Wednesday, December 21, 2022

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

Notes to a video lecture on

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


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


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

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

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


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:
< 400Ultraviolet
> 740Infrared


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

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

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

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

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


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.

AB + h·f → A + B
A + h·f → B
A + B + h·f → AB + C
A + BC + h·f → AB + C
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


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 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 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

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.


Energy of a single photon of an incident light h·fmin should be, at least, equal to Ebinding.
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
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

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.


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.
fmin = Ebinding/h =
= 8.282·10−19J /
/(6.626·10−34m²·kg/s) =
= 1.250·1015 J·s/(m²·kg)

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.