## Wednesday, December 21, 2022

### Fresnel Lenses: UNIZOR.COM - Physics4Teens - Waves - Properties of Light

Notes to a video lecture on http://www.unizor.com

Fresnel Lenses

One of the very clever inventions that simplified the process of making large convex lenses was made by French physicist Augustin-Jean Fresnel.
This invention allowed to make and use large lenses at the lighthouses that saved lives of many sailors.

Here is a diagram that explains this invention called Fresnel lens. Instead of a full convex lens on the top of a picture above Fresnel lens on the bottom cuts off all the parts that do not have any useful functionality (light blue) leaving only the base (red) and prismatic pieces (dark blue) that do the job of refracting the light.

The light blue pieces are only letting the rays of light through, so, by cutting them off, Fresnel was able to construct relatively large lenses of the same functionality as full convex lenses but much lighter and fit to be installed in the lighthouses.

On the picture above, if the parallel rays of light enter the Fresnel lens from below, the lens will focus the rays at a focal point above it, exactly as the full convex lens would do.
Conversely, if the point light source is at Fresnel lens focal point above it, the lens would let the light go down as parallel rays of light.
This type of Fresnel lens is called a positive Fresnel lens, it refracts the light.

Consider now a similar construction with reflected light.
As we know, parabolic mirror reflects the parallel rays of light into its focal point or, conversely, reflects a point light at its focal point into parallel rays of lights.

Similar idea as with Fresnel lenses, can be used to make a reflective mirror acting like a parabolic. It's schematically represented below and called a negative Fresnel lens. This construction is significantly more compact with practically the same functionality.

## Saturday, December 17, 2022

### Spectroscopy: UNIZOR.COM - Physics4Teens - Waves - Photons and Matter

Notes to a video lecture on http://www.unizor.com

Spectroscopy

In the previous lectures we have discussed in details the interaction between photons and atoms of hydrogen, mentioned the energy levels of hydrogen electrons En=13.6 /eV (electron-volts) and investigated the colors of light emitted by hydrogen electrons, when excited electrons jump from higher energy shell down to lower energy one, emitting light.

Not surprisingly, analogous situation is with any other element.
Any element has its electrons positioned at some distinct shells around a nucleus, with each shell having a specific for this element energy level. When extra energy is infused into electrons, they can jump to higher energy level shell and, later on, they relax into a lower level, emitting extra energy as electromagneting oscillations of specific frequency.

Since energy levels of shells are fixed and distinct for each element, the amount of energy released by an electron during the process of relaxation is also fixed and distinct, depending only on the difference in the level of energy between the shells, specific for each element.

As we know, the frequency f of electromagnetic oscillations emitted by an electron, when it jumps from a higher energy level shell to a lower energy level (the relaxation process), is directly proportional to the amount of energy E released by this electron during this process of relaxation
E = h·f
where h is Planck's constant.

Considering the energy levels of shells for different elements are, generally speaking, different, the light emitted by electrons jumping from a higher energy level to a lower one for different elements will, generally speaking, have different frequencies and, therefore, different color, if it falls into a visible spectrum.

Most likely, each element has some levels of energy that correspond to some visible light and, since they are, most likely, different for different elements, we can identify the elements by the light they emit, if their electrons relax to a lower energy level after being excited.

Spectroscopy is the field of science dealing with analyzing the spectrum of emitted light by different elements, identifying the composition of complex objects by the light emitted from them and making judgements towards the properties of these objects.

While spectroscopy, as a branch of science, deals with many aspects of interaction between electromagnetic oscillations and matter, we will only address the analyzing the emission of light when electrons are relaxing after being exposed to energy that exited them.

There are many elements, each having certain number of shells on different energy levels. That makes a job of identifying a particular element by a light emitted by it quite a complicated task.
By now the scientists have a pretty good picture about most of the elements as far as what kind of light frequency and wave length they can emit.

Just as an example, the iron can emit the light of wave lengths 516.891 nm, 495.761 nm, 466.814 nm, 438.355 nm, 430.790 nm, 382.044 nm, 358.121 nm and 302.108 nm (the last three are in ultraviolet part of a spectrum, not visible by a naked eye).

Oxygen emits light with wave lengths 898.765 nm, 822.696 nm, 759.370 nm, 686.719 nm and 627.661 nm (the first three are in infrared part of a spectrum, not visible by a naked eye).

Knowing all the potential energy differences between shells of some element allows to make a judgement not only about the presence of this particular element by the light it emits, but also about its temperature.

The hotter the element - the more wavelengths with greater energy (shorter wave length, higher frequency) will be in the light it emits. This is how the temperature of stars can be evaluated.

In the above example with iron there are 8 different wave lengths observable, but if the shorter ones dominate, it means the iron is heated to a greater temperature.
As we know, hot iron starts emitting red glow first, but, when its temperature is rising, the color shifts from red to white, which means that lights with shorter wave lengths are added.

There are many other important aspects of spectroscopy, but they are beyond the scope of this course.

## Thursday, December 15, 2022

### Laser: UNIZOR.COM - Physics4Teens - Waves

Notes to a video lecture on http://www.unizor.com

Laser

We all are familiar with laser scanners used in stores to scan the product code.
Many of us know about laser pointers that direct a small spot of light on objects quite far away.
Compact discs technology is based on lasers.
Laser printers are nowadays the dominant printing device used in computers.
There are many other applications of lasers, so, let's examine the laser's principles of working.

First of all the term laser is an abbreviation of "Light Amplification by the Stimulated Emission of Radiation". This name fully characterizes the process at the base laser's functionality, and here is why.

Atom Model

We will use a familiar model of an atom with a nucleus in a center and electrons circulating around it on different distinct orbits or, to better describe their place in three-dimensional space, in different distinct shells, with each shell containing certain number of electrons up to some maximum, specific for each shell of each element.
All electrons within any particular shell have the same level of energy.

The composition of shells is modeled as concentric spheres around a nucleus, each characterized by a specific energy level of electrons in it.
Contemporary view based on Quantum Theory, states that radii of these spheres can take only discrete values specific for each element.

When an electron absorbs energy from some outside source, it jumps from a lower energy level shell to a higher energy level one.
When it moves in an opposite direction, jumping back to a shell of a lower energy level, it emits radiation.

Atom of each element can absorb or emit energy in small but finite chunks, quanta, the size of which depends on the element's characteristics, in particular, on the discrete energy level differences between its different shells.

Absorption of a single quantum of energy by an atom manifests itself in one of its electrons jumping to a higher energy level shell with the shell energy level increment equal to the absorbed quantum of energy.

Emitting a quantum of energy is related to an electron changing its position from a higher energy level shell to a lower energy level one, with the decrease in energy level corresponding to an amount of energy emitted.

Atom emits energy as electromagnetic oscillations of certain frequency. In some cases this frequency for some elements corresponds to a frequency of a visible light of some color, in other case it might fall in the ultraviolet, infrared or other part of a spectrum.

If an electron of some element jumps from a shell with a higher energy level Ehigh to a shell with a lower energy level Elow, it emits the extra energy as a quantum of electromagnetic oscillations called photon of a specific frequency f, related to a difference in the levels of energy as
Ehigh − Elow = h·f
where energy is measured in Joules (J),
h = 6.62607015·10−34(J·s) is the Planck's constant and
f is a frequency in hertz (Hz).

The Idea

Generally speaking, all electrons of an atom of any element in a stable state occupy certain shells called ground level shells.
The atom of hydrogen with 1 electron has one ground energy level shell. The atom of sodium has 11 electrons positioned in 3 ground energy level shells: (2+8+1). Let's assume that we supply some constant external energy to atoms of some element that causes some of its electrons on the outer ground energy level shells to get excited and move to a higher energy level shells.
By supplying energy in quanta that exactly equal to a difference in the levels of energy between a ground energy level shell (a stable state of electrons), characterized by the energy level Eground, and a higher energy shell (excited state of electrons), characterized by the energy level Ehigh, we force many electrons to absorb this energy, get excited and move from the ground level shell to the higher energy level shell (energy must be conserved!), eventually overpopulating that higher energy level (not that stable) shell. The trick is to force these electrons, that overpopulate the higher energy level shell, to synchronously relax and jump back to a ground level shell, returning to a stable state.
If this is achieved, they synchronously emit a photon of radiation of the same frequency and phase producing a coherent (synchronized in frequency and phase) light.

Theoretical Solution

Assume, we have an overpopulated higher energy shell of energy level Ehigh.

Experiments showed that, when a single photon of frequency f carrying an energy
E = h·f = Ehigh−Eground
falls onto one of the excited electrons in the overpopulated higher energy shell, it is not absorbed by an already excited electron, but stimulates this electron to relax by going down to a ground level shell of the energy level Eground, emitting a photon of corresponding frequency
f = (Ehigh−Eground)/h
(energy must be conserved!)
which is the same as the frequency of the incident photon and is also synchronized with it in phase. The above is a description of Stimulated Emission of Radiation, the second part of the full name that produced an abbreviation laser.

Now we have two identical in frequency and phase photons, the incident one and the emitted by an electron that switched its position from the high energy level shell to the ground shell.

If we can redirect these two identical photons back to electrons on high energy level shell, they will cause two new electrons to relax, emitting two more photons fully synchronized with incident ones, thus enhancing the light, while maintaining the same frequency and phase for all photons.

We can repeat this redirection of new photons back to excited electrons, and on each repetition the number of emitted photons would, in theory, double.
Eventually, when this light is strong enough, we can release it and use for some purpose.

If there is an external source of energy that constantly excites electrons, which, in turn, forces them to jump to a high energy level shell and, at the same time, we can stimulate these electrons to relax, emitting extra energy in a form of electromagnetic waves of certain frequency and phase (the same for all relaxing electrons in all atoms of an element), we will obtain a flow of coherent (all of the same frequency and phase) electromagnetic oscillations.
If the frequency of these oscillations corresponds to a visible spectrum, we will get a coherent light of some color.

Practical Aspects

Exact details of implementation of the above idea are beyond the scope of this course, but a few possible technical details of the implementation can be suggested.

Imagine a tube with some gas and two electrodes in it connected to a battery. One of the most commonly used gas laser uses a mixture of helium (He) and neon (Ne) gases. The battery should be sufficiently powerful to establish a flow of electric current through this gas.

If the voltage of a battery is sufficient, this construction allows the gas atoms to constantly absorb the energy and, after a while, the gas might heat up (the molecules will increase their chaotic movement within a tube) and might emit some visible light to compensate for overloaded with energy electrons, that, after being excited to a limit, relax and release the extra energy as randomly emitted photons.

Let's make one end of this tube fully reflective and the opposite - partially reflective, that is, it's reflective for low level energy light, but the high energy ray of light will break through and go out through this end of a tube. Now, if we inject into this device a single photon of proper frequency level that corresponds to a difference in energy levels of electrons between excited and stable condition or just wait until some electron will spontaneously jumps down, emitting a needed photon, this photon will cause one of the excited electrons to relax, jump to a ground level shell and emit the photon identical to an incident one. That's what stimulated emission means.

The two photons, original and emitted, will not have enough energy to go through a partially reflective end of a tube and will just hit two previously excited electrons, causing emission of two new photons of the same frequency and phase. We have four photons now.

Continuing this "chain reaction" of producing more and more photons, the total energy of light grows until it's strong enough to go through a half-reflective end of a tube as the bright coherent light.

Actually, there are many different ways to implement the laser. Gas tube is not the only one. There are lasers built with liquid substance, solid substances, semiconductors etc.
The energy source might not only be an electric battery, but some bright light as well.

One of the applications of laser technology is military weapon system to destroy drones and other flying objects with high precision and minimum expenses.

## Sunday, December 11, 2022

### Problems on Photons: UNIZOR.COM - Physics4Teens - Waves - Photons and Ma...

Notes to a video lecture on http://www.unizor.com

Problems on Photons and Matter

Problem 1

A hydrogen atom has one electron that is located in a certain shell.

It can be in a "normal" state on the ground level of energy (energy level #1 or shell #1).
When it's excited, it can absorb certain amount of energy and jump to level #2 or #3 etc. to shells for higher energy electrons.

The increase in energy level number corresponds to absorption of more energy to jump to this level.
If electron emits some energy in a form of radiation, it jumps from higher energy shell to a lower energy.

The energy level number N for a hydrogen atom can contain an electron that carry an amount of energy equaled to
EN = Eground /
where Eground = 13.6 eV
(electron-volts)

Calculate the energy differences between shells, restricting the calculations only to the first four shells with energy levels 1, 2, 3 and 4.

Solution

To jump from a shell Li with energy level i to a shell Lj with energy level j electron needs to absorb (positive) or emit (negative) a photon with energy
13.6·(1/i² − 1/j²) eV

According to this formula, the table with calculated energy differences is

 i\j L1 L2 L3 L4 L1 0 10.20 12.09 12.75 L2 −10.20 0 1.89 2.55 L3 −12.09 −1.89 0 0.66 L4 −12.75 −2.55 −0.66 0

As you see, the energy difference between two consecutive levels rapidly diminishes with an increase in energy level.
It equals to 10.20 eV between levels 1 and 2, 1.89 eV between levels 2 and 3 and 0.66 eV between levels 3 and 4.

Problem 2

For all cases when the radiation is emitted by electrons of a hydrogen atom (consider only the first four shells, like in Problem 1) determine the frequency and wave length of electromagnetic oscillations and the color of light emitted.

Here are the data you might need:
(a) charge of an electron:
e = −1.60217663·10−19 C
(b) Planck's Constant:
h=6.62607015·10−34 J·s
(c) speed of light:
c=2.99792458·108 m/s
(d) colors for wave lengths in nanometers, according to one (out of many different) sources on the Web:
 < 400 Ultraviolet 380-440 Violet 440-485 Blue 485-510 Cyan 510-565 Green 565-590 Yellow 590-625 Orange 625-740 Red > 740 Infrared

Solution

Based on energy levels derived in Problem 1, we can obtain the frequencies of emitted light based on the formula
E = h·f
from which follows
f = E/h
Considering the energy in Problem 1 was calculated in electron-volts, to convert it to Joules we have to multiply the given energy by a charge of an electron e getting
f = e·E/h
For a frequency f the wavelength λ is calculated based on the speed of light
λ = c/f

Using the results of Problem 1 above, we obtain the frequencies of emitted light, depending on the difference in energy level between shells of the hydrogen atom:
 i\j L1 L2 L3 L2 2466·1012 L3 2923·1012 457·1012 L4 3083·1012 617·1012 160·1012

Converting this to wave length (in nanometers), we obtain
 i\j L1 L2 L3 L2 122 L3 103 656 L4 97 486 1878

Electrons transitioning to the energy level #1 from any other level emit ultraviolet light (wave length is too short, invisible).
Electrons transitioning to the energy level #2 from level #3 emit red light of 656 nanometers wave length.
Electrons transitioning to the energy level #2 from level #4 emit cyan light of 486 nanometers wave length.
Electrons transitioning to the energy level #3 from level #4 emit infrared light (wave length is too long, invisible).

## Thursday, December 8, 2022

### Photons and Matter: UNIZOR.COM - Physics4Teens - Waves

Notes to a video lecture on http://www.unizor.com

Photons and Matter

In previous lectures we discussed different forms of interaction between matter and electromagnetic oscillations.
In this lecture we will present some scientific foundation of these interactions.

Contemporary view on electromagnetic field oscillations is based on the fact that energy carried by these field oscillations is not infinitely divisible in however small pieces, but is delivered in chunks called quanta (this is a plural form, a singular form is quantum) or photons.

Quantitatively, these chunks of energy depend only on the frequency of oscillations f and are equal to h·f, where h=6.63·10−34J·s is Planck's constant.
Notice that the amount of energy in one photon depends only on the frequency of electromagnetic field oscillations, not on an amplitude of these oscillation. What does depend on an amplitude is the density of photons in space and time: the higher amplitude of oscillations at the source of radiation - the higher density of photons per unit of space during a unit of time.

Let's talk now about matter or, more precisely, about the structure of an atom.
The model of an atom, as we understand it today, consists of a nucleus and electrons orbiting around a nucleus on different orbits or, better said, in different shells around a nucleus. Electrons of any particular shell have the same energy. Every shell has its unique energy level shared by all electrons populating this shell. Shells can be viewed as concentric spheres characterized by their energy level.

The most important about these shells is that they can be only at discrete energy levels shared by electrons within it.

Since shells are at discrete energy levels, any exchange of energy between an atom and an electromagnetic field oscillations is possible only if photons of electromagnetic oscillations fit by their energy amount to a difference in energy level between different shells around a nucleus of this atom.

Consider an absorption of radiation by matter.
The absorption of a single photon means that some electron jumps from its shell with a lower energy level to a shell with the higher one.

This photon's energy is ΔE=h·f.
An atom can absorb energy in chunks only equal to a difference in energy levels between its shells.

An obvious consequence of this is that, if electromagnetic oscillations have such a frequency f that each photon's energy is equal to the energy difference between energy levels of two shells around a nucleus, atom can absorb this photon by some electron getting excited and jumping to a shell with a higher energy level.

If there is a mismatch between frequency of electromagnetic oscillations and difference between energy levels of shells around a nucleus, atom cannot absorb the photon. Most likely, an atom increases its own oscillations that results in heating.

A particular case of a mismatch between the energy of a photon and energy levels of the atom's electron shells is when a photon carries energy higher than the difference between the levels of energy of the atom's shells.
In this case excited electron cannot just jump to a higher energy shell, held there by a nucleus' attraction, but, instead, flies free. This happens in the photoemission. In this case excess of energy beyond maximum absorbable by an electron goes into kinetic energy of the electron that flies free.

Now consider an emitting light (that is, initiating electromagnetic oscillations) by previously excited electrons that return back to normal state by jumping from a higher energy shell to a lower one.
Since we deal with discrete amounts of energy between the shells, the frequency of emitted light will also have only discrete values.

Therefore, radiation emitted by any element, as a result of releasing previously absorbed energy, has certain number of possible discrete frequencies that depend on the properties of an element - the energy levels of electron shells of its atom.

Thus, for example, if there are 6 shells where electrons can potentially jump to, there are 5 different frequencies, when electron jumps from the 6th shell onto any of 5 others, releasing some photons of a corresponding frequency, plus there are 4 destination shells for an electron in the 5th shell, plus there are 3 shells to jump to from the 4th one, plus 2 shells to jump down the energy scale for an electron in the 3rd shell, plus 1 destination for an electron in the 2nd shell.

Altogether we have for this element k=5+4+3+2+1=15 different jumps with 15 corresponding frequencies of emitted radiation.
This simple result can be also obtained by using Combinatorics, as the number of combinations of 2 elements from a set of 6, which is
k=6!/(4!·2!)=6·5/2=15.

Consider, for example, an atom of hydrogen with one proton in its nucleus and one electron in some shell.
The normal state of this electron is called a ground state.
We can excite this electron by infusing it with energy of E1=10.2eV, which will cause it to raise to a first excited state.

When this electron with electric charge of −1.602·10−19C returns back to a ground state from the first excited state, it will emit a photon with the same energy that required to excite it - E1=10.2eV.
This amount of energy in a photon corresponds to its frequency
f1 = E1/h =
= 10.2·1.602·10−19/6.626·10−34
= 2.466·1015
Hz

and wavelength
λ1 = c/f1
where c=3·108m/sec is the speed of light.
That gives
λ1 = 3·108/2.466·1015 =
= 1.216·10−7 ~= 122
nm

Therefore, when an electron jumps from the first excited state to the ground one, it will release 10.2eV of energy, which corresponds to a photon with frequency f=2.466·1015Hz and wavelength 122 nanometers (ultraviolet part of a spectrum).

Any jumping of an electron from a shell with higher energy level to a lower one will emit certain radiation. Since each element has specific for this element energy levels, it also can emit only specific frequencies of radiation. By analyzing a spectrum of radiation we can determine what elements emitted this radiation.

## Tuesday, December 6, 2022

### Photochemistry: UNIZOR.COM - Physics4Teens - Waves

Notes to a video lecture on http://www.unizor.com

Photochemistry

Photochemistry is so important that we can authoritatively say that, if not for photochemistry, there would be no life on Earth.

After such a dramatic statement, let's get into the real science.

Light is a source of energy in terms of individual photons absorbed by material. These photons excite material's electrons. This is the beginning and a necessary condition for subsequent chemical reactions.

The chemical reactions caused by this process of absorbing photons of light is photochemical reaction.
The one of the most important such photochemical reaction is photosynthesis - a process that is at the heart of everything that grows on Earth.

Nutrients a tree gets from the ground are components of this chemical reaction, but without light these components will not significantly interact with each other to produce new leaves and branches. Light supplies the energy needed for this reaction, and not just any light. Visible light spectrum is necessary for photosynthesis.
Some other sources of energy will not do. Not heating, nor some static electric field or mechanical oscillations.

Different photochemical reactions need different light to trigger them. Photosynthesis is an extremely complex photochemical reactions that produce some live materials, like leaves on a tree, from simple minerals the tree receives from the ground, carbon dioxide from the air and plain water.
We still are not in possession of real details of this process, it's one of the mysteries of Life.

Here are some other examples of photochemical reactions.

Somehow the Vitamin D is formed in a human body, when sunlight falls on a skin.

Plastic degrades under sun light, its molecules break down into smaller components and it loosens its physical qualities, like translucence.

Our vision is a photochemical reaction.

Solar cells, where light is converted into electricity, accomplish this via photochemical reactions.

Types of Photochemical Reactions
(from Wikipedia Photochemistry)
with h standing for Planck's constant, f - for light frequency, A and B are two substances getting into chemical reaction, when the photons are falling on them.

Photo-dissociation:
AB + h·f → A + B
Photo-rearrangements:
A + h·f → B
A + B + h·f → AB + C
Photo-substitution:
A + BC + h·f → AB + C
Photo-redox:
A + B + h·f → A + B+

All the above reactions require the presence of photons (referenced as h·f above) to actually happen.

## Monday, December 5, 2022

### Luminescence: UNIZOR.COM - Physics4Teens - Waves

Notes to a video lecture on http://www.unizor.com

Luminescence

Luminescence is an effect of emitting visible light without such obvious source of energy as heat.
The exact mechanism of luminescence is rather complex, but, in general, it involves some external source of energy that excites electrons of some material, which, in turn, follows by their normalization with emitting extra energy as photons of visible light.
The luminescence can be observed in many different cases. Here are a few examples.

Electric Luminescence

Electric luminescence can be observed when electric current runs through an object causing it to emit photons as visible light.
It was discovered in the beginning of 20th century.

This is not the result of heating an object having certain electric resistance by a strong electric current, like in incandescent lamp, but the result of an impact of exciting the material's electrons and, as a result, emitting photons by these excited electrons.

The usual materials that have the capacity to produce electric luminescence are certain semiconductors, and the source of energy to excite their electrons is a relatively weak electric current or, in some cases, an electric field.

An example of the electric luminescence is light emitting diodes (LED) - semiconductors that emit light when an electric current flows through them.
The first LED was created in 1927 in Russia, but no practical usage of this was done until later by numerous European researchers.
A well familiar example is CRT screen for TV and computers, where phosphorus lights up under the electrons' bombardment.
A backlit clock can be constructed from two flat electrodes with one them being transparent and phosphorus layer in between, which lights up if the conductors have some voltage between them.

Chemical Luminescence

Chemical luminescence occurs as a result of certain chemical reactions, sometimes accompanied by emission of heat.
One of the substances that can produce light is luminol, which, if mixed with hydrogen peroxide, produce blue light.

Examples of a chemical luminescence are a glow stick we see as a party decoration or emergency lights.

Photoluminescence

Photoluminescence is emitting light from a substance previously exposed to light. During the stage of exposure to light this substance absorbs electromagnetic radiation (photons of light) that excite its electrons.
After the source of external light stops, these exited electrons gradually return to normal state, emitting extra energy (as photons of light) for some time.
The time delay between absorption of electromagnetic radiation (photons of light) and emitting it is different and depends on many factors. It can vary from milliseconds to hours.

Phosphor is an example of a material that absorbs visible light and, later on, will emit the light back. It might be seen in some watches that glow in the dark showing hands and numbers for quite some time.

Mechanical Luminescence

Mechanical luminescence is emitting light as a result of mechanical action on a solid material.
Some materials emit light after being exposed to such mechanical activities as pressure, deformation, oscillation (for example, by ultrasound), friction (rubbing) etc.

Termoluminescence

Termoluminescence is related to emission of light by some crystalline substances after, first, irradiating them and, second, heating them.
During the stage of irradiating the electrons absorbs the energy, but do not immediately emit it back. It's stored in some deformations of a crystalline lattice. Heating is needed to restore the defects in crystalline lattice and release this energy in a form of light.

## Sunday, December 4, 2022

### Problems on Photoelectricity: UNIZOR.COM - Physics4Teens - Waves - Photoelectricity

Notes to a video lecture on http://www.unizor.com

Problems on Photoelectricity

Problem A

Energy needed to tear a particular electron from the attraction of its nucleus is called binding energy of this electron.
Knowing it, we can calculate the minimum frequency of incident light to initiate photoemission of this type of an electron.

Express the minimum frequency of incident light fmin as a function of a binding energy of an electron Ebinding.
Find the corresponding period of oscillations and the wave length.
Assume, the medium of light propagation is vacuum.

Solution

Energy of a single photon of an incident light h·fmin should be, at least, equal to Ebinding.
Therefore,
Ebinding = h·fmin
where h is the Planck's constant.
From this
fmin = Ebinding/h
Angular frequency
ω = 2π·f = 2π·Ebinding/h
Period of oscillations is inverse of a frequency
τ = 1/f = h/Ebinding
The wave length λ depends of speed of light c and a period τ:
λ = c·τ
Therefore,
λ = c·h/Ebinding

Problem B

Using the results of Problem A above, calculate the minimum frequency of light required to start photoemission from a plate made of gold.
Also find the corresponding wave length of this light.
Perform calculations to three decimal places.
Consider the value of binding energy of a particular electron in gold to be
Ebinding=5.17eV (electron-volt)
and the value of Planck constant
h=6.62607015·10−34m²·kg/s

NOTE about electron-volt (eV) as a unit of energy and its relation with SI unit of energy joule (J).
1eV is an amount of kinetic energy gained by a single electron moving within an electrostatic field from point A to point B with the difference in electric potential between these points equal to 1V (volt).
Because the charge of an electron in coulombs (C) is 1.602176634·10−19C,
and 1V·1C=1J,
1 eV = 1.602176634·10−19J.

Solution

Ebinding = 5.17eV =
= 5.17·1.602·10−19J =
= 8.282·10−19J

Light frequency, as described above is related to the energy of its photons, it can be calculated from the formula
Ebinding = h·fmin
where h is the Planck's constant.
Therefore,
fmin = Ebinding/h =
= 8.282·10−19J /
/(6.626·10−34m²·kg/s) =
= 1.250·1015 J·s/(m²·kg)

Notice,
J = N·m = kg·m²/s²
Therefore, the units of this result are
J·s/(m²·kg) = N·m·s/(m²·kg) =
= kg·m²·s/(m²·kg·s²) = 1/s

which is the right units for frequency (number of oscillations per second).
So, we express the frequency in usual units 1/s:
fmin = 1.250·1015 1/s

Wave length calculation is based on the speed of light, approximately, 3·108 m/s.
λ = c·τ = c/f =
= 3·108/(1.250·1015) m·s/s =
= 2.4·10−7m = 240nm

This wavelength is below the visible spectrum from about 400 to about 700 nanometers and belongs to ultraviolet segment.

## Sunday, November 20, 2022

### Photoelectricity: UNIZOR.COM - Physics4Teens - Waves - Photoelectricity

Notes to a video lecture on http://www.unizor.com

Photoelectricity

In the previous lecture "Energy of Light" of this chapter we investigated the amount of energy carried by electromagnetic waves.
A single wavelength λ of harmonic oscillations of electromagnetic field carried an energy
Wλ = (λ/4)·[ε0·E0²+(1/μ0)·B0²]
where
ε0 is the electric permittivity of vacuum,
E0 is an amplitude of electric component of the field oscillations,
μ0 is the magnetic permeability of vacuum,
B0 is an amplitude of magnetic component of the field oscillations.

We can shorten this formula using a dependency between electric and magnetic amplitudes E0=c·B0 and expression of the speed of wave propagation c in terms of permittivity and permeability of vacuum c²==1/(ε0·μ0) described in the previous lecture.
Wλ =
= (λ/4)·
[ε0·E0²+(ε0/c²)·B0²] =
= (λ/4)·
[ε0·E0²+ε0·E0²] =
= (λ/2)·ε0·E0² =
= (c·τ/2)·ε0·E0²

where τ is a period of oscillations, that is the time a wave propagates by the distance equal to its wave length.

As you see, it's proportional to a wave length and to a square of amplitude of oscillations.

When these electromagnetic waves (like visible light) fall on some surface, some of this energy is absorbed by the material it falls on. Obvious consequence of this is that this energy is converted into heat, and the temperature of the surface should rise.

While this is, generally, true, some experiments showed that not only the temperature of the surface is increasing, but some electrons are flying off that surface and can be detected.
This was discovered by Heinrich Hertz, a German physicist, in 1887 and is known as Hertz effect or photoelectric effect or photoemission and these flowing away electrons were called photoelectrons.
. With proper arrangement these electrons, kicked out from their orbits around an atom's nucleus by falling waves of electromagnetic oscillation, can form an electric current and used to detect some events related to light.

A simple example of this arrangement can have a negative pole of a battery connected to a metal surface A, while a battery's positive pole connected to some other metal plate B positioned opposite to A on a small distance. If a light that causes photoelectric emission falls on a surface A, it kicks off some electrons, which will be immediately attracted to surface B, thus creating a current in a circuit, while the light is on. This can, for example, be used to count parts on a conveyer going across some light beam with the beam interrupted with each part passing by.

Numerous experiments were conducted to research this process.
Different materials were used (primarily, metals), different intensities and frequencies of light were experimented with and very important observations were made.

The fact that the beam of light kicks off some electrons from a surface of a metal was not surprising.
Expected results of experiments were that the brighter light (higher amplitude of electromagnetic oscillations) should cause electrons to be flying off a surface in larger quantities and higher speed and, therefore, higher kinetic energy. Also, even the low intensity beam of light should gradually supply sufficient amount of energy for electrons to start flying off a surface of a metal.

What was surprising was that experiments did not produce these expected results.

Most importantly, for any kind of metal there was a specific frequency of electromagnetic waves (ω - angular frequency, f - regular frequency, that is the number of oscillations per second and ω=2π·f) or, equivalently, specific wave length (λ=2π·c/ω=c/f, where c is a speed of light) such that with this and higher frequency (or, equivalently, this and shorter wave length) the photoelectric effect was observed, but lower frequencies of waves (or longer wave lengths) did not cause this effect, no matter how long or how bright the light was shining on a metal surface, so the low frequency light energy was converted into internal heat with no photoelectrons emitted.

Without pretending that the following is a true explanation of why a specific frequency (or higher) is needed to kick off photoelectrons from a surface of a particular metal plate, we consider the following analogy helpful to intuitively feel the underlying mechanism of photoemission.

Consider you are in a car riding on a bumpy road. Assume, all bumps are of the same height, but different length, some short and steep, some long and gradually changing the height. A car can go over a long bump, when it slowly rising to its maximum height and slowly going down or it can be a short bump of the same height, but the car will have a steeper rise to its peak and steeper way down.
If you ride inside a car on this bumpy road, you will, probably, feel almost nothing going over a long bump, but a short one will cause a jerk and you will be thrown up by a couple of inches.
Steep forceful rise of a car causes it to sharply accelerate upwards, and whoever is inside will feel a strong but short lived force pushing upwards and, maybe, even exceeding the force of gravity.

Similar thing happens when after washing your hands you try to get rid of excess of water by shaking it off. Slow movement of hands with any amplitude will not do a trick because the acceleration of the water drops would be insufficient to shake these drops off, but short and abrupt motion will be effective, the drops will fall off.

Well, similar thing happens with electrons. A long electromagnetic wave length (that is, a wave of a low frequency) rises the energy of electrons inside a metal slowly and the energy dissipates among other electrons, not actually rising in any particular electron high enough to tear it off a nucleus against a force of a nucleus attraction.
If, on the other hand, a short wave hits an electron, it acts like a short bump to a car, it quickly lifts it, kicking off an orbit, so it becomes a free photoelectron.

So, for light to cause the photoelectric effect it's necessary to have to abruptly "shake" an electron with a short wave length (that is, high frequency) of light.

Albert Einstein has received a Nobel Prize in Physics for a really scientific theory of photoelectric effect based on a concept of light energy carried by discrete packets called light quanta or photons.
This and works of other physicists, like Max Planck, was the beginning of quantum theory, one of the major developments of the 20th century Physics.

It also showed a duality of electromagnetic waves that in some circumstances act like waves, while in other cases demonstrate quantum properties similar to particle theory of the past.

The wave theory of electromagnetic field developed by Maxwell and reflected in Maxwell's equations (see the chapter "Electromagnetic Field Waves" in this part "Waves" of a course) considers electromagnetic field as the space with continuous distribution of energy.
Einstein's approach was to consider the light (and electromagnetic field in general) as a space with discrete distribution of energy, with each "particle" of light carrying certain finite amount of energy - a photon.

According to this theory, each light quantum (or photon) carries an energy Ephoton that is proportional to its frequency
Ephoton = h·f
where
h=6.63·10−34 J·s (Joules·seconds) is Planck's constant and
f is the wave frequency, units of which are 1/s (Hertz, oscillations per second).

Each electron is attracted to some nucleus. It means, it needs some energy Efree to fly free, same as a spaceship needs an energy to break off the Earth gravitation.
An electron can absorb only one photon and either is kicked off its orbit, if the frequency of incident radiation is sufficiently high, or stays near its nucleus, if the frequency of radiation is not high enough.

If a photon hits an electron and the photon's energy Ephoton exceeds than what's needed to set free an electron Efree, the electron becomes a free photoelectron and, for example, can be used in some electronic device.
Excess energy Ephoton−Efree is transformed into kinetic energy of a photoelectron, which results in a simple equation
Ephoton−Efree = ½me·V²
or
Ephoton = Efree + ½me·V²
where
me is a mass of an electron,
V is the speed of an electron kicked off its orbit by a photon.

The concept of a photon leads to a new understanding of a term intensity, as applicable to light and other electromagnetic radiation. Intensity of radiation is defined as a quantitative characteristic of radiation proportional to the number of photons produced per unit of time.

At the same time the number of photoelectrons emitted from a surface per unit of time as a result of an incident radiation with sufficiently high frequency is also proportional to a number of photons falling on this surface.
Therefore, the photoelectric current, that is proportional to a number of photoelectrons produced in a unit of time, is proportional to intensity of incident radiation.

## Tuesday, November 15, 2022

### Energy of Light: UNIZOR.COM - Physics4Teens - Waves - Photoelectricity

Notes to a video lecture on http://www.unizor.com

Energy of Light

As was presented in the lectures "Electric Energy" and "Magnetic Energy" of a chapter "Energy of Waves" in this part of a course "Physics 4 Teens", electromagnetic field carries energy, electric and magnetic.

The density of the field's total energy PE+M(t,x,y,z) at any time t at any point {x,y,z} of this field can be expressed as a sum of electric energy density
PE(t,x,y,z) = ½ε·E²(t,x,y,z)
where E(t,x,y,z) represents the intensity of an electric field
and magnetic energy density
PM(t,x,y,z) = ½(1/μ)·B²(t,x,y,z)
where B(t,x,y,z) represents the intensity of a magnetic field.

Constants ε and μ in the above expressions stand, correspondingly, for electric permittivity and magnetic permeability of an electromagnetic field.

For our purposes we assume the light propagates in the vacuum, so ε=ε0 and μ=μ0.

So, the total energy density of electromagnetic field is
PE+M(t,x,y,z) =
= PE(t,x,y,z) + PM(t,x,y,z)
.

Light is an oscillating electromagnetic field. Therefore, light carries energy.

Let's evaluate the amount of energy in one wave length λ of a single light ray propagating in vacuum in the direction of the Z-axis with speed c with sinusoidal harmonic oscillations of field intensities.

Assume, electric field intensity oscillates around the X-axis, depending on time t and distance from the source of light z, according to a formula:
E=E(t,z)=E0·cos(ω(t−z/c)),
where E0 is an amplitude and
ω is an angular frequency of oscillations.
The magnetic field intensity synchronously oscillates around the Y-axis with the same angular frequency ω:
B=B(t,z)=B0·cos(ω(t−z/c)),
where B0 is its amplitude.

The expression inside a cos() function for both (electric and magnetic) components of an electromagnetic field represents the fact that a wave on a distance z from the origin is similar to the wave at the origin (that is, when z=0), but reaches location z with a time delay of z/c needed to reach that point with the speed of wave propagation c.

Then the electric field energy in one wave length of this ray is
WE = 0λ½ε0E²(t,z)dz
which, considering the expression for electric field intensity, is
(1/2)·ε0)0λE0²·cos²(ω(t−z/c))dz

Analogously, the magnetic field energy in one wave length of this ray is
WM = 0λB²(t,z)/(2μ0)dz
which, considering the expression for magnetic field intensity, is
(1/(2μ0))0λB0²·cos²(ω(t−z/c))dz

Let's calculate the common for these two expression integral
0λcos²(ω(t−z/c))dz
Then we will use it for calculation of WE and WM.

Substitute
u = ω(t−z/c))
z = c·t−(c/ω)·u
Then
du = −(ω/c)·dz
dz = −(c/ω)·du
We have to change the limits of integration for a new variable u:
if z=0, u=ωt,
if z=λ, u=ωt-(λω)/c.

Recall the following relationships between angular frequency ω, speed of light propagation c, wave length λ, period τ and frequency f, which were all discussed in the lecture "Rope Energy" of the chapter "Energy of Waves" in this part "Waves" of the course "Physics 4 Teens":
λ = c·τ
τ = 1/f
f = ω/(2π)
from which follows that
λ = c·τ = c/f = 2πc/ω

Therefore, upper limit of integration for variable u is
ωt−(λω)/c = ωt−2π.

With all the above substitutions integral
0λcos²(ω(t−z/c))dz
can be expressed in terms of variable u as
ωtωt−2π cos²(u)·(−(c/ω)·du) =
= (c/ω)
ωtωt−2π cos²(u)·du

Using trigonometric formula
cos²(φ) = ½[cos(2φ)+1]
and standard integration techniques, the result of integration is
(c/ω)ωtωt−2π cos²(u)·du =
= π·c/ω

Based on the above, the energy of one wave length λ of a single ray of light is
Wλ = (1/2)(ε0πc/ω)·E0² +
+ (1/(2μ0))(πc/ω)·B0² =
=
[π·c/(2ω)]·[ε0·E0²+(1/μ0)·B0²]

Let's evaluate the amount of energy falling on some surface of area A during time T from a set of synchronous monochromatic parallel rays of light (flat waves) that are directed perpendicular to this surface based on speed of light c and its angular frequency (that we perceive as color) ω, assuming the light is harmonic synchronous oscillations of an electromagnetic field with electric amplitude E0 and magnetic B0.

During time T on the flat surface area A falls perpendicularly to it all the energy of an electromagnetic field in the volume V=A·c·T Assume that the direction of the propagation of light (red lines on the picture above) is vertically down along the Z-axis, while X- and Y-axes are on the bottom surface of area A.

The height of a cylinder on a picture above is the distance light covers in time T, that is, the height equals to c·T.
In this height we can fit certain number of wave length of waves. If the wave length is λ, the number of waves we can fit into the height c·T is
c·T/λ = c·T·ω/(2πc) = ω·T/(2π)

Taking into consideration the area A the light falls on, the total amount of energy falling on this area during time T is
WA,T = [ω·T·A/(2π)]·Wλ =
=
[ω·T·A/(2π)]·
·
[π·c/(2ω)]·[ε0·E0²+(1/μ0)·B0²] =
=
[T·A·c/4]·[ε0E0² + (1/μ0)B0²]

Density of energy falling per unit of area per unit of time that can be called the average intensity of light I can be obtained by dividing the above expression by A and by T getting
I = [c/4]·[ε0E0² + (1/μ0)B0²]

An important addition to these calculations allows to express this average intensity of light only in terms of electric field intensity.
In the lecture "Speed of Light" of the chapter "Electromagnetic Field Waves" of this part of a course "Physics 4 Teens" we have derived a simplified form of the Fourth Maxwell Equation as
By(t,z)/z = μ0·ε0·Ex(t,z)/t
where indices x under E and y under B correspond to a model we discuss in this lecture, that electric field intensity oscillates around X-axis, magnetic field intensity oscillates around Y-axis and the light propagates along Z-axis.
So, for our purposes we can safely drop these indices, which results in the following differential equation
B(t,z)/z = μ0·ε0·E(t,z)/t

In that same lecture "Speed of Light" we have expressed the speed of propagation of electromagnetic waves in vacuum as
c² = 1/(μ0·ε0)
Using this, we can rewrite the differential equation with partial derivatives above as
−c²·B(t,z)/z = E(t,z)/t

We have assumed that the ray of light is described by harmonic oscillations
E(t,z )= E0·cos(ω(t−z/c)),
B(t,z) = B0·cos(ω(t−z/c)),

Let's take the partial derivatives in the differential equation above for these field intensities.
B(t,z)/z =
= B0·sin(ω(t−z/c))·(ω/c)

E(t,z)/t =
= −E0·sin(ω(t−z/c))·ω

For these harmonic oscillations the differential equation above looks like
c·B0·sin(ω(t−z/c)) =
= E0·sin(ω(t−z/c))

from which follows
c·B0 = E0

Now the above expression for the amount of energy falling per unit of area per unit of time (average intensity of light) can be transformed into
I = [c/4]·[ε0E0² + (1/μ0)B0²] =
=
[c/4]·[ε0E0² + (1/(c²·μ0))E0²]

From c²=1/(μ0·ε0) follows
that 1/(c²·μ0)=ε0.
Therefore,
I = [c/4]·[ε0E0² + ε0E0²] =
= c·E0²·ε0 /2

Simple transformations allow to express the same average intensity of light in terms of magnetic field intensity
I = c·B0²/(2μ0)
or in terms of both electric and magnetic field intensities
I = E0·B0/(2μ0)

An important consequence from the above formulas is that the density of light energy, that is the average amount of energy falling on a unit of area during a unit if time or average light intensity is proportional to a square of its amplitude.

## Friday, November 4, 2022

### Speed of Light: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

Speed of Light

Based on Maxwell equations, we can derive separate differential equations for vectors of electrical (E) and magnetic (B) components of a electromagnetic field and the speed of its propagation in a simple case of vacuum as a medium and no additional sources of electric charges, moving or stationary, present.

As we know, oscillations of electromagnetic field are transverse, vectors E and B are perpendicular to each other and both are perpendicular to the direction of waves propagation.

Assume that the wave propagation is along the Z-axis of some system of coordinates, electric component E is changing along the X-axis and magnetic component B is changing along the Y-axis. For this type of the electromagnetic field we can state
E = {Ex,Ey,Ez}
with Ey=0 and Ez=0,
while Ex(t,z) being a function of time t and coordinate along the Z-axis z.
Analogously,
B = {Bx,By,Bz}
with Bx=0 and Bz=0,
while By(t,z) being a function of time t and coordinate along the Z-axis z.

Using the above format of vectors of electrical (E) and magnetic (B) components of the electromagnetic field, we can analyze the form of the Maxwell equations for them.

The third Maxwell equation that relates induced electric field to a changing magnetic field - the Faraday's Law - is

E = −B/t

Recall the definition of a vector (cross) product of a pseudo-vector
∇={∂/∂x,∂/∂y,∂/∂z}
by any vector
V(x,y,z) = {Vx,Vy,Vz}
using unit vectors i, j and k along the coordinate axes:
V =
= (
Vz/y − Vy/z)·i +
+ (
Vx/z − Vz/x)·j +
+ (
Vy/x − Vx/y)·k

(you can refresh this in the lecture "Curl in 3D" of this chapter of a course on UNIZOR.COM)

Considering a special characteristics of vector E with only Ex(t,z)≠0,
E =
= (
Ez/y − Ey/z)·i +
+ (
Ex/z − Ez/x)·j +
+ (
Ey/x − Ex/y)·k =
= (
Ex/z)·j

Considering a special characteristics of vector B with only By(t,z)≠0,
B/t = −(By/t)·j

Therefore, the third Maxwell equation in this case of a simple electromagnetic field takes the form
(Ex/z)·j = −(By/t)·j
Hence,
Ex/z = −By/t

The fourth Maxwell Equation looks like

∇⨯B = μ·ε·E/t + μ·σ·E

Here the variables participating in this equation are:
μ is a magnetic permeability of the media,
ε is an electric permittivity of the media,
σ is a material conductivity of the media.

Considering our medium is vacuum, σ=0, and the fourth Maxwell equation looks like
∇⨯B = μ·ε·E/t
Since only By(t,z)≠0,
B =
= (
Bz/y − By/z)·i +
+ (
Bx/z − Bz/x)·j +
+ (
By/x − Bx/y)·k =
= −(
By/z)·i

The vector E on the right side of the fourth Maxwell equation has only X-component not equal to zero.
Therefore, this equations looks like
−(By/z)·i = μ·ε·(Ex/t)·i
Hence,
By/z = μ·ε·Ex/t

As a result, the third and fourth Maxwell equations for an electromagnetic field in vacuum with no additional charges take the short form:
Ex/z = −By/t
By/z = μ·ε·Ex/t

To separate electric and magnetic components into separate single variable equations, let's differentiate the first equation by space coordinate z and the second equation - by time t.
∂²Ex/z² = −∂²By/tz
∂²By/zt = μ·ε·∂²Ex/

From these two equations we can construct one for Ex:
∂²Ex/z² = μ·ε·∂²Ex/

We can drop an index x for an electric component of an electromagnetic field since the other two components are zero.
In vacuum magnetic permeability μ is a known constant, approximately equal to
μ0=1.25663706212·10−6
(Henry/meter)

and an electric permittivity ε is also a known constant, approximately equal to
ε0=8.85418781281·10−12
.

In these terms the last equation for the electric component of an electromagnetic field is
∂²E(t,z)/z² = μ0·ε0·∂²E(t,z)/
This equation is the general wave equation for the direction of the wave propagation along the Z-axis.

Incidentally, one of the solutions to it, considered in the lecture "Wave Equation 2" of this chapter of the course on UNIZOR.COM is the wave function
E(t,z) = A·sin(ω·(t−z/v))
where ω is the frequency of oscillations and v is the speed of wave propagation.

Indeed, differentiating E(t,z) by z gives
E(t,z)/z =
= −(ω/v)·A·cos(ω·(t-z/v))

∂²E(t,z)/z² =
= −(ω/v)²·A·sin(ω·(t-z/v))

Differentiating E(t,z) by t gives
E(t,z)/t =
= ω·A·cos(ω·(t-z/v))

∂²E(t,z)/t² =
= −ω²·A·sin(ω·(t-z/v))

Therefore,
∂²E(t,z)/z² = (1/v²)·∂²E(t,z)/

If 1/v²=μ0·ε0, the wave equation for E(t,z)=A·sin(ω·(t-z/v)) is satisfied.

The last statement is extremely important, since it defines the speed of wave propagation in terms of magnetic permeability and electric permittivity of the medium.

Using the constants of magnetic permeability and electric permittivity of vacuum listed above, the speed of propagation of electromagnetic oscillations for vacuum can be calculated as follows:

1/v² = μ0·ε0 =
= 11.1265005604·10−18

v² = 1/(μ0·ε0) =
= 0.0898755178·1018

v = √0.0898755178·109 =
= 299,792,458
m/sec

The speed of light in vacuum is usually denoted as c, so
c = 299,792,458 m/sec

All the above calculations were performed by James Maxwell at the end of 19th century.
At the same time the speed of light was measured relatively precisely, and it was almost exactly the same as the speed of electromagnetic waves propagation calculated above.
That led to a hypothesis that light is the oscillations of electromagnetic field.

The magnetic permeability and electric permittivity of vacuum are not dependent on the frequency of oscillations. That's why the speed of light of any color in vacuum is the same as calculated above.
When the light goes through some material substance (like air or glass), the magnetic permeability and electric permittivity of a medium do depend on the frequency of oscillations and, therefore, the speed of light of different colors varies in material substance, which causes such effects as interference, diffraction or dispersion of light.

### Magnetic Field Energy: UNIZOR.COM - Physics4Teens - Waves - Energy of Waves

Notes to a video lecture on http://www.unizor.com

Magnetic Field Energy

Our task is to determine the energy carried by a magnetic field. More precisely, a density of a potential energy as a function of a local characteristic of a magnetic field - the value of its intensity vector B.

First, we have to choose a device where a uniform magnetic field can be created and evaluate the total amount of energy we have to spent to create this uniform magnetic field. This energy will be stored in a certain volume of space inside this device occupied by a created magnetic field as its potential energy.
Then, dividing the total amount of energy of the magnetic field inside this space by the space volume, we will get a density of the potential energy of the magnetic field.

Recall the description of an infinitely long solenoid as a device that produces relatively uniform magnetic field. It was described in the lecture "Electromagnetism - Magnetism of Electric Current - Solenoid" of this course "Physics 4 Teens" on UNIZOR.COM.
The absolute value of an intensity of a uniform magnetic field B inside such a solenoid, was expressed as
B = μ·n·I
where
μ is permeability of the medium inside a solenoid,
n is the density of the wire loops making up a solenoid and
I is the electric current running through a solenoid.

As physicists usually do, we assume that our real solenoid, though not infinite, is long enough to use this formula for absolute value of a magnetic field intensity at all points inside a solenoid and it's zero outside.

While this methodology depends on the fact that the magnetic field we analyze is inside a solenoid, the final formula for energy density will depend only on the field's intensity and, as a field's local characteristic, will be the same, no matter what is the outside source of a magnetic field, whether it's a solenoid or a few permanent magnets, or the result of electromagnetic oscillations far from the source of these oscillations.

Consider a battery of voltage V that we connect to a solenoid of resistance R to create a magnetic field inside it. Electric current I(t) through a solenoid in the beginning equals to zero and, gradually, rises to some value Imax=V/R (Ohm's Law).

Since the electric current varies with time from 0 to Imax, the above formula for magnetic field intensity is time-dependent:
B(t) = μ·n·I(t)

Recall that changing electric current in a wire causes changing magnetic field around it that adversely affects the current, resisting its change. This is a self-induction effect described in the chapter "Electromagnetism - Self-Induction" of this course.

Therefore, the process of rising of an electric current through a solenoid requires some work performed by the battery against self-induction of the solenoid.
This work is converted into energy of the magnetic field inside a solenoid, and we will quantitatively evaluate it.

According to Faraday's Law of Induction, the electromotive force (EMF) U generated against the increasing electric current in a wire loop is proportional to a rate of change of the magnetic flux Φ1 flowing through this single loop:
U(t) = −dΦ1(t)/dt
Magnetic flux is related to a field intensity in a single loop of wire of area A as
Φ1(t) = B(t)·A
For a solenoid with total number of wire loops N the flux will be N times greater:
ΦN(t) = B(t)·A·N

Using the above stated dependency between field intensity B(t) and the electric current I(t), we can express the magnetic flux through a solenoid as
ΦN(t) = μ·n·I(t)·A·N

Obviously, if the length of a solenoid is l and the density of the wire loops is n, the total number of wire loops is
N = n·l

Therefore,
ΦN(t) = μ·n²·I(t)·A·l
Incidentally, the expression A·l is a volume of space inside a solenoid Vspace, so we write the above formula as
ΦN(t) = μ·n²·I(t)·Vspace

The power P(t) at any moment of time t needed to run the electric current (that is, the amount of work per unit of time) depends on the voltage, that is EMF U(t) needed for it and the electric current at the time I(t) as
P(t) = U(t)·I(t)
Using the expression above for the EMF, we derive an absolute value of power as a function of time
P(t) = dΦN(t)/dt·I(t) =
=
d/dt
[μ·n²·I(t)·Vspace]·I(t) =
= μ·n²·Vspace·
[dI(t)/dt]·I(t)

The constants in front form a characteristic of a solenoid called inductance L=μ·n²·Vspace, so we simplify the expression for power as
P(t) = L·I(t)·[dI(t)/dt]

During the time period from t to t+dt the electric current will rise by dI=[dI(t)/dt]·dt and the work performed will be dW=P(t)·dt.
Therefore, to increase the electric current from I to I+dI we have to perform work equal to
dW = L·I·dI

The total amount of work W we have to perform to rise the electric current from 0 to Imax can be obtained by integrating the above expression by I:
W = [0,Imax]L·I·dI = ½L·I²max

When the electric current reaches Imax, the magnetic intensity B reaches its maximum value, and the potential energy accumulated in the magnetic field will be equal to work performed.

From the above formula for magnetic field intensity B=μ·n·I follows that
I = B/(μ·n)
Substituting this into an expression for total work performed,
W = ½L·B²/(μ·n)² =
= ½·μ·n²·Vspace·B²/(μ·n)² =
= ½·Vspace·B²/μ

Dividing the total work performed W (that is, the total potential energy accumulated by a solenoid during the process of rising the electric current) by the total volume inside a solenoid Vspace, we obtain the potential energy density PM of a magnetic field as a function of a local field characteristic - its intensity B:

PM = B²/(2μ)

We have used a simple kind of a magnetic field to derive the formula above - the static uniform finite field inside a solenoid. Notice, however, that the formula contains only the local property of the field - its intensity at any point B.
Even if the field is of a more complex kind (variable, non-uniform, infinite etc.), since only its intensity participates in the above expression for a potential energy density, the formula should be valid.
In particular, it's valid for a magnetic component of an oscillating electromagnetic field.

Combining the result for an electric component of electromagnetic field presented in the previous lecture with the formula above for magnetic component, we obtain a total potential energy density of an electromagnetic field

PE+M = ½[ε·E² + B²/μ]

Final comment about an oscillating electromagnetic field.
Once formed at the source, variable electric field possess certain potential energy.
Where is it going if the source stopped producing it?

Since variable electric field produces variable magnetic field, this energy is transferred to a newly formed variable magnetic field into its potential energy.

In turn, this potential energy of a variable magnetic field is used to generate a new variable electric field with its potential energy, which it uses to generate a new variable magnetic field etc.

That's how energy is delivered by an electromagnetic oscillations to some final destinations, where it might be used to convert it to some other forms.

## Saturday, October 29, 2022

### Electric Field Energy: UNIZOR.COM - Physics4Teens - Waves - Energy of Waves

Notes to a video lecture on http://www.unizor.com

Electric Field Energy

Let's calculate the potential energy density PE of an electric field inside a capacitor as a function of the field's intensity E, which we assume to be uniform between the plates of a capacitor.

While the methodology will depend on the fact that this electric field is between the plates of a capacitor, the final formula will depend only on the field's intensity and, as a field's local characteristic, will be the same, no matter what is the outside source of an electric field, whether it's plates of a capacitor or a few point charges, or the result of electromagnetic oscillations far from the source of these oscillations.

Consider a battery used to charge a capacitor from initial no electric charge between plates with the difference of potential between the plates being equal to zero to some charge Qmax with corresponding difference of potential between the plates (voltage) being equal Vmax. This process of charging requires some work performed by the battery.
The total energy accumulated inside a capacitor as a result of charging a capacitor should be equal to this amount of work.

This work performed by a battery to charge a capacitor from 0 voltage to Vmax can be considered a function of accumulated charge W=W(Q) and it grows from zero to some maximum Wmax - the energy accumulated by a capacitor as it accumulated a charge of Qmax.
During the charging process the voltage between the plates of a capacitor can also be considered as a function of accumulated charge Q, that is V=V(Q), and it grows from zero to Vmax.

Recall that the difference in electric potential between two points of an electric field (voltage) is the amount of work needed to transfer a unit of electric charge (1 coulomb) from one point to another.
Therefore, to transfer an additional infinitesimal amount of electric charge dQ from one plate of a capacitor to another, when there is already transferred amount of electric charge Q that creates a voltage V(Q) between the plates, the battery has to spend an additional amount of work
dW(Q) = V(Q)·dQ

Further recall that the electric charge on each plate of a capacitor (positive on one plate and negative on another) Q and the voltage between the plates V are proportional to each other with a capacitor's capacitance C being a factor:
Q = V·C or V = Q/C

Therefore, the above expression for an additional amount of work can be written as
dW(Q) = [Q/C]·dQ

To calculate a total amount of energy Wmax needed to charge a capacitor from zero to Qmax, we have to integrate dW from 0 to Qmax:
Wmax = [0,Qmax][Q/C]·dQ =
= ½·Q²max/C

Of course, the amount of charge Qmax can be any from zero to some practical maximum, so we can drop an index max, and the work spent by a battery will be
W = ½·Q²/C = ½V²·C

Let's approach the same problem from a different viewpoint that involves the intensity E of an electric field between the plates of a capacitor charged with Q amount of electricity to a voltage V between its plates.

The capacitance C of a capacitor was discussed in a lecture "Electromagnetism" - "Electric Field" - "Capacitors" of this course and, as was shown there, depends on the area of each plate A, distance between plates d and electric permittivity of a medium between the plates ε:
C = ε·A/d
with the vacuum having the permittivity ε0 and any other medium having it as
ε = εr·ε0
where εr is relative permittivity of a medium.

Using this expression for a capacitance C, the total work to charge a capacitor with Q amount of electricity to a voltage level V can be written as
W = ½·Q²·d/(ε·A) = ½V²·ε·A/d

The definition of the electric field's intensity E is the force with which a field acts on a unit charge, while voltage V between plates is an amount of work needed to move a unit of charge from one plate to another.

Therefore, using the principal
Work = Force ⨯ Distance,
we can write
V = E·d

Substituting this into a formula for the total work yields
W = ½E²·d²·ε·A/d = ½ε·E²·(A·d)

Notice that A·d is a volume of the space between the plates of a capacitor, occupied by an electric field.
Therefore, the expression W/(A·d) characterizes the density PE of a potential energy of an electric field between the plates of a capacitor.

Hence, the potential energy density of an electric field is a local characteristic that depends on the field's intensity and the permittivity of a medium where the field propagates
PE = ½ε·E²

Final comment is related to the fact that we used a simple kind of an electric field to derive the formula above - the static uniform finite field between two plates of a capacitor. The formula contains only the local property of the field - its intensity at any point E.
Even if the field is of a more complex kind (variable, non-uniform, infinite etc.), since only its intensity participates in the above expression for a potential energy density, the formula should be valid.
In particular, it's valid for an electrical component of an oscillating electromagnetic field.