called stopping voltage $$U_{\text G}$$ (see illustration 5).

When an electron moves from one charged electrode to the other, it gains or loses potential (electric) energy depending on whether accelerating voltage or stopping voltage is applied. The electron with elementary charge $$e$$ gains / loses the following electrical energy $$W_{\text e}$$ if it passes through the voltage $$U$$:

As can be seen from Eq. 4, an electron loses/gains more energy when a larger voltage $$U$$ is applied.

• Electron gains energy $$e \, U_{\text B}$$ when accelerating voltage $$U_{\text B}$$ is set between electrodes. The electron accelerates to the opposite electrode.

• Electron loses energy $$e \, U_{\text G}$$ when the stopping voltage $$U_{\text G}$$ is set between the electrodes. The electron is decelerated by the opposite electrode.

## Ejected electrons generate photocurrent

To determine whether electrons are being ejected, we need an ammeter that measures the electric current between the two electrodes. If the electrode is now illuminated with a sufficiently high frequency of light, the ammeter will indicate a non-zero current. This is exactly the electric current that is used to prove that the light ejects the electrons.

If we switch off the light, then the photocurrent drops to zero. When we switch on the light again, the ammeter shows a value different from zero. The photocurrent $$I_{\text p}$$ between the two electrodes can be controlled in two ways:

1. Vary the light intensity: For example, you can use an aperture to regulate the intensity of the incident light and thus vary the number of photons that hit the electrode. This also changes the number of electrons ejected. Doubling the light intensity, and thus doubling the number of photons, results in doubled photocurrent.

2. Change voltage: With the help of a higher accelerating voltage, you can accelerate slower electrons so that they also reach the opposite electrode and thus contribute to the higher photocurrent. With the help of a larger stopping voltage, on the other hand, you can slow down faster electrons, so that they do not reach the opposite electrode and thus the photocurrent decreases.

If the voltage source is switched off, you can still measure a photocurrent with a proper light frequency, because some electrons fly straight towards the opposite electrode after being released. However, keep in mind that not all of the released electrons - with the voltage switched off - land on the opposite electrode. The measured photocurrent is not maximized! The reason for this is that some electrons exit the illuminated electrode at an angle and miss the opposite electrode.

How can you make sure that ALL the electrons ejected end up on the electrode? Or in other words: How can you maximize the photocurrent? This is where the previously learned accelerating voltage comes into play! Increase the accelerating voltage $$U_{\text B}$$. This makes the opposite plate even more positively charged. Of course, this results in a greater attractive force on all the electrons that are knocked out. Increase the voltage as long as the photocurrent $$I_{\text p}$$ also increases. If you increase the voltage so that the photocurrent at the ammeter does not change any more, i.e., goes into saturation, then all electrons land on the opposite electrode. The electrical attraction of the opposite electrode on the electrons becomes so strong that even the obliquely emitted electrons are 'sucked in'. The photocurrent reaches a maximum possible value $$I_{\text{max}}$$.

## Overcome work function of the illuminated electrode

In the photoelectric effect experiment, it is found that not all light is able to eject electrons. The electrons can move freely within the metallic electrode, but they cannot move out of the electrode because they are bound to it. This binding of the electron to the electrode can be overcome if you add enough energy to the electron so that it can be knocked out of the bond. But if you don't add enough energy to the electron, then of course it will continue to be bound and will not come out of the electrode.

The work function $$W$$ differs depending on which material the electrode is made of. An electrode made of nickelium has a larger work function than an electrode made of aluminum. Electrons are more strongly bound to the nickelium electrode than to the aluminum electrode. Consequently, it will be more difficult to release electrons from the nickelium material. Here are a few examples of different materials:

Table : Examples of work function of various materials from which the illuminated electrode is made.
Material Work function in joules Work function in electron volts
Cesium (Cs) 3.1 × 10-19 J 1.94 eV
Natrium (Na) 3.6 × 10-19 J 2.28 eV
Aluminium (Al) 6.7 × 10-19 J 4.20 eV
Zinkium (Zn) 7 × 10-19 J 4.34 eV
Platinium (Pt) 8.6 × 10-19 J 5.36 eV

You probably noticed from Table 2 that the work function was not only given in Joules (J), but also in Electron volts (eV). This is a typical compact energy unit in which the work function in the photoelectric effect and other quantum mechanical effects is given.

• Divide the energy in Joules (J) by the value of the elementary charge $$1.6 \cdot 10^{-19}$$ to express it in Electronvolts (eV).

• Multiply the energy in Electronvolts (eV) by the value of the elementary charge $$1.6 \cdot 10^{-19}$$ to convert it to Joules (J).

How do you supply the electron with this required energy? So how do you overcome this work function and thus also the binding of the electron to the electrode? Now, photons come into play. You have learned that the photon carries an energy $$W_{\text p}$$, which is determined by the light frequency $$f$$ or wavelength $$\lambda$$. Now there are two possibilities which can occur:

1. Photon energy is smaller than the work function: $$W_{\text p} < W$$.
In this case, the photon cannot eject an electron.

2. Photon energy is greater than the work function: $$W_{\text p} \geq W$$.
In this case, a photon ejects an electron.

### Express work function with cutoff frequency or cutoff wavelength

Let us assume that we use photons with energy $$W_{\text p}$$ smaller than the work function $$W$$. So no electrons can be ejected. What do you have to do to make the photon energy greater than the work function? Look at the equations 2 and 3. You just have to use the light with a larger frequency $$f$$ (smaller wavelength $$\lambda$$).

If you increase the light frequency $$f$$, you will eventually arrive at the so-called cutoff frequency $$f_0$$.

According to 2 the photon energy $$h \, f_0$$ is exactly equal to the work function $$W$$.

If you manage to set the light frequency $$f_0$$ (cutoff frequency), then you can use the equation 7 to find out an important property of the illuminated material, namely its work function.

Obviously, the cutoff frequency $$f_0$$ is different depending on which material is illuminated. Different materials have different work function. For nickelium (Ni), the cutoff frequency is larger than for aluminum (Al) because the electrons in nickelium electrode are more bound (nickelium has larger work function).

Using the general relationship $$c = \lambda \, f$$ between the wavelength and the frequency, you can express the work function 7 using the cutoff wavelength $$\lambda_0$$.

Table : Cutoff frequency and wavelength for some materials of the illuminated electrode.
Material Cutoff frequency $$f_0$$ Cutoff wavelength $$\lambda_0$$
Cesium (Cs) 4.7 × 1014 Hz 641 nm
Natrium (Na) 5.5 × 1014 Hz 545 nm
Aluminium (Al) 1 × 1015 Hz 297 nm
Zinkium (Zn) 1.1 × 1015 Hz 285 nm

Equivalent to the above two conditions for photon energy and work function, two conditions for frequency $$f$$ and cutoff frequency $$f_0$$ can be established:

1. Light frequency is smaller than cutoff frequency: $$f < f_0$$.
In this case, the photon cannot eject an electron.

2. Light frequency is greater than cutoff frequency: $$f \geq f_0$$.
In this case, a photon knocks out an electron.

## Photoelectric effect equation as a consequence of conservation of energy

But what if you take a higher frequency of the light than the cutoff frequency $$f_0$$? Or equivalently: What if the photon energy is much greater than the work function: $$W_{\text p} \gt W$$? A part of the photon energy (namely work function $$W$$) is needed to overcome the bond of the electron to release it from the electrode. We subtract this work function $$W$$ from the photon energy. But there is still a remaining energy:

Where is this remaining energy $$\Delta W$$? After all, it cannot simply disappear according to the energy conservation law!

In the experiment, it is observed that if you use light with sufficiently large frequency, then the ammeter will show a non-zero electric current: $$I_{\text p} \neq 0$$. As it looks, electrons released still have a velocity in the direction of the opposite electrode. And there we have our remaining energy! The remaining energy $$\Delta W$$ is given to the electron in the form of kinetic energy $$W_{\text{kin}}$$. The remaining energy is the kinetic energy of the ejected electron: $$W_{\text{kin}} = \Delta W$$.

You can substitute the photon energy 2 into eq. 10:

To find out the concrete velocity $$v$$ of the released electrons, you use for $$W_{\text{kin}}$$ the classical formula for kinetic energy:

Here $$m_{\text e} ~=~ 9.109 \cdot 10^{-31} \, \mathrm{kg}$$ is the rest mass of the electron. Rearranged for the photon energy, you get the famous formula for the photoelectric effect, for which Albert Einstein got the Nobel Prize:

The velocity $$v$$ in 13 is not directly measurable, so we express it with the measurable quantity, using the stopping voltage $$U_{\text G}$$.

To determine the value of the stopping voltage, you must increase the stopping voltage very slowly in the experiment and observe how the photocurrent decreases. In this way, fewer and fewer electrons reach the opposite electrode. At some point, you will reach a value of the voltage at which the repulsive force of the opposite electrode is so large that not a single electron reaches the electrode. The photocurrent has now dropped to zero. You have slowed down the last fastest electrons. At this value of the stopping voltage the fastest electron has lost the energy $$e \, U_{\text G}$$. This lost energy just corresponds to its kinetic energy:

You may replace in this way the kinetic energy in 13 with $$e \, U_{\text G}$$:

Using Eq. 15, you can obtain the work function of the material for a given light frequency/wavelength and the stopping voltage read on the voltmeter. Or, for a given work function and stopping voltage, figure out the light frequency / wavelength.

Note, however, that you are not allowed to increase the stopping voltage further, so that the equation 14 is valid. Further increasing of the stopping voltage does not change the photocurrent, because it is already zero. You have to set exactly the reverse voltage value at which the photocurrent just becomes zero!

## Energy-frequency graph for the photoelectric effect

You can illustrate the equation 11 in a graph:

Kinetic energy $$W_{\text{kin}}$$ is plotted on the $$y$$ axis and light frequency $$f$$ is plotted on the $$x$$ axis.

If you look closely at the equation 16, you will see that it is a straight line equation of the form $$y = mx+b$$. In our case:

• $$y$$-axis is the kinetic energy: $$y = W_{\text{kin}}$$.

• $$x$$-axis is the light frequency: $$x = f$$.

• The slope $$m$$ of the straight line is the Planck's constant $$m = h$$.

• $$y$$-intercept $$b$$ is the negative work function: $$b = -W$$.

To get the intersection of the line with the $$x$$ axis, you must set the kinetic energy to zero in Eq. 16: $$W_{\text{kin}} = 0$$. After rearranging the equation, you will find that the intersection point corresponds to the cutoff frequency $$f_0$$ (see illustration 6).

With this knowledge you are able to determine the Planck's constant $$h$$ with the help of the photoelectric effect.

You hopefully know from mathematics how to determine the slope of a straight line:

In our case, the slope corresponds to the Planck's constant: $$m = h$$. Here $$y_{2}$$ and $$y_{1}$$ are two arbitrary values $$W_{\text{kin,1}}$$ and $$W_{\text{kin,2}}$$ for kinetic energy of electrons. And $$x_{2}$$, $$x_{1}$$ are associated light frequencies $$f_2$$, $$f_1$$. So you can rewrite equation 17 like this:

Unfortunately, the kinetic energy of electrons is not directly experimentally accessible. Since you perform the experiment in a plate capacitor, you can use it to rewrite the kinetic energy with an experimentally accessible quantity. This quantity is the stopping voltage $$U_{\text G}$$ between the two electrodes.

In this case, the kinetic energy corresponds to the electrical energy: $$W_{\text{kin}}=e \, U_{\text G}$$. Here, $$U_{\text G}$$ is the stopping voltage necessary to completely decelerate the electrons. Substituting electrical energy into eq. 19 yields:

Investigate the photoelectric effect with two different, known light frequencies $$f_1$$ and $$f_2$$, read off the corresponding stopping voltages $$U_{\text{G},1}$$ and $$U_{\text{G},2}$$. Use 19 to determine the Planck's constant $$h$$ experimentally.

## Photoelectric effect shows contradictions to wave theory

What is observed in the photoelectric effect experiment cannot be explained by the assumption that light is wave-like.

According to the classical idea, electrons should be able to be ejected with any light frequency $$f$$. Something like a cutoff frequency $$f_0$$ does not exist in the wave theory of light. However, this contradicts what is observed in the experiment.

According to classical wave theory, you could simply illuminate the electrode with an arbitrary light frequency $$f$$ for longer to supply more energy to the electrode. The photoelectric effect would then occur after a time delay. In the experiment, however, we observe no such behavior: No matter how long you illuminate the electrode, no electrons can escape if the light frequency is too small.

According to the classical wave theory, the kinetic energy of the electrons $$W_{\text{kin}}$$ should decrease with increasing light frequency $$f$$, because in the classical wave picture the following proportionality holds: $$W_{\text{kin}} \sim 1/f^2$$. The kinetic energy should even decrease quadratically. But what one finds in the experiment to the photoelectric effect is an increase and no decrease of the kinetic energy.

Moreover, according to the classical idea, the kinetic energy should increase quadratically with the amplitude of the incident light: $$W_{\text{kin}} \sim A^2$$. However, the experiment shows a different behavior: Increasing the amplitude $$A$$ (increasing light intensity) has no effect on the kinetic energy of the electrons.

Now you have learned all the important points about the photoelectric effect. In the next lesson, we will turn to another important experiment that has made quantum mechanics so successful, the Franck-Hertz experiment.

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

### Exercise #1: How Large Is the Photocurrent?

During the photoelectric effect, you illuminate one of the plates of the plate capacitor with light of intensity $$I_1 ~=~ 10^6 \, \frac{ \text{photons} }{ \mathrm{s} \cdot \mathrm{m}^2 }$$ (photons per second per square meter). While doing so, you measure a photocurrent $$I_{\text P1} ~=~ 2 \, \mathrm{mA}$$ with the ammeter connected to the plate capacitor.

If you instead illuminate the capacitor plate with a different light intensity $$I_2 ~=~ 2.5 \cdot 10^6 \, \frac{ \text{photons} }{ \mathrm{s} \cdot \mathrm{m}^2 }$$, what photocurrent $$I_{\text P2}$$ will you measure?

Assumption: The photocurrent increases proportionally with the light intensity.

Hint: Proportional reasoning. If the current is proportional to the intensity, then it means: $$I_{\text P} ~=~ k \cdot I$$, where $$k$$ is a constant, whose value does not matter here, and $$I$$ generally denotes the intensity of light.

#### Solution to Exercise #1

Since the photocurrent increases linearly with the light intensity, the ratio between photocurrent and intensity is always the same: $\frac{ I_{\text{P1}} }{I_1} ~=~ \frac{ I_{\text{P2}} }{I_2}$

Rearrange the equation for the desired photocurrent: $I_{\text{P2}} ~=~ \frac{I_2}{I_1} ~\cdot~ I_{\text{P1}}$

Just substitute the values given in the task to calculate the photocurrent: \begin{align} I_{\text{P2}} &~=~ \frac{ 2.5 \cdot 10^6 \, \, \frac{ \text{photons} }{ \mathrm{s} \cdot \mathrm{m}^2 } }{ 10^6 \, \frac{ \text{photons} }{ \mathrm{s} \cdot \mathrm{m}^2 } } ~\cdot~ 2 \, \mathrm{mA} \\\\ &~=~ 5 \, \mathrm{mA} \end{align}

### Exercise #2: Determine Work Function and Electron Velocity

You irradiate a metal plate with blue light of frequency $$6.9 \cdot 10^{14} \, \text{Hz}$$. Using the stopping potential method, you bring the photocurrent to $$I_{\text P} = 0 \, \text{A}$$ and measure a stopping potential of $$0.59 \, \text{V}$$.

1. What is the work function $$W$$ of the metal plate in joules (J) and in electron volts (eV)?
2. What is the velocity of the fastest electrons?

#### Solution to Exercise #2.1

Use the Einstein equation: 1 $$h \, f ~=~ e \, U_{\text G} ~+~ W$$

Rearrange the equation to solve for the unknown work function: $$W ~=~ h \, f ~-~ e \, U_{\text G}$$

Substitute the values for light frequency $$f$$, stopping potential $$U_{\text G}$$, Planck's constant $$h = 6.626 \cdot 10^{-34} \, \mathrm{Js}$$, and elementary charge $$e = 1.6 \cdot 10^{-19} \, \mathrm{C}$$ to calculate the work function $$W$$ specifically: \begin{align} W &~=~ 6.626 \cdot 10^{-34} \, \text{Js} ~\cdot~ 6.9 \cdot 10^{14} \, \text{Hz} ~-~ 1.6 \cdot 10^{-19} \, \text{C} ~\cdot~ 0.59 \, \text{V} \\\\ &~=~ 3.626 \cdot 10^{-19} \, \text{J} \end{align}

To convert the work function, given in joules, into electron volts, we simply divide the work function by the value of the elementary charge: \begin{align} W &~=~ \frac{ 3.626 \cdot 10^{-19} \, \text{J} }{ 1.6\cdot10^{-19} } \\\\ &~=~ 2.27 \, \text{eV} \end{align}

So, the photon must have at least this energy (work function) to eject an electron!

#### Solution to Exercise #2.2

The emitted electrons migrate to the opposite metal plate, which is negatively charged in the stopping potential method. This means: The negatively charged electrons are repelled by the negatively charged plate. Using the stopping potential, we can vary negative charge on the plate and thus increase or decrease the repulsion force. In the stopping potential method, our goal is to adjust the stopping potential $$U_{\text G}$$ such that the repulsion force is just enough (neither smaller nor larger) that even the fastest electrons can no longer reach the positive plate. This value of stopping potential is used to determine the velocity of the fastest electrons. And this value is given in the problem!

Set the electrical energy $$e \, U_{\text G}$$ that an electron with charge $$e$$ gains when it traverses the potential $$U_{\text G}$$ equal to the kinetic energy $$\frac{1}{2} \, m \, v^2$$: $$e \, U_{\text G} ~=~ \frac{1}{2} \, m \, v^2$$

Solve for velocity $$v$$: $$v ~=~ \sqrt{ \frac{ 2 e \, U_{\text G} }{ m } }$$

The mass of the electron is provided in the formula collection: $$m = 9.109 \cdot 10^{-31} \, \mathrm{kg}$$. Substitute given values: \begin{align} w &~=~ \sqrt{ \frac{ 2 ~\cdot~ 1.6 \cdot 10^{-19} \, \text{C} ~\cdot~ 0.59 \, \text{V} }{ 9.109 \cdot 10^{-31} \, \mathrm{kg} } } \\\\ &~=~ 455 \, 266.45 \, \frac{\mathrm m}{\mathrm s} \end{align}

### Exercise #3: Calculate Stopping Voltage

A sodium plate with work function $$W = 2.3 \, \mathrm{eV}$$ is irradiated with light of wavelength $$\lambda = 300 \, \mathrm{nm}$$.

What stopping voltage $$U_{\text G}$$ do you need to set to completely stop the photoelectric current?

#### Solution to Exercise #3

Use the energy balance formulated by Albert Einstein, where the total energy $$h \, f$$ of a photon is transferred to the electron, of which a part, namely the work function $$W$$, is used to remove the electron: $$h \, f ~=~ e \, U_{\text G} ~+~ W$$

Rearrange the equation to find the unknown stopping voltage: 2 $$U_{\text G} ~=~ \frac{h \, f ~-~ W }{ e }$$

Now you just need to substitute the correct values, where you either need to convert the wavelength $$300 \, \mathrm{nm} = 300 \cdot 10^{-9} \, \mathrm{m}$$ into frequency $$f$$, or rearrange the relationship $$c = \lambda \, f$$ to solve for $$\lambda$$ and substitute it into 2. Let's choose the first method as an example: 3 \begin{align} f &~=~ \frac{ c }{ \lambda } \\\\ &~=~ \frac{ 3 \cdot 10^8 \, \frac{\mathrm m}{ \mathrm s} }{ 300 \cdot 10^{-9} \, \text{m} } \\\\ &~=~ 10^{15} \, \text{Hz} \end{align}

We convert the work function, given in electron volts (eV), into the SI unit joules (J). To do this, we simply multiply the given value in electron volts by the value of the elementary charge $$e ~=~ 1.6 \cdot 10^{-19} \, \text{C}$$: 4 \begin{align} W &~=~ 2.3 \, \text{eV} \\\\ &~=~2.3 ~\cdot~ 1.6 \cdot 10^{-19} \, \text{C} \, \text{V} \\\\ &~=~ 3.68 \cdot 10^{-19} \, \text{J} \end{align}

Here, the unit $$\text{C} \, \text{V}$$ corresponds to the unit joules. Then we have the fundamental constant (Planck's constant) $$h$$, which has the value $$6.626 \cdot 10^{-34} \, \text{Js}$$. Substituting everything into 2 yields: \begin{align} U_{\text G} &~=~ \frac{ 6.626 \cdot 10^{-34} \, \text{Js} ~\cdot~ 10^{15} \, \text{Hz} ~-~ 3.68 \cdot 10^{-19} \, \text{J} }{ 1.6 \cdot 10^{-19} \, \text{C} } \\\\ &~=~1.84 \, \text{V} \end{align}

### Exercise #4: Determine Planck's Constant with Stopping Voltage

A photocathode (one of the two capacitor plates) is irradiated with monochromatic light in two experiments to trigger the photoelectric effect.

In the first experiment, the photocathode is irradiated with the wavelength $$\lambda_1 = 231 \, \mathrm{nm}$$. The ejected electrons migrate to the opposite plate, generating a photocurrent. You nullify this current by applying a stopping voltage of $$U_1 = 0.96 \, \mathrm{V}$$.

In the second experiment, the photocathode is irradiated with the wavelength $$\lambda_2 = 150 \, \mathrm{nm}$$. You compensate for the resulting current by applying a stopping voltage of $$U_2 = 3.85 \, \mathrm{V}$$.

1. Determine the fundamental constant using these measurement results: Planck's constant $$h$$ in joule-seconds (Js).
2. Determine the work function $$W$$ of the photocathode in joules (J) and in electronvolts (eV) using the determined Planck constant.
3. Determine the cut-off frequency $$f_0$$ and the cut-off wavelength $$\lambda_0$$.

#### Solution to Exercise #4.1

You conducted two independent experiments. This means: you can set up two equations that you can use to determine Planck's constant $$h$$.

First Experiment: Irradiating a metal plate with the wavelength $$\lambda_1$$ ejected electrons from the plate, which were stopped using stopping voltage. Substitute the wavelength $$\lambda_1$$ and the corresponding stopping voltage $$U_1$$ into the formula for the photoelectric effect: 1 $$h \, f_1 ~=~ e \, U_1 ~+~ W$$

Kinetic energy was expressed as $$e \, U_1$$. The light frequency $$f_1$$, used to irradiate the photocathode, is not known, but that's not a problem because we can convert light frequency $$f_1$$ to light wavelength $$\lambda_1$$ using the formula $$c = \lambda_1 \, f_1$$. Rearrange this formula for frequency: $$f_1 = \frac{c}{ \lambda_1 }$$ and substitute it into Eq. 1: 2 $$h \, \frac{c}{ \lambda_1 } ~=~ e \, U_1 ~+~ W$$

And there's our first equation. If you look closely, you'll see that it has TWO unknowns, namely the constant $$h$$ as well as the work function $$W$$ of the photocathode. If you were to solve the equation for $$h$$, you couldn't do so definitively because you don't know the work function $$W = h \, f_0$$. Here, $$f_0$$ is the cut-off frequency, which is also not given. Therefore, a second equation from the photoelectric effect experiment is urgently needed.

Second Experiment: Proceed in this case analogously to the first experiment. Substitute the wavelength $$\lambda_2$$ and the corresponding stopping voltage $$U_2$$ into the photoelectric effect formula. Additionally, express the frequency $$f_2$$ with the given wavelength: 3 $$h \, \frac{c}{ \lambda_2 } ~=~ e \, U_2 ~+~ W$$

Now, subtracting Eq. 3 from Eq. 2 eliminates the unknown work function $$W$$: \begin{align} h \, \frac{c}{ \lambda_1 } ~-~ h \, \frac{c}{ \lambda_2 } &~=~ e \, U_1 ~-~ e \, U_2 ~+~ W ~-~ W \\\\ h \, \frac{c}{ \lambda_1 } ~-~ h \, \frac{c}{ \lambda_2 } &~=~ e \, U_1 ~-~ e \, U_2 \\\\ h \, c \, \left( \frac{1}{ \lambda_1 } ~-~ \frac{1}{ \lambda_2 } \right) &~=~ e \, \left( U_1 ~-~ U_2 \right) \end{align}

In the last step, $$h \, c$$ was factored out on the left side and the elementary charge $$e$$ on the right side. Now, rearrange for the unknown Planck constant $$h$$: $$h ~=~ \frac{e}{c} \, \frac{ U_1 ~-~ U_2 }{ \frac{1}{ \lambda_1 } ~-~ \frac{1}{ \lambda_2 } }$$

Just substitute concrete values. Here, $$c = 3 \cdot 10^8 \, \frac{\mathrm m}{\mathrm s}$$ and $$e = 1.6 \cdot 10^{-19} \, \mathrm{C}$$: \begin{align} h ~&=~ \frac{ 1.6 \cdot 10^{-19} \, \mathrm{C} }{ 3 \cdot 10^8 \, \frac{\mathrm m}{\mathrm s} } \, \frac{ 0.96 \, \mathrm{V} ~-~ 3.85 \, \mathrm{V} }{ \frac{1}{ 231 \cdot 10^{-9} \, \mathrm{m} } ~-~ \frac{1}{ 150 \cdot 10^{-9} \, \mathrm{m} } } \\\\ &~=~ 6.59 \cdot 10^{-34} \, \mathrm{Js} \end{align}

If you look at the Planck constant in your formula collection, you'll see that the determined value matches well with the exact value $$h ~=~ 6.626 \, 070 \, 15 \,\cdot \, 10^{-34} \, \text{Js}$$.

#### Solution to Exercise #4.2

Since we have determined the Planck constant $$h$$ in (a), we can use it to determine the work function $$W$$. There are various ways to achieve this. The simplest way is to use the photoelectric equation established from the first experiment: $$h \, \frac{c}{ \lambda_1 } ~=~ e \, U_1 ~+~ W$$

And then rearrange this equation for $$W$$: $$W ~=~ h \, \frac{c}{ \lambda_1 } ~-~ e \, U_1$$

Substite concrete values: \begin{align} W ~&=~ 6.59 \cdot 10^{-34} \, \mathrm{Js} ~\cdot~ \frac{ 3 \cdot 10^8 \, \frac{\mathrm m}{\mathrm s} }{ 231 \cdot 10^{-9} \, \mathrm{m} } ~-~ 1.6 \cdot 10^{-19} \, \mathrm{C} ~\cdot~ 0.96 \, \mathrm{V} \\\\ &~=~ 7.02 \cdot 10^{-19} \, \mathrm{J} \end{align}

To convert the joule value to electron volts (eV), we just need to divide the result by the elementary charge $$e = 1.6 \cdot 10^{-19} \, \mathrm{C}$$: $$W ~=~ 4.39 \, \mathrm{eV}$$

#### Solution to Exercise #4.3

To calculate the cutoff frequency $$f_0$$, we use the relationship between the work function $$W$$ and the cutoff frequency: $$W ~=~ h \, f_0$$

Rearranging for $$f_0$$ and substituting concrete values: \begin{align} f_0 ~&=~ \frac{W}{h} \\\\ &~=~ \frac{ 7.02 \cdot 10^{-19} \, \mathrm{J} }{ 6.59 \cdot 10^{-34} \, \mathrm{Js} } \\\\ &~=~ 1.07 \cdot 10^{15} \, \frac{1}{\mathrm s} \\\\ &~=~ 1.07 \cdot 10^{15} \, \mathrm{Hz} \end{align}

This corresponds to the following cutoff wavelength \begin{align} \lambda_0 ~&=~ \frac{c}{f_0} \\\\ &~=~ \frac{ 3 \cdot 10^8 \, \frac{\mathrm m}{\mathrm s} }{ 1.07 \cdot 10^{15} \, \mathrm{Hz} } \\\\ &~=~ 2.8 \cdot 10^{-7} \, \mathrm{m} \\\\ &~=~ 280 \, \mathrm{nm} \end{align}