Reflection of straight and circular waves occurs when waves meet circular or straight reflectors.
When plane waves hit a surface at an oblique angle, they are reflected. This reflection follows the laws of reflection. All waves can be reflected.
Water waves are reflected from obstacles in their paths the same way as light and sound waves. All reflections obeys the laws of reflection.
See the figure below.
The laws of reflection states that:
The angle of incidence i equals the angle of reflection r.
The incident ray, reflected ray, and normal at the point of incidence all lie on the same plane.
Reflection of waves obeys the laws of reflection.
Plane waves normal to the reflecting surface
Plane waves incident onto a straight reflector at 90o .to the surface will be reflected such that they are perpendicular to the reflecting surface. see the figure below:
straight and circular waves: reflection of plane waves by curved reflectors
When plane waves falls onto a concave reflector, they converge to a point in front of the reflecting surface. This is the same way all rays of light parallel and close to the principal axis converge after reflection. The plane waves will be reflected as circular waves that seems to change direction after the converging point. see the figure below:
straight and circular waves: plane waves incident to convex(diverging) reflector
When plane waves meets a convex reflector. they are reflected such that they appear to diverge diverge. from a point behind the convex surface. The waves reflected from convex reflector has virtual principal focus.
circular waves against a straight reflector.
Circular waves incident to a straight reflector will be reflected as circular waves. These waves seem to have a converging point behind the plane reflector. see the figure below:
circular waves incident to the concave reflector straight up and moves as plane waves after reflection. See the figure below.
Electrostatics is a branch of physics that deals with behavior and properties of charges that are not flowing. When we subject materials to mechanical friction force against other materials, the electrons near the surface jump out from one material and become lodged to the other material. In other word, when materials rub each other, electrons are transferred. The transfer of electrons is what is referred as charging of the material.
Materials are made from matter and matter is made of atoms. Atoms are considered to be very tiny particles whose size is in the order of 0.1 nanometers and that cannot be divided further. Atom is considered as the blue print of every matter whether it is a gas, liquid or solid. They are the basic structures that are joined together to make molecules that composes matter.
Electrostatics: Structure of an atom
Atom is made up of two parts, a central core called nucleus and outer orbits where electrons goes around the nucleus. The nucleus contains particles called protons and neutrons closely and tightly packed inside.
Protons carries a positive charge whereas electrons carries negative charges. Neutrons carries no charge.
The number of protons and electrons in an atom are equal in number such that the resultant charge is zero. This is because there are equal number of positive charge as there are negative charge so that they cancel out each other making the overall charge in an atom to be zero.
Causes of electrostatics charging
In some materials , electrons are not tightly bund to the nucleus and so when given some little energy, they tend to jump out of the atom. When two materials are rubbed against each other, the heat energy developed due to friction may cause some loosely held electrons from one material to move and be transferred to the other material. Some materials easily losses elecrons whereas others readily accepts electrons during friction.
Materials that losses electrons are said to be positively charged because they have overall more positively charged protons compared to electrons.
Materials that gains electrons are said to be negatively charged because they have overall more negatively charged electrons as compared to the protons. As an example, when polythene is rubbed against flannel clothe, it gains electrons and becomes negatively charged . Consequently, flannel clothe becomes positively charged because it looses some of its negatively charged electrons to polythene.
Glass will loose electrons to silk when they are rubbed together making the glass to gain positive charge and silk to be negatively charged.
The following has been observed when materials have been charged by friction.
Excess negative charge on one body is equal to excess of positive charge on the other body and so no new charges is ever created. In electrostatics charges are never created, they are only transferred.
Some materials will always acquire they same type of charge during charging and so it may be possible to predict the charges on materials after you rub them together.
The quantity of charge in some cases maybe small and in some cases charges may escape before they are detected. When charging by friction, the idea environment is a dry atmosphere and clean charging bodies to avoid discharge.
Some Experiments to explain electrostatic charges
Take a polythene strip and rub it against silk and then take the strip near a thin stream of flowing tap water as shown:
When a charged strip is brought near a thin stream of water, the of water is strongly attracted to the polythene as shown.
when a plastic comb, pen or plastic ruler is rubbed against your clothe or hair, it is observed to attract small pieces of paper as shown.
A household mirrors and windows attract dust and other particles when wiped with a dry clothe because of electrostatic charges.
All the above observations are as a result of electrostatic charges.
There are two types of charges namely negative and positive charges. The SI unit of charge is the coulomb(c).
1 Coulomb = 1000 millicoulombs
millicoulomb = 1000 microcoulombs
1 coulomb = 1000 000 microcoulombs
The basic law of charges
The basic law of charges states that like charges repel, unlike charges attract. In this lesson, we will discuss physics experiments that can verify this basic law.
Experiments to verify the law of charges
To investigate what happens when two charges bodies are brought together, you may need the following apparatus:
glass rods
silk cloth
Silk Thread
Stand
Bunsen burner
polythene rod
duster
To investigate the law of charges in electrostatics, use the following procedure:
Dry glass rod by running it over a Bunsen flame a few times.
rub the dry rod with a silk and then suspend it by a thread on a stand
Dry a second glass rod over bunsen burner and rub it with silk cloth.
Hold the second glass rod close to the first suspended glass rod as shown.
With the glass rod still suspended, bring a polythene rod rubbed with fur close to it as shown.
Observations from experiments on law of charges
when a charged glass rod is moved close to a suspended charged glass rod, they were observed to repel each other.
When a charged polythene rod is moved close to a suspended charged glass rod, they were observed to repel each other.
Explanation
The glass rods were rubbed with the same material and so they acquired same positive charge . The repulsion between them implies that like charges repel each other.
When polythene rod was rubbed with fur, it acquired negative charge. When the charged polythene rod attracts the positively charged glass rod, it shows that opposite charges attracts each other. The above experiment and observations brings us to conclusions on charges with the basic law of charges that states that like charges repel while unlike charges attract.
The basic law of charges states that like charges repel, unlike charges attract. In this lesson, we will discuss physics experiments that can verify this basic law.
Experiments to verify the law of charges
To investigate what happens when two charges bodies are brought together, we need the following apparatus:
glass rods
silk cloth
Silk Thread
Stand
bunsen burner
polythene rod
duster
To investigate the law of charges, use the following procedure:
Dry glass rod by running it over a Bunsen flame a few times.
rub the dry rod with a silk and then suspend it by a thread on a stand
Dry a second glass rod over bunsen burner and rub it with silk cloth.
Hold the second glass rod close to the first suspended glass rod as shown.
With the glass rod still suspended, bring a polythene rod rubbed with fur close to it as shown.
Observations from experiments on law of charges
when we moved a charged glass rod close to a suspended charged glass rod, we observe them to be to repelling each other.
When a charged polythene rod is moved close to a suspended charged glass rod, they were observed to repel each other.
Explanation
The glass rods were rubbed with the same material and so they acquired same positive charge . The repulsion between them implies that like charges repel each other.
When polythene rod was rubbed with fur, it acquired negative charge. When the charged polythene rod attracts the positively charged glass rod, it shows that opposite charges attracts each other. The above experiment and observations brings us to conclusions on charges with the basic law of charges that states that like charges repel while unlike charges attract.
Atomic structure describes how an atom is built from protons, neutrons, and electrons. At the center of the atom is the nucleus, containing positively charged protons and neutral neutrons. Negatively charged electrons orbit the nucleus in shells, with their negative charge attracting the positive protons to hold the atom together. Atoms are electrically neutral because they have an equal number of protons and electrons.
The nucleus of an atom has a specific number of protons and neutrons. The number of protons in the nucleus is called the atomic or proton number. When the number of protons and the number of neutrons in the nucleus are summed up, the resultant number is known as the mass number. Mass number is also known as the nucleon number.
Different atoms has different mass number. For example, hydrogen atom has mass number of 2, meaning it has 1 neutron and 1 proton in it’s nucleus. A neon atom has mass number as 20 having 10 protons, 10 neutrons and 10 electrons. similarly, helium atom has mass number 4 with 2 protons, 2 neutrons and 2 electrons.
describing the mass number in atomic structure
If a certain atom X has atomic number Z with N neutrons and mass number A, then we can express it as:
$$^{A}_{Z}X \ \ \ where A = Z + N$$
Thus neon, helium and hydrogen atom will be represents as:
where Ne is neon atom, He is the helium atom and H the hydrogen atom.
There exists atoms that have the same atomic number but with different mass numbers. Such atoms are said to be isotopes. For example carbon-14 and carbon-12 has mass number 14 and 12 respectively but both has atomic mass 6.
The two will be represented as shown:
$$^{12}_{6}C \ \ and \ \ ^{12}_{6}C $$
Stability of the nuclear in atomic structure
A nuclear is said to be stable when a ratio of it’s proton to neutron number is 1 or close to 1. that is
$$\frac{\text{mass of proton}}{\text{mass of neutron}}=1$$
As atoms gets heavier, there is a marked deviation from this ratio, with the neutron number exceeding that of protons. This causes the nucleus to be unstable and hence increases chances of the nuclear disintegrating to gain stability. A graph of number neutrons N against number of protons Z for different nucleus is illustrated below.
From the graph, it is observed that the stable nuclides are outside the stability line.
Nuclides above the stability lines have too many neutrons. Such nuclides decays in such a way that the number protons increases.
Nuclides below the stability line have too many protons . Therefore, they decay to decrease the number of protons.
Exams questions on X-rays often cover topics like the production, properties, and applications of X-rays. They also address their interaction with matter. Common questions ask about the continuous and characteristic spectra produced in an X-ray tube. They inquire about the effects of changing the anode voltage or using filters. Questions also explore the mechanisms behind X-ray production. Other areas include the dual nature of X-rays (wave-particle duality), their relationship to electromagnetic radiation, and their interaction with biological tissues, including both beneficial and harmful effects:
Exam questions on X-rays
1. Figure 1. Shows part of an x-ray tube
(i) Explain how x-rays are produced in the tube. (1mk)
2. (a) Explain why an x-ray tube is evacuated. (1mk)
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(b) Distinguish between ‘hard and soft’ x – rays (1mk)
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3. X-rays are passed through the air surrounding a charged electroscope.
State what is observed. (1 mk)
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4. Figure 11 shows the circuit of a modern X-ray tube.
(i) Indicate the path of the X-ray beam supplied by the tube. (1 mk)
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(ii) Name the part labeled C and state its function. (2 mks)
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(iii) Name a suitable metal that can be used for the part labeled B and
give a reason for your choice. (2 mks)
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(iv) How can the intensity of x-rays in the tube be increased. (1 mk)
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Exam questions on X-rays
4. The diagram in figure 10 below shows an X-ray tube
(i) What is the nature of the voltage across AB? (1 mk)
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(ii)Give the name of this X-ray tube (1 mk)
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(iii) How can the intensity of the X-rays be reduced (1 mk)
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(iv) The potential difference between the anode and the cathode is 40KV. what would be the maximum velocity of the electron hitting the target? Take the mass of an electron to be 9.1 x 10-31kg and the charge of an electron as 1.6 x 10-19C (3 mks)
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(v) State what happens to thee quality of the X-rays if this voltage is increased (1 mk)
Energy of the x-rays are energy possessed by x-rays radiations. X -rays are electromagnetic waves of very short wavelengths of order 10-10 m. Therefore x-rays carries energies according to plank’s theory. The wavelength of the x-rays overlaps with that ultraviolet and gamma rays in electromagnetic spectrum. This is because X-rays are classified in the range 10-11 to 10-8 in electromagnetic spectrum.
Determining Energy of the x-rays
If a bombarding electron is stopped in a single collision, some or all it’s energy is converted to X-ray energy. At a given accelerating potential difference, the X-rays produced will have varying wavelengths . The minimum wavelength (λmin) corresponds with a collision in which all the kinetic energy in the motion is converted to x-rays.
The kinetic energy (K.E) of the bombarding electron is practically equal to eV. This energy is expressed as. K.E = ev where V is the accelerating potential and e the charge on an electron.
When all the kinetic energy of an electron is used to produce x-ray waves, the frequency of the x-rays produced will be maximum. This maximum frequency is expressed as fmax.
Using Plank’s theory, the X-ray energy is given by: E= hf. where h is the Plank’s constant and f the frequency of the radiation.
eV = hfmax
remember that speed of a wave is given by the general equation v=fλ. But x-rays being part of electromagnetic spectrum moves at a speed of light represented by c.
Hence the speed of the x-ray waves will be c= fλ, c being the speed of light.(c=3.0 x 108 ms-1)
Therefore the speed of the x-ray waves produced will be given by: c = fmaxλmin.
we now express frequency of the x-ray produced in terms of speed and wavelength:
From the equation, we see that the most energetic X-rays have the shortest wavelength.
Example problem on energy of the x-rays
Find the frequency and the energy of a type of X-rays whose wavelength is 10-10 m. (velocity of light c = 3.0 x 108 ms-1, Plank’s constant h = 6.63 x 10-34 Js).
solution
c=fλ
$$ f = \frac{c}{\lambda} $$
$$ f = {3.0 \times 10^8}{10^{-10}} = 3.0 \times 10^{18}Hz $$
E=hf
E = 6.63 x 10-34 x 3.0 x 1018 = 1.989 x 10-15J
practice Questions
1. The frequency of X-rays ranges from 3.0 x 1017 to 3.0 x 1019Hz. Determine
(a) the range of the wavelengths.
(b) the maximum energy of the X-rays. (velocity of light c= 3.0 x 108ms-1 and plank’s constant h= 6.63 x 10-34Js)
2. An X-ray tube has an accelerating potential difference of 100 Kv. what is the shortest wavelength in its X-beam? Take plank’s constant h=6.63 x 10-34Js, charge on an electron e = 1.6 x 10-19C and velocity of light c = 3.0 x 108ms-1.
hard x-rays
X-rays are usually classified as hard or soft. Hard x-rays are described as having high frequency hence shorter wavelengths and so high penetrating power. Hard X-rays are produced by use of high accelerating voltage that causes the electrons move at very high speed towards the target in the x-ray tube.
soft x-rays
Soft X-rays are produced by electrons moving at relatively lower velocities . They are of less energy and lower frequency. Specifically, they fall within the 100-3,000 eV energy range. Due to their lower energy, soft X-rays are more readily absorbed by matter and have limited penetrating power, requiring measurements to be conducted in a vacuum. They are valuable for studying surfaces, near-surface interfaces, and for spectroscopic techniques like X-ray Absorption Spectroscopy (XAS) and X-ray Photoelectron Spectroscopy (XPS)
Examination questions on photoelectric effect involves key concepts like work function, threshold frequency, and the relationship between light frequency, photon energy, and electron kinetic energy. The photoelectric effect involves the emission of electrons from a metal surface when light shines on it. Here are some example exam questions:
Examination questions on photoelectric effect
1. (a) What is meant by threshold wavelength? (1 mark) (b) How does intensity of radiation incident on a metal surface affect the photoelectrons emitted. (1 mark) (c) In an experiment using a photocell, light of varying frequency but constant intensity was made to strike a metal surface. The maximum kinetic energy of the photoelectrons for each frequency, f, was measured. The values obtained are shown in table 2 below
Maximum K.E (x 10-19)
2.8
5.4
7.4
9.0
10.0
11.0
Frequency (x 1015) Hz
1.5
1.9
2.2
2.42
2.57
2.75
(i) Plot a graph of maximum K.E against frequency. (5 marks)
2. The fig. below shows a photo – cell.
What factor determine the kinetic energy of the electrons emitted, hence show the relationship. (2 marks)
3. (a) What is meant by the term photo – electric effect. (1 mark)
b) The figure below shows an arrangement used to investigate photo-electric effect.
(i) Name the parts marked P and Q. (2 marks)
(ii) State three measurable quantities in this set up. (3 marks)
(iii) State how the intensity of light affects the photo – current. (1 mark)
c) The results obtained for various monochromatic radiations of different colors are shown in the grid below.
(i) The graph indicates that there is a frequency below which no electrons are
emitted. explain why this is so. (1 mark)
(ii) From the graph determine;
(i) Plank’s constant, h (Take electron charge, e = 1.6 x 10-19C) (4 marks)
Exam Questions on work functions in photoelectric effect
(ii) The work function of the metal. (3 marks)
(iii) Sketch on the same graph, the expected graph of another metal which has a lower work function than the metal used. (1 mark)
4. The graph in Figure shows the variation of frequency of radiation f with the greatest kinetic energy of the emitted electrons.
From the graph determine
(i) Plank’s constant (4mks)
(ii) Hence or otherwise calculate the work function of the metal. (3mks)
5. (a) What do you understand by the term photoelectric effect? (1mk)
(b) Name one factor that determines the velocity of photoelectrons produced
on a metal surface when light shine on it. (1mk)
(c) In a photoelectric effect experiment, a certain surface was illuminated with
radiations of different wavelengths and the stopping potential determined
for each wavelength. The table in figure 9 below shows the results obtained.
Stopping potential (Vs) V
1.35
1.15
0.93
0.62
0.36
Wavelength (x 10-7m)
3.37
4.04
4.36
2.42
5.46
On the grid provided plot a graph of stopping potential (Y-axis) against frequency. (7mks)
6. The table below shows the kinetic energy of photoelectrons for various radiations for a given photo emissive surface.
kinetic energy (x 10-19)
0.42
1.47
2.10
2.73
3.36
3.78
Frequency x 1014 Hz
4.50
6.00
7.00
8.00
9.00
9.75
i) Plot a graph of kinetic energy (y axis) against the frequency (x – axis) (5mks)
ii) Use the graph to find h the Planck’s constant (2mks)
iii) From the graph determine the work function of the surface (2mks)
iv) State 3 factors that affect the photoelectric effect (3mks)
(v) State two applications of photoelectric effect (2 marks)
X-rays are produced when fast moving electrons are suddenly stopped by matter. An X-ray (Röntgen radiation) is a form of high-energy radiation with a wavelength shorter than those of ultraviolet rays and longer than those of gamma rays. X-rays were discovered in science as a type of unidentified radiation emanating from discharge tubes by experimenters investigating cathode rays. They were accidently noticed by a German scientist known as Wilhelm Röntgen. Röntgen, a professor of physics discovered X-rays while experimenting with Crookes tubes and began studying them on November 1985.
How x-rays are produced
X-rays are produced in x-ray tube where highly accelerated electrons are suddenly stopped by a metal target. Figure below shows the structure of an X-ray tube.
figure 1: an x-ray tube
How x-ray tube works
When a current flows through the filament in the cathode, electrons are produced by thermionic emission. Thermionic emission is the release of charged particles, usually electrons, from a heated surface, such as a metal. Electrons are then accelerated towards the target by a high potential difference of about 100Kv between the cathode and the anode. The metal target is the anode. Cathode is concave shaped in order to focus the electron beam produced towards the metal target. The metal target is usually composed of tungsten embedded onto copper.
Tungsten is preferred because of it’s high melting point. Copper is also preferred as it is a good conductor of heat. This ensures efficient dissipation of heat. A lot of heat is produced during x-ray production. This is because most of kinetic energy of the electrons are converted to heat when they heat a metal target. Only about 0.5 % of the kinetic energy is converted to x-ray radiations. Because heat produced is a lot, cooling is enhanced by the cooling fins on the outside of the tube and circulation of oil through the channels in the copper block(node).
modern x-ray tube
In an improved recent model of an x-ray tube, the target is designed to rotate during operation so as to change the point of impact. This reduces wear and tear resulting from frictional forces. The figure below shows the more modern x-ray tube.
figure 2: improved x-ray tube with rotating anode
X-ray tube is made up of strong highly evacuated glass. This ensures that no gaseous particles are inside to collide with the fast moving electrons. Collisions with air particles would cause electrons to loose some of their energy before they reach their target hence lacking enough energy to produce x-ray.
The target is usually set at an angle of 45o to the electron beam so that to direct x-rays out of the tube through a window on the lead shield. see figure 1 above. Incase some x-rays are produced at an angle that would make them unable to exit the tube, then they will be safely absorbed by the lead shield surrounding the tube. A step-up transformer is used to increase voltage from main supply so as to provide the high voltage required to accelerate electrons.
x-ray tube in application
properties of X-rays
They are not deflected by magnetic or electric fields. This shows that they carry no charge.
X-rays penetrate matter according to the density of matters. High density materials like lead are least penetrated by x-rays.
Affects photographic emulsions. This makes x-rays useful in x-ray photography.
They ionizes air molecules
causes fluorescence in certain substances such as zinc sulphide
X-rays are of shorter wavelengths compared to visible light.
They can be plane-polarized, be diffracted and be reflected
The Einstein’s Equation of Photoelectric Effect describes how energy of an electromagnetic rays incident onto metal surface is distributed. When an electromagnetic ray falls on a metal surface , the energy absorbed by the electron is divided into two:
Energy used to dislodge an electron from it’s orbit to the surface.
Energy used as a kinetic energy to accelerate an electron from the surface
The energy of a photon can thus be expressed as sum of two energies:
Energy of a photon
=
energy needed to dislodge an electron from metal surface
+
maximum kinetic energy gained by the electron
The work function
The minimum energy needed to dislodge an electron from a metal surface is called the work function (Wo) of the metal. Work function energy varies from metal to metal.
Work function is expressed in a unit called electron-volt(eV) or joules (J).
1 eV = 1.6 x 10-19J.
Emission of photoelectrons occurs only when frequency incident onto the metal surface is above certain value. The minimum frequency to provide energy required to dislodge an electron from the metal surface is usually referred to as the threshold frequency(fo) of the given metal.
From the genera equation of the electromagnetic waves: c = fλ, the corresponding wavelength for the threshold frequency is known as threshold wavelength (λo) .
wo = hfo
which can also be expressed as :
$$W_o =h\frac{c}{\lambda_0}$$
when any radiation is of frequency f that is lower than fo , The energy it gives (hf) will be lower than Wo .
when hf= hfo, electrons are dislodged from their atoms but they just remain on the metal surface. An extra energy will be required to move them from the surface. That can happen if the radiation has extra energy besides that which is enough just to dislodged an electron and brings it to the surface.
Kinetic energy of a photoelectron comes from what remains after part of the electromagnetic waves energy has been used to extract an electron.
Thus when a wave is carrying an energy greater greater than the work function, the excess energy appears as the kinetic energy of the emitted electron.
from our earlier expressions of the wave energy, we see that:
$$hf-W_o = \frac{1}{2}m_ev^2$$
me being the mass of an electron and v the velocity gained in the acceleration.
Thus total energy hf in a wave is usually expressed as:
$$hf = W_o + \frac{1}{2}m_ev^2$$
Einstein’ s Equation in terms of wavelength
since Wo = hfo the Einstein’s equation can be written as:
The table below shows work functions of some metals:
Example problems involving The Einstein’s Equation
Question one
The minimum frequency of light required to cause photo electric emission from potassium surface is 4.92 x 1014Hz. When the surface is irradiated using a certain source, photoelectrons are emitted with a speed of 6.51 x 105ms-1. Determine
The work function of potassium
The maximum kinetic energy of the photoelectron
frequency of the source of radiation
(take h =6.63 x 10-34Js and me = 911 x 10-31Kg)
solution
(1) by application of The Einstein’s Equation Wo = hfo
The threshold wavelength of a photoemmisive surface is 0.55\µm. calculate:
(a) its threshold frequency
(b) the work function in eV
(c) the maximum speed with which a photoelectron is emitted if the frequency of the radiation is 9.5 x 1014Hz(Take plank’s constant h=6.63 x 10-34Js and me =9.11 x 10-31kg)
solution
(a) λo=0.55μm=5.5×10−7m
by using the general equation of speed of light:
$$c=f_o \lambda_o$$
$$\text{so that } f_o= \frac{c}{\lambda_o}$$
Assuming the area to be circular with a radius r, then we can express A in terms of r such that:
A = πr2 and the Intensity becomes
$$I= \frac{1}{A} \times P = \frac{P}{\pi r^2}$$
It can therefore be shown that, Intensity is inversely proportional to area.
consider the setup below:
Intensity of the radiation is varied by changing the distance between the source and the surface of the cathode (r). The corresponding values of current is then recorded.
From the experiment, one discovers that the current increases when distance r is reduced.
From the basic law of charges, current I is directly proportional to the number of electrons. That is I = ne. where n is the number of electrons and e is charge on a single electron.
Hence, the photocurrent is directly proportional to the number of photoelectrons emitted per second. We can therefore conclude that the number of photoelectrons produced is directly proportional to the intensity.
Photo electric effect is where we study behavior of metals when electromagnetic radiations falls on them. When we provide metals with some energy, the energy extracts electrons in their atoms causing them comes to the surface. This energy that extracts electrons from metal surface will come from heat energy or from electromagnetic radiations.
When Electromagnetic waves are incident onto a metal surface, some electrons gains enough energy to escape from the surface. For instance, when light energy falls on a negatively charged conductor, it becomes discharged.
The emission of electrons by a metal exposed to light is known as the photoelectric effect.
The electrons the metal emits by photoelectric effect we refer to as photoelectrons.
A material that exhibits photoelectric effects we say that it is photoemmisive.
Illustrating Photo electric effect
To experiment on photo electric effects, you may need the following materials.
An electroscope
zinc plate
Emery paper
glass rod
silk cloth
Ebonite rod
piece of fur
Table lamp
Ultraviolet lamp
sheet of glass
Investigating photo electric effect
proceed as the follow:
Clean the surface of the zinc plate with emery paper and lay it on the electroscope cap.
Rub the glass rod with silk and charge the electroscope positively.
Shine light from the table lamp onto the zinc plate and observe if the electroscope leaf falls or rises
Repeat the steps above using ultraviolet lamp and observe if the electroscope leaf falls or rises
Discharge the electroscope and lay the zinc plate on it again
Rub the ebonite rod with fur and then charge the electroscope negatively
Shine light from the table lamp onto the zinc plate making the light as intense as possible by holding it as close as possible to the surface and see if the electroscope leaf falls or rises
Repeat 6 and 7 using ultraviolet lamp and observe if the electroscope leaf falls or rises.
Recharge the electroscope negatively and try to discharge it by shining ultraviolet light on the zinc plate as shown
As the leaf begins to fall, insert a sheet of glass between the lamp and the plate.
Observations on Photo electric effect experiment
visible light from table lamp was shone onto the positively charged electroscope but the leaf did not fall.
ultraviolet rays from ultra violet lamp are directed on to the positively charged electroscope but the leaf did not fall.
When visible light was shone onto the negatively charged electroscope, the leaf of the electroscope did not fall.
When ultraviolet rays were directed towards the negatively charged electroscope, the leaf of the electroscope falls.
A sheet of glass is placed between the violet lamp and the discharging electroscope and it stops the discharge. Discharging continues when it is removed.
Explaining the experiments on photoelectric effects
Neither visible light nor ultraviolet rays were able to discharge positively charged electroscope. This is because there are no electrons to be dislodged from positively charged electroscope. This suggests that the effects of radiations is to eject electrons which has negative charges from the metal surface.
The fact that visible light is not able to discharge a charged electroscope while the ultraviolet rays does suggests that not all the radiations can eject electrons from a metal surface. Ultra violet rays has a higher frequency compared to visible light suggesting that a certain minimum energy is required to eject electrons from a metal surface.
A sheet of glass stopping the discharging process suggests that the glass block absorbs the ultraviolet rays and so they are not able to penetrate through and reach the zinc plate.
One reason a positively charged electroscope is not discharged by radiations is that photoelectrons emitted from the positively charged zinc plate do not escape as they are neutralized by the positive charges on the plate and so the leaf divergence remains the same.
Illustrating photoelectric effect using a photocell
A photocell contains two metal electrodes sealed in an evacuated tube. The figure below shows a photocell circuit.
Evacuation of the tube prevents electrons from being stopped by air molecules.
The tube is usually made of quartz material in order to allow ultraviolet rays pass through easily. Electrons after being ejected at the cathode, they are attracted to the anode by a potential difference across the electrodes.
Before any radiation falls on the cathode, no deflection is noted on the galvanometer. This means that no current is flowing through the circuit. When a radiation of given frequency falls on the cathode, deflection is observed on the galvanometer meaning that current has been flowed through the circuit.
Electrons after being ejected from the cathode, are accelerated towards the anode by the potential difference hence the circuit is complete.
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