# The Photoelectric Effect

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19.2The Photoelectric Effect

In this Physics tutorial, you will learn:

• Why it is not always the same thing doing things in pieces and all at once?
• What is the Photoelectric Effect?
• What are the Laws of Photoelectric Effect? How were they discovered first?
• What is the stopping voltage? Why is it applied?
• Why the theoretical explanation of Photoelectric Effect was impossible within the framework of classical physics? How did Einstein overcome this issue?
• What is the work function? Where does it depend on?
• What is the Einstein's Equation of Photoelectric Effect?
• What is the theoretical explanation of the Laws of Photoelectric Effect?
• What is the quantic detachment caused by photoelectric effect?
• What is the intensity of this saturation current?

## Introduction

Can you make a 20 kg object move along a flat horizontal surface by pushing it by a 100 N constant force if the friction coefficient is 0.4?

What happens if you use the same force but in this case, you split it into 5 attempts of 20 N constant force each?

Ten people are behind a 2 m high wall. Can they climb over the wall?

What if there are more people but the wall is also higher? For example, if there are five times more people and the wall is five times higher, can anyone climb over the wall without any aiding tool? What if there were ten thousand people behind a 10 m high wall? (Recall what happens in prisons to answer this question).

Obviously, in all the above examples, we referred to situations that apparently are similar, but which have fundamental differences in the outcome. Thus, a 100 N force can move the 20 kg object if the friction coefficient is 0.4 as based on the Newton's Second Law of Motion we have

Fnet = F - f
= F - μ ∙ m ∙ g
= 100 N - (0.4 ∙ 20 kg ∙ 10 N/kg)
= 100 N - 80 N
= 20 N

This means the object accelerates at

a = Fnet/m
= 20 N/20 kg
= 1 m/s2

This cannot occur when we apply 5 times a 20 N force as Fnet is negative (Fnet = 20 N - 80 N = - 60 N). Hence, since the object is initially at rest, it cannot decelerate further, so it cannot move.

The example regarding people climbing over the wall is more evident. If the wall is 2 m high, all people can climb it over as one of them can carry the other people on its shoulders and so on. But if the wall is 10 m high, none of them can climb it over, even if there are thousands of people behind the wall. That's why people cannot escape from prisons.

The above examples are given purposely, to demonstrate that some phenomena do not work equally when split into smaller pieces. In the previous tutorial, we saw that light is discrete, as it is made by a large number of small "sparks" known as photons. Now, we will see what happens when a prolonged weak light falls on a metal surface and then we will compare the outcome to another case, where a strong light falls on the same metal surface. The, we will explain the phenomenon observed.

## Laws of the Photoelectric Effect

By definition, the photoelectric effect is the phenomenon of electrons detachment from the surface of a metal when light falls on this metal surface. This phenomenon was first discovered by Hertz, in 1887.

We can study the laws of photoelectric effect using the electric circuit shown in the figure below.

A glass balloon containing high vacuum and two metal electrodes (cathode, C and anode, A) are connected to the source (cathode is connected to the negative pole of the source and anode to the positive one). This results in the application of a potential difference between the electrodes. The experiment shows that when a light beam of a suitable frequency falls on the negative electrode (i.e. on the cathode C), a number of electrons detach from cathode and move towards the positive electrode (anode, A) due to the potential difference produced by the source. The milliammeter, A is used to measure the current, I flowing in the circuit and the voltmeter, V measures the potential difference between the electrodes. The voltage applied at the electrodes can be changed through the variable resistor (rheostat) represented through the vertical arrow in the circuit.

The I vs ΔV graph below shows the dependence of current in the circuit from potential difference (voltage). It is obtained for two different situations, where in the second case is used a light with a higher flux (energy per second falling on the cathode surface) than in the first one. However, the wavelength is the same for both cases.

The increase of voltage (starting from zero) is accompanied with the increase of current up to a maximum value Isat, at which the saturation regime is set. This saturation current does not increase anymore with the increase in voltage as the number of electrons detached from cathode reach immediately the anode for higher voltages applied. The value of saturation current depends on the light flux and it is higher from greater fluxes.

From the graph you can see that there is some current flowing through the circuit even when the voltage in the circuit is zero, albeit very small. This phenomenon occurs because some electrons detached from cathode move directly towards anode, closing thus the circuit. To prevent this undesired current, a (negative) stopping voltage ΔV0 is applied in the circuit by changing the polarities of electrodes. Obviously, for higher negative values of stopping voltages, the current in the circuit remains zero. The graph indicates that this stopping voltage has the same value for both light beams, so it does not depend by the amount of light flux. Indeed, experiments show that stopping voltage depends only by the light wavelength (and frequency therefore). This fact is demonstrated in the figure below, in which the current vs voltage graph for two beams of different wavelengths (λ1 > λ2) is shown.

All these experiments and graphs derived from them, led to the following conclusions, known as the four laws of photoelectric effect:

1. Photoelectric effect occurs only when the frequency f of the incident light is greater or at least equal to a frequency f0, which is a characteristic of the metal used. This characteristic frequency is known as the hreshold frequency of photoelectrict effect. Mathematically, we write
2. f ≥ f0 ⟹ photoelectric effect does occur
3. The stopping voltage ΔV0 depends only by the frequency f of the incident light in a proportional fashion. Mathematically, we write:
∆V0∝f
4. The number of photoelectrons detached from the metal plate in every second is proportional to the light flux falling on the cathode surface. Mathematically, we have:
N(e)/tΦ/A
5. The photoelectric effect is a phenomenon that practically does not have any inertia. It occurs simultaneously with the light incidence on the cathode surface.

Despite the experimental discovery of the four laws regarding photoelectric effect, their theoretical explanation was impossible within the framework of classical physics. This fact was another alert for the scientists of that time to review their understanding about the continuous nature of light.

## Einstein's Equation of Photoelectric Effect. The Particle Nature of Light

The phenomenon of photoelectric effect was explained by Einstein in 1905, who accepted as true the Planck's hypothesis on the discrete nature of light. Einstein elaborated further this idea by asserting that light is emitted, propagated and absorbed in specific portions (quanta) of energy. He supported the idea of the discrete nature of light and that light is composed by microscopic particles (photons) having the energy Ephoton = h · f and impulse p = E / c.

Here, c represents the speed of light particles or photons. Einstein considered the proton-electron interaction to explain the photoelectric effect and he applied the law of conservation of energy for this. Thus, the electron near the metal surface "absorbs" the photon and "consumes" part of photons energy (E = h · f) during its detachment from metal. This energy is known as Work Function, Φ and represent the minimum work necessary to detach the electron from metal. When the energy of incident photon absorbed by the electron during the photoelectric effect is greater than the work function, the electron gains kinetic energy and thus, it can escape from metal. In this way, the Einstein's Equation of Photoelectric Effect derived from the law of conservation of energy, is

Energy of Photon = Work Function + Kinetic Energy of Electron

or

h ∙ f = Φ + KE
Remark! Do not confuse the light flux Φ with the work function Φ in the photoelectric effect. They are completely different things. The first (light flux) is measured in Lumen, lm, which is candela times steradian (1 lm = 1 cd · sr), while work function Φ has the unit of energy (Joule).

The table below shows the work function in electronvolts, eV (1eV = 1.6 × 10-19 J) for a number of metals.

It is an interesting fact that Einstein won the Nobel Prize for explaining the phenomenon of photoelectric effect but did not win any Nobel Prize for his discoveries in relativity. This is because nobody from the Nobel Prize committee was able to understand in-depth his idea on relativity at that time.

### Example 1

Calculate the wavelength of photon needed to make an electron move at 6 × 106 m/s towards the anode during the photoelectric effect. The cathode is made of tungsten. Take me ≈ 9.1 × 10-31 kg.

### Solution 1

From the table above we see that the work function of tungsten is

Φ = 4.50 eV
= 4.50 ∙ 1.6 × 10-19 J
= 7.2 × 10-19 J

First, we must find the frequency of the incident photon, fph. From the Einstein's Equation of Photoelectric Effect, we have

h ∙ fph = Φ + me ∙ ve2/2
fph = Φ + me ∙ ve2/2/h
= (7.2 × 10-19 J) + (9.1 × 10-31 kg) ∙ (6 × 106 m/s)2/2/6.626 × 10-34 J ∙ s
= 2.58 × 1016 Hz

Giving that for all EM waves we have

c = λ ∙ f

where c = 3 × 108 m/s, we obtain for the wavelength λph of the incident photon:

λph = c/fph
= 3 × 108 m/s/2.58 × 1016 Hz
= 1.163 × 10-8 m

When the electron-photon interaction occurs at the surface of metal, the kinetic energy resulting from the photoelectric effect has a higher value than when this interaction occurs inside the metal. This is because electron transmits some of its kinetic energy to the metal atoms during collisions taking place when it in on the way to move to the metal surface. The kinetic energy of electron in Einstein's Equation of Photoelectric Effect represents the maximum kinetic energy of electron, i.e. when applying this equation we assume the electron is on the surface of metal, not inside it.

## Explanation of the Laws of Photoelectric Effect

From Einstein's Equation of Photoelectric Effect, it is obvious that this effect takes place only when photon's energy is greater than or equal to the work function (1st Law of Photoelectric Effect). Otherwise, electron could not be able to detach from metal surface. Therefore, the condition for the photoelectric effect to occur is that

hf ≥ Φ

or

f ≥ Φ/h

The quantity Φ/f depends on the type of metal and it represents the threshold frequency for the photoelectric effect to occur. We denote this threshold frequency by f0, so we have:

f0 = Φ/h

We can calculate the value of threshold frequency for every metal shown in the table of work function provided in the previous paragraph. For example, the threshold frequency for iron (Φ = 4.36 eV,/sub>), is

f0 = Φ/h
= 4.36 ∙ 1.6 × 10-19 J/6.626× 10-34 J · s
= 1.05 × 1015 Hz

During the stoppage of photoelectrons by the electric field of stopping voltage, the work done by the electric forces is equal to the kinetic energy of the fastest photoelectrons, i.e.

KE = e ∙ ∆V0

where e is the elementary charge or the charge of electron (e = -1.6 × 10-19C) and ΔV0 is the stopping voltage. Thus, we have

h ∙ f = Φ + e ∙ ∆V0

Therefore, we obtain for the stopping voltage:

∆V0 = h ∙ f/e - Φ/e

From the above formula, we see that there is a linear relationship between the stopping voltage and light frequency, i.e. the stopping voltage increases with the increase of light frequency (2nd Law of Photoelectric Effect). This relationship is shown in the graph below.

Photoelectric effect is a quantic phenomenon and as all the other quantic phenomena it is characterized by the probability of events occurrence. This means not all incident photons can cause photoelectric effect on a metal by detaching an electron from its surface. Some incident photons are absorbed by atoms of metals without producing any photoelectric effect while some other photons are reflected back by the metal surface. Thus, a photon can produce one or no photoelectrons when it falls on a metal.

If we denote by Nph the number of incident photons on the metal surface in one second and by Ne the number of photoelectrons produced in the same time, we have

Ne/Nph ≤ 1

This ratio (known as "quantic detachment caused by photoelectric effect") shows the probability for the photoelectric effect to occur. It is denoted by α and is a dimensionless quantity like all types of probability. Thus, we have

α = Ne/Nph

If α = 0, this means no electrons are detached from the surface of metal as the energy of all incident photons has been smaller than the work function.

If α = 1, this means each of the photons has detached one electron from metal. Theoretically, this occurs when all photons have a higher energy than work function. However, practically, this is impossible.

Photoelectrons that detach from cathode and reach the anode produce the photocurrent, which reaches the saturation value Is for specific values of accelerating voltage. The intensity of this saturation current is

Is = e ∙ Ne
= e ∙ α ∙ Nph

The number of photons incident on the cathode in every second is determined through the light flux Φ, and is given by

Nph = Φ/h ∙ f

Hence, the saturation current is proportional to the light flux, as

Is = e ∙ α ∙ (Φ/h ∙ f)

This outcome represents the third law of photoelectric effect.

The fourth law states that photoelectric effect is a phenomenon without inertia. This is evident, based on the fact that the photon-electron interaction practically occurs at instant.

In this way, based on the quantum hypothesis on the particle nature of light, we managed to explain the laws of photoelectric effect found experimentally.

### Example 2

What is the stopping voltage when a 500 nm photon falls on a sodium cathode (Φ = 2.26 eV)? Take the charge of electron e = 1.6 × 10-19 C.

### Solution 2

Clues:

λ = 500 nm = 5 × 10-7 m
Φ = 2.26 eV = 2.26 × 1.9 × 10-19
J = 3.616 × 10-19 J
e = 1.6 × 10-19 C
(c = 3 × 108 m/s)
(h = 6.626 × 10-34 J · s)

First, we calculate the frequency of photon. Thus, giving that

c = λ ∙ f

we obtain for the frequency f of photon:

f = c/λ
= 3 × 108 m/s/5 × 10-7 m
= 0.6 × 1015 Hz
= 6 × 1014 Hz

Now, using the equation

∆V0 = h ∙ f/e-Φ/e

or

∆V0 = h ∙ f - Φ/e

we obtain for the stopping voltage on the sodium cathode for the given frequency:

∆V0 = (6.626 × 10-34 J ∙ s) ∙ (6 × 1014 Hz)-(3.616 × 10-19 J)/1.6 × 10-19 C
= 3.976 × 10-19J - 3.616 × 10-19 J/1.6 × 10-19 C
= 0.225 V

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