NUCLEAR PHYSICS

Atomic Structure
The proton and neutron reside inside the nucleus of the atom. The electron revolves round the nucleus on the shell. The force of attraction that exist between the positive charged nucleus and the electron produces the electrostatic force that help to keep the electron moving round the nucleus on the shell.

Atomic Number
Atomic Number is the number of proton that are present in the nucleus of an atom.
It is denoted by letter Z.

Mass Number / Nucleon Number:
Mass Number is the number of neutron and proton that are present in the nucleus of an atom. The mass number of an atom determines the mass or the weight of the atom. It is denoted by letter A.

Neutron Number:
Neutron Number is the number of neutron that are present in the nucleus of an atom. It is denoted by letter N.

Formula for calculating mass, atomic, and neutron number:
Mass number A = atomic number Z + neutron number N
Mass number A = Z + N
From this relation or formula, you can calculate any one of them when the other two are given.
Worked Examples:
Example 1:



Example 2:


α β γ
RADIOACTIVITY:
Radioactivity is the spontaneous decay or disintegration or break down of the nucleus of a radioactive elements with the emission of alpha particle, beta particle, gamma radiation and energy.
The rate of decay of the nucleus of a radioactive elements depend on the nature or type of radioactive elements and it varies from one element to another.

Examples of radioactive elements:
Plutonium, polonium, Radom, uranium, thorium, actinium, radium

Instruments for Detection of radiation:
Radiation can be detected using the following instruments:
• Dekatron counter
• Photographic plate
• Rate meter / scaler
• Pulse electroscope or wulf
• Geiger Muller tube
• Diffusion cloud chamber
• Crookes sprinthariscope / scintillator

See the note for explanation s:


Types of radioactivity
• Natural Radioactivity :
This is the natural/spontaneous disintegration of the nucleus of unstable radioactive element.

Types of natural radioactivity:
Types of radioactivity means the type of particles that are emitted during the decay process.
 Alpha (α ) emission :
In alpha emission, the atomic number of the element is decreased by 2 while the mass number is decreased by 4. Alpha emission is the emission of helium nucleus.
General Equation of alpha emission:

Example 1:

From the equation, Ra decay and emit alpha particle to form Rn, He and energy.

Example 2:



 Beta ( β )emission:
In beta emission, the atomic number is increased by 1 while the mass number remain unchanged. Beta emission I is the emission of electron.

General equation of beta emission:





Exampl 1:

Explanation:


Example 2:

 Gamma ( γ)emission:
In gamma emission, the emitted particles are not indicated or reflected an as such they can not be accounted for.
Example:

• Artificial / induced Radioactivity:
Artificial radioactivity occur when a radioactive and non radioactive element is bombarded with a fast moving particles such as proton, neutron electron, etc , so that the nucleus  become unstable and decay.

General formula of artificial radioactivity:



Example 1:


Explanations:
Nitrogen is bombarded with helium particle to yield unstable..... Which further decay to form hydrogen, oxygen and energy.

Example 2:

Example 3:
 Difference between natural and artificial radioactivity:


Similarities between natural and artificial radioactivity:
1. In both cases, new elements are formed
2. Energy is released in both cases
3. The masses of the new elements formed are different from the masses of the parents elements

Properties of Alpha  (α) particle:
 They are helium nucleus He.
 They are positively charged particle
 They deflect slowly towards negative plate in electric field
 They have the highest ionization energy
 They are massive
 They have less penetrating power.
Properties of Beta ( β ) emission:
 They are stream of electron
 They are negatively charged particles
 They deflect towards positive plate in electric field
 They are lighter
 They deflect faster
 They have low ionization energy
 They have higher penetrating energy
 They are represented by ...

Properties of gamma (γ)emission:
 They are electromagnetic in nature
 They do not require material medium for their propergation.
 They do not deflect in magnetic field / electric field
 They are neutral
 They have the highest penetrating
 They travel at the speed of light.

Isotopes:
Isotopes are atoms of the same elements that have the same atomic number but different mass number.
Isotopes have the same chemical properties but different physical properties. The reason is, the R.A.M of most elements are not whole number.
Examples of elements that are isotopes:
1. Graphite is isotope of carbon
2. Deuterium, tritium are isotopes of hydrogen
3. Ozone is isotope of oxygen
4. Chlorine 37 is isotope of chlorine 35. Etc.


Radioisotopes:
Radioisotopes are artificial isotopes that are obtained by firing neutron, proton, deuterium at nucleus of a radioactive element so that it become unstable and disintegrate/ decay and emit alpha particle, beta particle, gamma rays and energy.

How to calculate the R.A.M of isotopses
R.A.M = sum of the % by mass of each elements
Example
Example :
Calculate the R.A.M of chlorine if the percentage by mass of the isotopes are 25% of chlorine 37 and 75% of chlorine 35 respectively.
Solution:
R.A.M = 25/100 x 37 + 75/100 x 35
R.A.M = 9.25 + 26.26 = 35.5.    R.A.M of chlorine = 35.5
 Uses of Radioisotope:
Radio isotopes are used in the following:
• It is used In radio therapy for the treatment of cancer.
• It is used in radio graphy in studying defect in metals, measuring the thickness of materials
• It is used as treacer in industries to detect wear parts of machine ( gear box), detect leakages in pipes and tanks.
• It is used in argriculture to produce fertilizer and preservatives
• It is used in radio sterilization to sterilize fabric in textile industries and medical supplies. Bacteria are destroyed on exposure to the rays.
• It is used in dating techniques. Carbon dating to trace the activity of isotopes in ancient world and carbon materials that can provide information about the age of the material.
Carbon is formed when neutron collides with nitrogen in the air. The neutron is produced through the cosmic rays which enters the upper atmosphere and react with the air molecules.

Half Life:
It is the time taken by  the atom of a radioactive element to decay to half of it initial mass.

Decay Constant:
It is the number of atoms that decay for one second.
The number of atoms that decay is directly proportional to the number of atoms that are present initially.

Therefore, dN/dt α N
dN/dt = - ʎ N, where dN/dt is the rate of decay, -ʎ is the decay constant. The negative sign shows that the decay is decreasing with time, N is number of initial atom present.
If we integrate both sides of the equation, we have


Formula for calculation:
N =N₀e ⁻ᵞᵗ
Where N is the present mass after decay, No is the initial mass of atom before the decay, t is the time and ʎ is the decay constant.
Example:
 Calculate the number of decay in a second. ii) how many seconds will elapse before 10 expo 4 atoms remain. iii) what is the count rate at a certain instance when a piece of radioactive contains 10 expo 12 atoms and the half life of the material is 30 days?
Solution:






Decay curve:
A decay curve is obtained by plotting rate of decay of a radio active element on the vertical axis against time on the horizontal axis. A curve line graph is obtained as shown below:
Decay curve:






From the graph, t is the half life.
N = No/2, remember that N = Noe⁻Πt
Therefore,
           No/2 =N₀e ⁻ᵞᵗ.,   ½ = Noe⁻Πt / No ,  ½ = eΠt,  0.5 = e⁻Πt
Take the natural log of both sides,
Log0.5 = LogeΠt ,  log0.5 = - At,  -0.693 = -Πt,
Therefore,    -0.693 / -t = Π

Application of formula in solving problems:
An element source has a decay constant 1.36 x10⁻¹¹ sec. Calculate the half life of the element.

Solution:
T½ = 0.693 / Π = 0.693/ 1.36 x 10⁻¹¹  = 1614 years.

More exercise:

6

Nuclear Energy:
This is the energy that is released during nuclear activities.
In nuclear reaction, the masses of the products are usually different from the masses of the 4 reactants.


Nuclear energy equation :
 ∆E = ∆mc–²
 where ∆E is energy released or energy change
∆m = is mass defect / change in mass and c is the velocity.

Electron Volt (EV):
Electron volt is the energy pocess by an electron that is moving under the influence of a potential difference of one volt. It is the unit of energy in nuclear energy.

Value of electron volt (ev):
1ev = 1.602 x 10⁻¹⁹  Joule
1Mev = 1.602 x 10⁻¹⁹  Joule x 10⁻6  =  1.602 x 10⁻¹³ Joule

Atomic mass unit ( à.m.u.)
The unit of energy is also expressed in a.m.u.

Value of a.m.u:
1a.m.u = 931Mev which will be 931 x 1.602 x 10⁻¹³ Joule
              = 1491.462 Joules

Worked Examples:
1. Calculate the energy of 2.25ev.
Solution:
1ev = 1.602 x 10⁻¹⁹  
Therefore,
   2.25ev will equal 2.25 x 1.602 x10⁻¹⁹  = 3.6045 x 10⁻¹⁹  Joules

2. What is the value of 3.43 Joules in a.m.u?

Solution:
1 a.m.u = 931Mev = 931 x 1.602 x 10⁻¹³  = 3.6045 = 10⁻¹⁹ Joules
Therefore,
6874.23 x 10⁻¹⁹  Juole = 3.43 X 10⁻¹⁹  / 3.6045 x 10⁻¹⁹ 
                             = 0.952 a.m.u

Binding Energy:
It is the amount of energy that is required to split nucleon
( neutron and proton) in the nucleus of an atom.

When the proton and neutron are spitted, the total mass is usually less than the mass of the nucleus.
Therefore,
Binding energy can also be called mass difference of nucleus and nucleon.

FORMULA FOR CALCULATING BINDING ENERGY
Binding energy = mass of nucleus – mass of nucleons

Worked Examples:


TYPES OF NUCLEAR REACTIONS
Nuclear Fussion:
This is the type of nuclear reaction that involves the combination of two or more light or smaller nuclei at high temperature to form a heavier or bigger  nucleus and energy is released.
Examples of nuclear fusion: thermonuclear fusion equation
• Magnesium is bombarded with helium to form silicon and energy is released
• Boron is bombarded with hydrogen to form carbon and energy is released
• Tritium is bombarded with hydrogen to form helium and energy is released
• Tritium is bombarded with deuterium to form helium, proton and energy is released.
• Remember to write the typical equation for each
Nuclear Fission:
This is the type of nuclear reaction that involves the splitting of heavy nucleus of a radioactive element into smaller nuclei of different masses by bombarding it with fast moving particle such as neutron, proton, electron, etc.
Examples of nuclear fission equation
• Na + n ➡  Na ➡ Mg + e + Energy
Sodium is bombarded with proton to form unstable sodium  with mass numbe 24, which further split to magnesium, electron and energy is released
•
•
• U +  n  ➡ U (unstable) ➡ Ba + Kr + 2n + energy
When Uranium is bombarded with proton, it form unstable uranium with mass number 236, which further split to barium, kripton, two atoms of proton
Chain Reaction:
Tis is the type of nuclear reaction whereby nucleus of a radioactive element is split into tow or more nuclei and each nuclei in turn are split into many nuclei and the process keep on repeating itself, by bombardment with fast moving proton.
It is the multiple action of nuclear fission.
Diagram to illustrate chain reaction

Application of chain reaction:
Chain reaction is used in principle of operations of bomb.

Diagram of a nuclear reactor

Explanation of parts:
Graphite:
It is used to moderate the speed of the neutron so that the chain reaction is prevented from dying out.
Boron steel rod:
It absorb excess neutron and control the number of neutron so that too many neutron would not increase the reaction and make to be out of control.
Electric motor:
The electric motor help to move the rod in and out of the reactor to a suitable position. If there is electrical failure, the rod would fall and shot down the reactor automatically.
Concrete Biological Shield:
It prove t the operator from intense radiation inside the core.

Characteristics of Nuclear Activity:
• There is always a slight loss of mass in the reaction
• Alpha, beta ,gamma particles and energy are emitted
• New elements are always formed
• The properties of the elements are always different from that of the parents elements
• Radioactive elements decay with a specific half-life
Peaceful application of atomic energy:
• It is used in Tue generate electricity
• It is used in the produce nuclear weapons
• It is used in the detect leakages or holes in pipes
• It is used to detect hidden weapons
• It is used in radiotherapy to detect broken bones
• It is used to power nuclear plants
• It is used as fuel
• It is used in food preservation
Advantages of nuclear fusion over nuclear fission:
In nuclear fusion,
• Power can be generated
• Lightest elements can be used
• The raw materials are cheap
• Less dangerous by-products are produced
• There is no maximum limit of hydrogen that can be obtained
Disadvantages of nuclear fusion over nuclear fission:
• The binding energy needed for the reaction is greater than the energy produced or released
• It is cost to run
• It takes longer time
More problems:





Alternating Current (A.C.)
A.C. means alternating current. It is a current that flow in the positive and negative direction periodically.

Waveform of Alternating current:
The waveform of alternating current is sinusoidal waveform.
Diagram of a.c waveform:

Alternating Current Circuit:
Equation of A.C current
I = I₀ Sin θ
In the production of A.C voltage or current, the coil is in a magnetic field and sweep through angle θ.
Therefore,
I =  I₀ Sin θ
W = θ/t. If we make  θ the subject, we will have, θ = w*t. Understand?
 If we substitute for θ  in the equation, I =  I₀ Sin θ
Then the equation will become,
I =  I₀ sin wt.
The next thing is:
When the coil sweep through one revolution or one circle, angle
θ = 360°. We will change 360° to radian
Therefore,

θ = 360° = 2π radian. Therefore,
W = θ/t = 360°/t = 2π/t. Therefore, w = 2π/t.
At this point, remember the formula that connect t and f,
t = 1/f. You will now substitute for t in the formula, w = 2π/t thus:
W = 2π/t = 2π ÷ 1/f = 2π x f/1 = 2πf, therefore
W = 2πf, where f is frequency. It is measured in hertz.
At this point, we will substitute for w in I = I₀ Sin wt.
Therefore,
I =  I₀ sinwt =  I₀ sin 2πf * t. That is, I =  I₀ sin 2πft
Therefore, the general equation for a.c current is
I = Io sin 2πft, where
 f is the frequency. it is measured in hertz
 t is the time. It is measured in second
 Io is the peak current. It is measured in ampere
 I is the instantaneous current, measured in ampere.
 W is angular velocity measured in radian per second.
Equation of A.C current
Every thing that we did above is applicable to a.c voltage. Therefore the general equation for a.c voltage is the same as that of a.c current. The only difference is that letter V is used instead of Letter I.
Therefore,
V =  V₀ sin θ , V = sin wt,  V =  V₀ sin 2πft. That is all.

Application of general equation of a.c. current/voltage to solve problems
An a.c. voltage is represented by the equation, V = 4Sin900πt.
Calculate the: I) the peak voltage. II) the frequency. III) the angular velocity.
Solution:
You have to compare each parts of the give equation with the general equation thus:
General equation is I = I₀ Sin 2πft
The given equation is V = 4 Sin 900πt
I) Peak voltage is  V₀ = 4volt
II) To find frequency, you compare by saying,
Sin 2πft = Sin 900πt
Sin will cancel Sin, π will cancel π and t will also cancel t.
Therefore we have,
2f = 900, make f the subject of formula, then
f = 900/2 = 450 hertz
III) To find angular velocity, w, you will use
W = 2πf, therefore,
W = 2 x π x f = 2 x π x 450 , w = 900π radian per second.
Or, w = 2 x 22/7 x 450 = 2828.57 rad/sec

Peak Current, I₀ / peak Voltage, Vo
It is the maximum value of the a.c current or a.c Voltage.

Root Mean Square (r.m.s) of current/ voltage
It is the steady part or d.c part of the alternating current that will produce the same quantity of heat if it flow through the same resistance in the same amount of time.

Relationship between peak current (I₀) and I(r.m.s)
Let us derive the formula that connect I₀ and I(r.m.s).
Firstly,
The rate at which heat is produced is directly proportional to the square of the current.
The I(r.m.s)² is the square of the average of I₀².
Therefore,
I(r.m.s)² = average of I₀²
I(r.m.s)² = I₀²/2.  We will square both sides . then we will have,
I(r.m.s)² = I₀²/2, if we make I(r.m.s) the subject, we will get,
I(r.m.s) = square root of I₀²/2 = √(I₀²/2) = √I₀²/√2 = I₀ /√2
Therefore,
I(r.m.s) = I₀ /√2. Or I(r.m.s) = 0.7070*I₀ , I₀ = √2*I(r.m.s)
Io is the peak current while I(r.m.s) is the root mean square of the current.

Relationship between peak voltage(V₀) and V(r.m.s)
Every thing we did above is also applicable to voltage. Therefore, the formula is the for voltage. The only difference is that letter V is used instead of letter I.
Therefore,
V(r.m.s) = V₀/√2. Or V(r.m.s) = 0.7070*V₀ Or, V₀ = √2*V(r.m.s)
Where Go is the peak voltage, V(r.m.s) is the root mean square of the voltage.

Note:
Why is it that when a voltmeter or ammeter is connected across a.c source, the voltmeter or ammeter give a steady reading?
Answer:
When a voltmeter or ammeter is connected across a.c source, the voltmeter or ammeter give a steady reading because the instrument is only measuring the steady or d.c Part of the a.c current or voltage and not the a.c part.
It is the d.c part of the a.c current/voltage that is powering the electrical devices, not the a.c part. If not so, the devices would power on and off repeatedly for as long as it is powered by ac current/voltage.

Alternating Current Circuit
These are circuit which are powered by alternating current or alternating voltage.

Types of a.c circuits
Resistive a.c circuit:
Circuit diagram:

A resistive circuit is an a.c. circuit that contains only resistor.

Resistance of a R- circuit:
The circuit therefore obey ohm’s law. Therefore,
V = I*R ( ohm’s law)
Remember that for a.c, voltage and current,
V = Vo Sin wt,  and I = I₀ Sin wt . we will put (substitute) this into the ohm’s law that we stated above.

Therefore,
V = I*R will become,
Vo Sin wt = (I₀ Sin wt)* R. We will make R the subject, then we will get,
R = (Vo Sin wt)/(I₀ Sin wt). Sin wt will cancel sin at, then we will get
R = Vo/I₀ (for peak voltage and current)
R = V(r.m.s)/I(r.m.s)  ( for r.m.s voltage and current)

Phase Relationship between Current and Voltage in Resistive Circuit
In a resistive circuit, the current is in phase with the voltage. This means that current and voltage start their journey from the same point at the same time, reach their maximum and minimum points at the same time.

Diagram that illustrate

Vector or Phasor diagram of R- Circuit

Worked Examples:

Capacitive a.c Circuit
Diagram of C- circuit:

A capacitive circuit is an a.c circuit that contains only capacitor as the active components.

Phase Relationship between Current and Voltage in Capacitive Circuit
In a capacitive circuit, current leads the voltage by 90°or π/2 radian or ¼ circle. Or voltage lag behind current by 90° or π/2 radian or ¼ circle. The reason is that voltage always wait behind to charge the capacitor before it will start to flow in the circuit while current start to flow immediately in the circuit.
Current and voltage are out of phase. This means that current and voltage start their journey from different points at different time and reach their maximum and minimum points at different time.

Diagram to show phase relationship:


Vector or phasor diagram of C- Circuit:




Phase Difference between voltage and current in c- circuit
This is the amount in degree by which the current leads the voltage in a capacitive circuit. The phase difference is 90° or π/2 or ¼ circle.
Therefore,
V = Vo Sin wt , and   I = I₀ Sin ( wt + π/2 ) .
Note that π/2 is added because current leads the voltage in capacitive circuit.

 Capacitive Reactance ( Xc) of a capacitor in a capacitive circuit
The resistance of a capacitor in a  capacitive circuit is called capacitive reactance. The symbol is Xc. Its unit is ohm.
Definition of capacitive reactance Xc:
Capacitive reactance is the resistance that a capacitor offers to the flow of an a.c. current.
Formula for calculating capacitive reactance Xc
Xc = 1/wC.
Remember that w = 2πf ( we derived this formula above)
Therefore,    Xc = 1/2πfC .
where f is the frequency of current or voltage, measured in hertz. C is the capacitance of the capacitor, measured in microfarad.
At this point, we will put ( substitute for ) Xc in the formula of ohm’s law. Therefore,
From ohm’s law, V = I*R. Where R is represent capacitive reactance Xc.
Therefore,
V = I*R = I * 1/2πfC = I/2πfC
Worked Examples:
A 2microfarad capacitor is connected across 150V(r.m.s.), 60 Hz a.c source. Calculate: I) the r.m.s value if the current   ii) the peak value.

Solution:
From the question, V(r.m.s) = 150Volt, f = 60Hz,
C = 2microfarad = 2x10~6Farad
Note that:
 we are to calculate the r.m.s value of the current. i.e I(r.m.s)
The only formula that we can use to calculate the r.m.s value of current is Xc = V(r.m.s)/I(r.m.s)
Xc is not given in the question. You will notice that C and f are given in the question. So we must first of all calculate Xc and then use the calculated value of Xc to calculate I(r.m.s). Therefore, we use the formula:
I. Xc = 1/2πfC,  to calculate Xc
Xc = 1/ 2*3.142*60*0.000002 = 1/ 0.00075408
Xc = 1326.11924 ohms
At this point, we use the calculated value of Xc to calculate the current. Therefore,
Xc = V(r.m.s) / I(r.m.s) ,   make I(r.m.s) the subject of the formula,
I(r.m.s) = V(r.m.s) / Xc = 150 / 1326.11924 = 0.113112 Amps
II. The peak value of the current.
To calculate the peak value of the current, we use the formula:
I(r.m.s) = I₀/√2 . Let us substitute
0.113112 = Io/√2, 0.113112 * √2 = I₀
Io =  0.113112 * 1.4142 = 0.1600 Amps.


A.C Circuit tha contain only Inductor


Circuit diagram of inductive circuit



An inductive circuit is an a.c. circuit that contains only inductor. The resistance of the inductor is called  inductance. It is measured in Henry. The inductance of an inductor is represented by letter L.


Phase Relationship between Current and Voltage in Capacitive Circuit
In an inductive circuit, the voltage leads the current by 09° or π/2 radian or ¼ circle. Or we say the current lag behind the voltage by 90° , or π/2 radian or ¼ circle. The reason is that current always wait behind to build a magnetic field around the inductor before it flow in the circuit while voltage start to flow in the circuit immediately.

Diagram to show phase relationship between V and I in L- circuit
Phase Difference between voltage and current in L- circuit
This is the amount in degree by which the voltage leads the current or the amount in degree by which current lag behind voltage. The phase difference is 90° or π/2 radian or ¼ circle.

Now, since current lag behind voltage, the formula of a.c current will become,
I = Io Sin ( wt – π/2 ) while V = Vo Sin wt. ( no change).

Inductive Reactance XL of inductive circuit
This is the resistance that an inductor offer to the flow of a.c current. Its unit is ohm. Its symbol is L.

Formula for calculating inductive reactance Xc
XL = wL.
 Remember that w = 2πf. Therefore,
XL = 2πfL.
Where L is the inductance of the coil, f is frequency of current or voltage. Π is 22/7.
At this point, we will put ( substitute for ) XL in the formula of ohm’s law. Therefore,
From ohm’s law, V = I*R. Where R is represent inductive reactance XL.
Therefore,
V = I*R = I * 2πfL = I*2πfL. Therefore,
V = I*2πfL

Worked Examples
Find the impedance ( inductive reactance) across an inductor of 0.2H when an a.c voltage of 60Hz is applied across it, if the voltage is given by V = 150 Sin 120πt, calculate the r.m.s and the peak value of the current.
Solution:
Note: impedance is another name for inductive reactance.
From the question,
L = 0.2 henry, and V = 150 Sin 120πt.
 If you compare the equation with the general equation for a.c. voltage or current ( V = Vo Sin wt), you will agree that,
Vo = 150 volt,
If you look at the question, you will notice that frequency is not given. Ok! Then you have to calculate the frequency using the method I used in worked example above under general equation of a.c current or voltage.
Therefore, general equation is V = Vo Sin 2πft while in the question,
v = 150 Sin 120πt, then Vo = 150 V, compare the sin parts.
Sin 2πft = Sin 120πt. The sin,π and t will cancel one another.
Then,
2f = 120.
If we make f the subject of formula, then f = 120/2 = 60Hz. Let us bring all the data together:
Vo = 150V, f = 60 Hz and L = 0.2 Henry
i) Impedance Z: XL = 2πfL = 2*3.142*60*0.2
Impedance Z( XL) = 75.408 ohms
ii) We must first calculate Io before we can calculate I(r.m.s) because the value Vo is given in the question. Also, we can not calculate R.m.s value of current now because the value of V(r.m.s) is not given in the question. Also, note that you can’t use Vo and V(r.m.s) in the same formula. You can only use Vo and Io together in the same formula or V(r.m.s) and I(r.m.s).

We will use the formula, Vo = I₀ * XL,
Vo = 150 V , XL = 75.408 ohms, Io = ?
therefore, if we substitute the values in the formula, then we will get:
150 = I₀ * 75.408,  then, I₀ = 150/ 75.408 ,  I₀ = 1.989 Amps.

iii) Now that we have calculated peak current(I₀), we can now calculate r.m.s current by using the formula I(r.m.s) = I₀/√2.
Therefore,
I₀ = 1.989 Amps., then , I(r.m.s) = 1.989/1.4142 = 1.406 Amps.
Therefore, r.m.s current = 1.406 amps.

Series R-C circuit:
A series R-C circuit is an AC circuit that contains a resistor and a capacitor connected in series. The circuit connection is as shown below:
Circuit diagram:



Impedance of R-C Circuit
It is the opposition that the two components offer to the flow of an AC current. Its unit is ohm.
Vector or Phasor diagram of R-C circuit:


Formula for calculating the impedance of R-C circuit
From the vector/ phasor diagram, we will obtain the formula for calculating the impedance of R-C circuit. Thus:
Using Pythagoras theorem,
Impedance, Z² = R² + Xc²,  Z = √(R² + Xc²). Recall that Xc = 1/2πfc. Then,   Z = √(R² + (1/2πfc)²). Where Xc is the capacitive reactance. R is the resistance. c is the capacitance of the capacitor. F is the frequency of the ac voltage or current.
At this point, we will bring in the formula into ohm's law.
Therefore, V = I * R, where R represent Z. i.e R = Z. Then
V = I *R ; V = I * Z;
V = I*√(R² + (1/2πfc)²)
Using shocahtoa, tan A = opposite / adjacent, tan A = Xc / R. If we make A the subject of the formula, then we will have,
A = tan~ ¹(Xc/R). Where R is the resistance of the resistor measured in ohms. Xc is the capacitive reactance of the capacitor measured in ohms. A is the phase angle by which current leads the voltage.

Worked Examples:



Series R-L circuit:
Is an ac circuit that contains a resistor and an inductor connected in series. The circuit is as shown below.
Series R-L circuit diagram:

Impedance of R- L series circuit:
This is the opposition/ resistance that a resistor and an inductoroffer to the flow of an ac current or  in an ac circuit.

Vector / Phasor  diagram of R-L series circuit:




From the vector / phasor  above,
Tan A = XL / R.  If we make A he subject, then we will get,
A = tan~¹ (XL / R). Recall that XL = 2πfL. Therefore,
A = tan~¹(2πfL/R).  R is the resistance measured in ohm, f is the frequency of the ac voltage / current measured in hertz, L is the inductance of the inductor measured in henry. A is the phase angle between current and voltage.

Formula of impedance in R-L series circuit:
From the vector / phasor diagram above, we will obtain the formula for calculating the impedance of R-L series circuit thus:
Z² = R² + XL² ; Z = √(R² + XL²). Recall that XL = 2πfL, then we will get,
Z = √( R² + (2πfL)² ).
At this point, we will bring in the formula into ohm's law .
From ohm’s law ,
V = I * R. Where R represent impedance, Z.   I.e R =  Z. Therefore,
V = I * Z;  V = I * √(R²+XL²) ; V = I * √(R² + (2πfL)²).
Note: you can equally use the voltage vector diagram to get the impedance.


Worked Examples:


Series R-L-C circuit:
This circuit contains resistor, inductor and capacitor connected in series as shown in the diagram below.

Impedance of R-L-C series circuit:
This is the resistance that resistor, inductor and a capacitor offer to the flow of ac current in an a. Circuit.

 Vector or phasor diagram of R-L-C series circuit:


Phase angle of R-L-C series circuit:
From the vector diagram,
If we use voltage vector diagram, Tan A = (VL – VC) / Vr
Also, if we use  reactance vector diagram, we will get,
Tan A = ( XL – XC ) / R. If we makeanglew A the subject, we will get,
A = Tan~1 ( XL – XC ) / R. Where XL is the inductive reactance of the inductor, XC is the capacitive reactance of the capacitor and R is the resistance of the resistor. A is the phase angle between V and I.
Recall that XL = 2π*f*L and XC = 1 /2π*f*C. have it in mind that these formulae can be substituted for in the formula above. Then we will get, A = tan~1 ( 2πfL + 1/2πfC )

Formula for calculating impedance in R-L-C series circuit:
From the vector diagram above, if we use the voltage vector diagram and  we will use pythagoras theorem, then, we will get ,
(Vr.m.s)² = Vr² + ( VL - VC)²; Recall that V = I*R, then Vr.m.s = I*Z, VL = I*XL, VC = I*Xc and Vr = I*R. We will now substitute for all of them in the above formula . then we will get,
Vr.m.s² =Vr² + ( VL – VC)² ;
(I*Z)² = ( I*R)² + ( I*XL – I*XC)².if we factorise I², then we will get,
I²Z² = I² ( R² + ( XL – XC )² ). I² at the right hand side cancels I² at the left hand side. Then we will get,
Z² = R² + ( XL – XC )².if  we make Z the subject of the formula. Then we will get,
Impedance, Z = √( R² + ( XL – XC )² )
At this point, we will substitute for Z in ohm's law, V = I * , R = Z,
Then, V = I * Z ,
V = I * √( R² + ( XL – XC )² ). If we make  I the subject of the formula, then we will get,
I = V / ( √( R² + ( XL – XC )² )
Worked Example:
From the circuit below, calculate the capacitive, Xc,  inductive reactance XL, impedance, the current and power.
Circuit diagram:
Solution:
R = 100 ohms, L = 3 henry, C = 4 microfarad, V = 100 volt, f = 160/π hertz.
1. Inductive reactance, XL
XL = 2πfL , f = 160/π Hz, L = 3 H
XL = 2×π×160/π×3
XL = 2×160×3 ( the π cancels each other)
XL = 960 ohms.
2. Capacitive reactance, Xc:
 Xc = 1 / 2πfC
Xc = 1 / 2×π×160/π×0.000004
Xc = 1 / 2×160×0.000004
Xc = 1 / 0.00128
Xc = 781.25 ohms.
3. Impedance, Z
Z = √( R² + ( XL – xc )² )
R = 100 ohms, Xc = 781.25 ohms, XL = 960 ohms
Z = √( 100² + ( 960 – 781.25)² )
Z = √(100² + 178.75²)
Z = √(10000 + 31951.56)
Z = √ 41951.56
Z = 208.82 ohms

4. Current, I:
V = I × Z, V = 100 volt, Z = 208.82 ohms
Let us substitute for the variables, therefore,
100 = I × 208.82. Make I the subject, then we have
I = 100 / 208.82
I = 0.488 ampere

5. Power:
Power = I² × R = I² × Z
Power = 0.488² × 208.82
Power = 48.82 / 2 = 24.41 watt.

Example2:
From the circuit diagram above, calculate the :
I. Resistance of the resistor
II. Inductance of the inductor
III. Capacitance of the capacitor
IV. Voltage across the capacitor
V. Impedance

Solution:
From the diagram,
V = 240V, f = 50Hz, I =10A, VL = 50V, Vr = 140V, Vc =?,
Note that since the components are connected in series, the same value of current flows through each of them.
I. Vr = Ir × R
Vr = 140v, Ir = 10 A, R =?
140 = 10 × R,  make R the subject, Then,
R = 140 / 10 = 14 ohms.
II. Inductive reactance XL:
VL = IL × XL.  VL = 50v, IL = 10 A, then,
50 = 10 × XL. Make XL the subject of formula, then
XL = 50 / 10 , XL = 5 ohms
Now , we will calculate the inductance L of the inductor thus:
XL = 2πfL, f = 50Hz, XL = 5 ohms,
XL = 2 × 22/7 × L. Let us substitute for the values
5 = 2 × 22/7 × 50 × L. Let us make L the subject, then we will get
Therefore,
L = 5 × 7 / 2 × 22 × 50 , L = 35 / 2200, then,
L = 0.0159 Henry
III. Capacitive reactance Xc:
Vc = Ic × Xc.  Ic = 10 A, Vc = ?, Xc = ?.
We can not use this formula to calculate Xc because the value of Vc is not given in the question.

We must first of all calculate Vc before we can calculate C. Therefore, we will use the voltage formula relation which is ,
Vr.m.s = Vr + VL + Vc  ( the reason is ,it is the 240V that is shared across each of the  three components)
Vr.m.s = 240V, Vr = 140V, VL = 50V, Vc = ?
Therefore,
240 = 140 + 50 + Vc.  Make Vc the subject of the formula,
 then,
Vc = 240 – 50 – 140 = 50V
Vc = 50V.
Now , we can calculate the value of Xc using the formula,
Vc = Ic × Xc
50 = 10 × Xc. Make Xc the subject of the formula, then we get,
Xc = 50 / 10 .  Xc = 5 ohms.
Now that we have calculated Xc, we will now calculate the capacitance of the capacitor using the formula:

Xc  = 1 / 2πfC. F = 50Hz, Xc = 5 ohms. We substitute for the values,
5 = 1 / 2 × 22/7 × 50 × C. We make the subject of the formula, then,
C = 5 × 7 / 2 × 22 × 50.
C = 35 / 2200.  Then C = 0.0159 microfarad.

IV. Voltage across the capacitor Vc.
Note ! We have already calculated Vc in the working above because we needed the value of Vc to calculate the value of C.
V. Impedance of the circuit, Z:
Z = √( R² + ( XL – Xc )² ).
R = 140 ohms, XL = 5 ohms, Xc = 5 ohms, then
Z = √( 140² + ( 5 – 5)² )
Z = √(140² + 0² )
Z = √ 140² = √ 19600
Z = 140 ohms.


Parallel R-C Circuit:
In a parallel R-C Circuit, resistor is connected in parallel with a capacitor.
The current that flow through the resistor and the capacitor are different but the voltage across the resistor and the capacitor are the same.
Circuit diagram:


Impedance of R-C parallel Circuit:
It is the resistance/ opposition that R-C offer to the flow of ac current. It is measured in ohm.
Formula for Calculating impedance in R-C Parallel Circuit:
R is the resistance of the resistor, Xc is the capacitive reactance of the capacitor, and Z is the total resistance of the circuit called impedance.
   1 / Z = 1 / R + 1 / Xc .  Recall that Xc = 1 / 2πfC.
Let us make  the subject of the  by fraction concept, then we will get,
L.C.M = R × Xc, then, 1 / Z = (Xc + R ) / ( R × Xc ), then ,
Z = ( R × Xc ) / ( Xc + R ), therefore,
Impedance, Z = ( R × Xc ) / ( Xc + R ).
From ohm's law, V = I × R. Where R is the impedance of the circuit.
Therefore,
V = I × R,  R = Z, then,  V = I × Z ,   and Z = ( R × Xc ) / ( R + Xc ). then,
V = I × ( R × Xc ) / ( R + Xc ). , or , I = V / ( ( R × Xc ) / ( R + Xc ) )
Worked Example 1:
A 2 ohms resistor is connected in parallel with a 5 microfarad capacitor across 240 voltage source of frequency 50 hertz. Draw the circuit diagram and calculate the circuit impedance, the circuit  current and the in each components.
Solution:
From the question, R = 2 ohms, C = 5 microfarad, V = 240 volt, f = 50 Hz
I. Circuit diagram.





II. Impedance Z:
Z = ( R × Xc ) / ( Xc + R ).  We must first calculate Xc using the formula,
Xc = 1 / 2πfc. , then , Xc = 1 / 2 × 22/7 × 50 × 0.000005
Xc = 1 / 0.001571.     Xc = 636.54 ohms.
Now that we have calculated Xc, we can now calculate the impedance Z using the formula we stated earlier.
Therefore,
Z = ( Xc × R ) / ( R + Xc )
Z = ( 636.54 × 2 ) / ( 2 + 636.54 ),  Z = 1.994 ohms.

III. Circuit current:
V = I × Z. Then , 240 = I × 1.994. Make I the  subject of the formula.  I = 240 / 1.994.   I = 120.36 amperes.

IV. Current that flow in each components:
Note that since the components are connected in parallel, the voltage across them are the same. Therefore,

Current that flow through resistor R is:
Vr = Ir × R.   We substitute for the variables. Then,
240 = Ir × 2.   Ir = 240 / 2.   Ir = 120 amperes.

Current that flow through the capacitor C is :
Vc = Ic × Xc.  We substitute for the variables. Then,
240 = Ic × 636.54. We make Ic the subject of the formula. Then,
Ic = 240 / 636.54.   Ic = 0.38 ampere.
Worked Example 2:
From the circuit diagram, the current that flow in the circuit, the Capacitor and  the capacitive rectance  capacitor and impedanc








Energy level in atom
The energy level of different atoms are different because of the numbers of electron in their outermost shell. The energy of the electron can be change from one energy level to another by disturbing the atom.

Energy level in hydrogen
Ground state energy level Eo:
This is the lowest energy level of an electron in an atom.
 Excited state energy level En:
This is the energy level attain by an electron in an atom when electron is disturbed.

Excited energy or excitation energy:
This is the minimum amount of energy that is required to raise the energy level of an atom from the ground state to excited state.
Formula for calculating energy level in atom
E = En – Eo. Where Eo is the energy level of electron in ground state. En is the energy level of the electron in the excited state and E is the energy different between the energy level of the ground state and the excited state.

Energy level chart:


N = 1,2,3,4,5,.......
N represent energy level. Each number represent a particular energy level.

The value of energy level of an atom
The ground state energy level for hydrogen = - 13.6 eV.
Recall that 1 eV (electron volt) = 1.06 * 10⁻¹⁹  Joule.
Therefore, -13.6 eV = -13.6 * 1.06 * 10⁻¹⁹  Joule.
     = -21.76 * 10⁻¹⁹  Joule.

Formula for calculating other energy levels of electron in an atom
En = 13.6 eV / n² where n represent the electron quantum number
Work example:
 calculate the energy level of an electron when n = 2
solution
En = 13.6 eV / n². n =2, then,
En = 13.6 eV / 2².   En = 13.6 / 4.   En = 3.4 eV.  Let us convert this to joule. Therefore,
1 eV =  1.06 *10⁻¹⁹  Joule. Therefore,
3.4 eV = 3.4 *  1.06 * 10⁻¹⁹  Joule.
  = 3.604 joule
Example 2:
Find the energy level between En = 3 and 4

Solution
En = E4 – E3
En = (13.6 eV / 4²) – (13.6 / 3²)
 En = (13.6 eV / 16) – (13.6 / 9)
 En = 0.85 – 1.51.     En = - 0.66 eV

Emission of radiation or spectra
When an atom is heated or excited the electron in the atom gain energy and move from its ground state (lower energy level) to higher state (excited energy level). When the disturbance or heat is removed, the electron return back to its ground state and the electron releases energy in the form of radiation or light

Photon or quanta
It is a packet of radiation energy that is emitted by a body when its temperature is increased or lowered.
Formula for calculating energy of quanta or photon
E = hf.  F is frequency of radiation of photon. Then h is called planck constant.  h = .....
Work example
Spectra is the excess energy in the form of radiation emitted by a particle when the is returning from excited to ground state


Equation of radiation energy
Recall that,   E = hf and  E = En – Eo.
Now let us substitute these formulae into one another, then we will get,
Hf = En – Eo. Also recall that, v = hf (from wave equation)
Let us make f the subject and substitute in the formula above. Therefore,
V = hf.   F = v / h. Then,
Hf = En – Eo,   h* v/h = En – Eo. Then,
h =  h* v /  En – Eo.
Where h is  planck constant., v is velocity. Eo is ground state energy level and En is excited state energy level.


Questions on Electrolysis
*What is the part played by electrolytein electrolysis?
It is required to cover the surface of an oobject of total area 600 cm² with nickle of thickness 0.10 mm using a constant current of 5.0 A. Determine the time for which the rrent must flow if the mass of the nickel deposited per coulomb is 0.00030 g/c and its density is 9.0 g/ cm cube. If PD of 40 volt is employed, determine the cost at 2p per kilowatt-hour.
*State Faraday laws of electrolysis.
Describe an experiment to determine the mass of copper deposited per Coulomb.
State what precaution you would take to ensure that a reliable value. is obtained.
A potential difference of 6 V is maintained between the plate of a copper voltmeter and 0.66g of copper is deposited. Calculate the electrical energy consumed by the àq11q11ameter.( mass of copper deposited per  = 0.00033g/c).
*Describe an experiment to check the accuracy of the 1A  reading of ammeter, assuming that the value for the mass deposited  coulomb of copper is available.
What is the cost of depositing a layer of silver 2mm thick on an object of  total surface area 150 cm² if a current of 1.0A used for an hour s 2p? ( density of silver is 10.5g per cmsqr ; mass of silver deposited per coulomb is 0.00112g/c).
*Describe an experiment to determine how the mass of copper deposited in a copper voltameter varies with the quantity of electricity passing through it. Give à circuit diagram and full experimental details.
An electric current passes through two voltmeter in series, containing sulphate and silver nitrate respectively. What is the mass f silver deposited in a given time if the mass of copper deposited is in that time is 1g? ( R.A.M of cooper is 63, f silver 108, valency of copper is 2, valency of silver is 1).
*Describe an experiment to decompose water y electrolysis.
Calculate the current required to liberat 10 m cube of hydrogen per min in electrolysis. (1 liter of hydrogen weighs 0.09 g at the temperature and pressureat which the volumeis measured in this case. The mass of hydrogen deposited per Coulomb is 0.0000105g/).
*Explain the term electrolysis, ionization,electrochemical equivalent.
*State Faraday’s laws of electrolysis and describe how you would attenp to test the second law experimentally.
*Calculate the time for which a steady current of 0.2A Must pass through a water voltameter  in other to liberate a quantity of hydrogen which would occupy a volume of 50 cm cube at STP. Take the mass of hydrogen deposited per Coulomb to be 0.00001045g/c and its density at STP to be 0.00009g/cm cube).
*Explain how the ionic theory account for Faraday’s law of electrolysis.
* Give an account of the processes when dilute sulphuric acid is electrolysed between platinum electrodes.
*A cell with copper electrodes and containing copper sulphate solution is in series with one containing dilute sulphuric acid and platinum electrodes. When 3.175g of copper have been liberated, what volume of hydrogen measured at STP is released? ( rtm of copper = 63.5, hydrogen = 1.008, valency of copper = 2, density of hydrogen at STP , 0.0899g per ).
*State the laws of electrolysis and describe two experiments to illustrate these laws.
*what are the chief differences between the passage o electricity  through metals and through ionized solutions.
*A steady current pass through a silver voltameter and a 10 ohms resistance arranged in series. A high-resistance voltmeter placed across the terminals f the 10 ohms coil reads 6V. Given that 0.8048 g of silver is eposited on the cathode of the voltameter in 20 minutes, calculate the mass of silver deposited per Coulomb.
*State the relationship between the mass of a substance liberated during electrolysis and the charge that passed . Describe an experiment by which this relationship can be established for a metal such as copper.
*An ammeter and a copper voltameter are placed in eries in  circuit nd after 20 minutes the mass of copper deposited is 0.36 g. That ammeter reads 0.8 A. What can you deduce from the ammeter reading?
( Assuming mass of copper deposited per Coulomb = 0.0003g/c )

EXPLANATION OF FARADAY’S SECOND LAW OF ELECTROLYSIS
Diagram here:



The set up consists of a copper voltameter, silver voltameter,  and a water voltameter are connected in series as shown in e set up diagram.
The copper and the silver cathodes are weighed and their masses are recorded. The same quantity of electricity is passed through the three voltameters for about 30 minutes. The copper and silver are then rinsed with distilled , dried  reweighed to know the masses of copper and silver deposited. The volumes of hydrogen and oxygen are read andverted to their masses using their densities.
If their masses are compared, they will be in the ratio,
 Copper : Silver : Hydrogen : Oxygen = 32 : 108 : 1 : 8.
These ratios are the same is the same as the ratio of their chemical equivalents.

FORMULA FOR CALCULATION IN FARADAY’S SECOND LAW OF ELECTROLYSIS
Note that in Faraday’s second law of electrolysis, two substances are involved. Therefore, the formula is as follow:
Mass of A / mass of B =  z of A / z of B.
A and B are the two substances while Z is their electrochemical equivalents.
Also, the formula can be written as w,
Mass of A /mass of B = (rtm of A / valency of A) / (rtm of B/valency of B)

Worked Examples:
• A copper and water voltameter are in series, and at the end of a period of time 3.0 g of copper is deposited, calculate the mass of oxygen deposited.
Solution:
Mass of copper = 3g, mass of Oxygen = ?,rtm of Copper = 63, valency = 2, rtm of Oxygen =16,valency = 2.
Formula:
M of C/M of O =(rtm of C/valency of C)/rtm of O/valency of O)
Let us substitute for the variables,
3/M of O = (63/2) / (16/2) ➡ 3/M of O = 31.5 / 8. Make No the subject.
M of O = 3 * 8 / 31.5 ➡ M of O = 24 / 31.5. Then, M of O = 0.7619 g.
• A copper and a silver voltameter are connected in series, and at the d of a period of time, 4.0 g of copper was , calculate the Mass of silver deposited at the me time. Chemical equivalent of copper = 31.5, that of silver = 108
Solution:
M of C = 5g, M of S = ?, Z of C = 31.5, Z of S = 108.
Formula:
M of C/M of S = Z of C / Z of S. Let us substitute for the variables. Then,  
       5 / M of S = 31.5 / 108. Let us make M of S the subject of the formula.
   5 * 108 / 31.5 = M of S. ➡ M of S = 5 * 108 / 31.5.
   M of S = 17.1428 g.
Note that chemical equivalent = relative atomic mass / 2.

FARADAY’S CONSTANT F.
Faraday’s constant is the quantity of charge which will liberate one mole of a monovalent element. The symbol is F.
Faraday’s constant F = 96500 c/mol.

CALCULATION ON FARADAY’S CONSTANT
Worked Example:
The mass of silver deposited when a ready current of 2A flow for 30 minutes through a silver voltameter.
Solution:
Current = 2A, time = 30 minutes, ass of silver = ?
First, we will calculate the quantity of charge produced by 2A that flow for  30 minutes. then,
Quantity of charge  produced = I * t  ➡ quantity of charge = 2 * 30 min
Quantity of charge = 2 * 30 * 60 seconds
                                   = 3600 Coulomb.
According to Faraday’s law, 1 F will liberate one mole of a monovalent element. Silver is a monovalent element. Then
1 F will liberate mole of silver. One mole of silver is 108 g mass. Then,
   If 1 F = 96500 C liberate 108 g of .
   Then, 3600 C will liberate x g of silver.
96500 / 3600 = 108 / x. Let is make x the subject of the formula.
26.81= 108 / x  ➡ x = 108 / 26.81  ➡ x =   4.0283g
Therefore,  2A of current for 30 minutes will liberate 4.0283 g of silver.

CONDUCTION OF ELECTRICITY THROUGH GASS:
Gasses also conduct electricity when the gases are subjected to a very low pressure in a closed chamber.   The Conduction of electricity through gas is study by the use of discharge tube.
DISCHARGE TUBE:
A discharge tube t of a long glass tube with metal electrodes sealed to each end of the tube. A vacuum pump is connected to he tube for evacuating the tube of air.
Diagram of discharge tube:

CONDITIONS UNDER WHICH GASES CONDUCT ELECTRICITY:
Gases conduct electricity under low pressure and high voltage.
EXPLANATION OF OPERATIONS:
When the pressure inside the tube is reduced to a very low pressure ( say -0.01 mm Hg)  and a high voltage ( say 1000 volt)  is connected across the cathode and the anode of the tube, the gas in  tube breaks into ions ( I.e ionized ). The positive ions move toward the cathode while the negative ions and free electrons move toward the anode. At the cathode, the positive ions knock off electrons from the cathode. These electrons produced at the cathode are called cathode rays.

CATHODE RAYS:
Cathode rays are stream of electrons.
PRODUCTION OF CATHODE RAYS:
Cathode rays are produced when the pressure inside a cathode ray  tube is reduced to a very low pressure ( say -0.01 mm Hg)  and a high voltage ( say 1000 volt)  is connected across the cathode and the anode of the tube, the gas in the  tube breaks into ions ( I.e ionized ). The positive ions move toward the cathode while the negative ions and free electrons move toward the anode. At the cathode, the positive ions knock off electrons from the cathode. These electrons produced at the cathode are called cathode rays.

Properties of cathode rays:
• Cathode rays are stream of moving particles.
• Cathode rays are negatively charged particles.
• Cathode rays are attracted toward the anode in electric field.
• Cathode rays make glass and other materials to glow or fluores.
• Cathode rays are deflected toward the North pole in a magnetic field.
• Cathode rays e gas to ionized.
• Cathode rays turn wheel paddle placed in their parts.
• Cathode rays pocess energy.
• Cathode rays produce heart energy.
• Cathode rays have high penetrating power.
• Cathode rays affect photographic plates.
• Cathode rays can penetrate through some metals.

APPLICATION OF GAS DISCHARGE TUBE:
• Fluorescent lamps and neon signs:
The  pprinciple of gas discharge tube is used fluorescent tubes for lighting and display of signs.
Fluorescent tubes contains mercury vapour or sodium vapour which glow at low pressure when cathode rays pass through the vapour. The cathode rays produced inside the tube cause emission of ultraviolet rays when they bombard the vaporised atoms of mercury. The inside of the fluorescent lamp is coated with phosphor so that it gives off visible light when the ultraviolet rays strike it. Fluorescent tubes are also used as display signs. Neon gas is commonly use because it gives bright orange  light. Fluorescent amps are more efficient than filament lamps.

Production of neon sign:
Neon sign is produced when an evacuated discharge tube is filled with neon gas at low pressure of 5 mm of Hg.  The gas discharge between the electrodes and give its characteristic colour. The colour that the discharge tube produce depend n the nature of the gas used inside the tube and the nature of the coating inside the tube.

THE HOT CATHODE:
Hot cathode is another way of producing free electrons apart from what happened in a discharge  tube when electrons  are produced from a cold cathode.
Whenever a metal is heated to a sufficient high temperature, electrons are emitted from the surface of the metal.

Thermionic emission:
Thermionic emission is the process whereby electrons are emitted from metal surface.
 Diagram to explain thermionic emission:

Explanation of the operations:
When the cathode or hot filament is heated to a high temperature by a voltage source across its terminal, the filament emits electrons by thermionic emission. extra energy is given to the free electrons at the metal surface. The extra energy enable the electrons to brake away from the metal surface and exist as free elections. The electrons accelerate to the positively biased anode and therefore produce current which flow through the galvanometer.

Thermionic Devices:
These are devices that operate on the principle of thermionic emission. They devices :
• Diode valve
• X- ray tube
• Cathode ray tube

Explanation:
• Diode valve:
Diode value is a simple application of thermionic emission. It consist of an anode, filament made of tungsten wire and a cathode which surrounds the filament.
Diagram of valve:

Operations of diode valve:
When the filament is heated to a high temperature by a voltage source across its terminal, the filament emits electrons by thermionic emission. The electrons accelerate to the positively biased anode and therefore produce current which flow at the anode. The current can be detected by a galvanometer.

Diode valve characteristics:
The diode value characteristics is used to show and explain the action of  diode valve.
Diode valve characteristic circuit:

Operations of the circuit:
 When the filament is heated to a high temperature by a voltage source across its terminal, the filament emits electrons by thermionic emission. The electrons accelerate to the positively biased anode and therefore produce anode current which flow at the anode. The current can be detected by a galvanometer. A voltmeter is connected across the anode and the cathode to know the the anode voltage Va that produced the anode current Is. Different anode voltage Va are applied to obtain different anode current Ia. Each of the anode voltages Va are plotted its corresponding anode current  Ia to obtain the diode valve characteristics.

Diode valve characteristics curve:



Explanation of the curve:
The diode characteristics curve shows the behaviour of diode valve. The cure shows that diode current Ia is not directly proportional to diode voltage Va. Initially, as the anode voltage Va is increased, the diode current Ia increases until it reaches its maximum value. But with further increase in anode voltage , the anode current remained constant or decreases. This point of maximum value of Ia is called saturation point. The anode current this point is called saturation current. For this reason, diode valve does not obey ohms  and it is called a non ohmic conductor or substance. In diode valve, current flow only in one direction.

Similarities between thermionic emission and photoelectric emission:
• Both of them are surface phenomenon.
• Electrons are released in both cases.
• Energy is released in both cases.
• Energy transformation take place in both cases.
• Increase in temperate is involved in both cases.

Differences between thermionic emission and photoelectric emission
Thermionic emission Photoelectric emission
Electrons are released in the presence of heat in a vacuum.
Electrons are released at a particular temperature depending on the nature of the cathode. Electrons are released in the presence of sun light.
Electrons are released at threshold frequency depending on the nature if the metal surface.

Cathode ray oscilloscope:
Cathode ray oscilloscope is the application of thermionic emission. It is a vacuum tube that contains  electron gun at one end and  a fluorescent screen at the other end. Between these two are two pairs of reflectors that are used to control the electron unto the screan.
Diagram of cathode rays oscilloscope:


Explanation of operation of cathode rays oscilloscope:
The heated filament heats the cathode and cause it to produces stream of electrons by thermionic emission, which travel to the screen. The positive biased anodes accelerate the electron unto the screen. The X-plates are used  to move the electron beam  left and right while the y- plates move the electron beam up and down. The grid is used to regulate the amount of electrons that gravel to the screen and cause the screen to. Cooling fin is attached to cool the temperature of the tube.


DUALITY OF MATTER: