Lesson Plan and Lesson note on Physics
Energy quantization
Subject: Physics
Theme: Energy Quantization And Duality Of Matter
Topic: Energy quantization
Sub Topic: Energy Level of an Atom
Date: dd/mm/yyyy
Class: S.S 3
Average Age: 16 years and above
Duration: 40 Minutes
No of Learners: 40
At the end of the lesson, the students should be able to:
1. Explain what Energy quantization is.
When an atom or metal is heated and it absorb energy to particular intensity the electrons start to collide with one another and with the electron shell until the electrons are able to liberate themselves from the orbit of the atom. They create an atom with high energy level which teams up with low energy level. When this occurs, radiation is emitted.Energy quantization is the process by which a packet of radiation energy known as photons or quanta is emitted when heat is applied or when temperature of a substance is high.
It can also be defined as the definite amount of energy required by an electron to stay in the orbit of an atom.
The energy E of a photon or quantum is given by:
E = hf
En - Eo = hf
ΔE = hf
But v = λ/t ≡ c = λ/t
also, f = 1/t
:. v = λf ≡ c = λf
:. f = c/λ
Hence ΔE = hc/λ
where:
ΔE or E = energy of a photon or quantum
En - Eo is the difference in the energy of the photon or quantum between the initial and final state of the atom
h = plank's constant = 6.6 x 10-34 Js
c = velocity of electromagnetic waves in free space = 3 x 108 m/s
λ = wavelenght of electromagnetic wave (m)
f = frequency of electromagnetic wave (Hz)
2. Explain the energy level in an atom.
Electrons in atoms are arranged around their nuclei in position known as energy level or electron shell.The energy level in atoms refer to the energy of the orbital that an electron occupies.
Energy level is also called energy state. The higher the level, the further away the electron is.
Energy levels are represented by a series of horizontal lines numbered, n= 1, 2, 4, ... ∞, n is called quantum number, the outermost state is ionised state with energy level 0ev.
It requires more energy to remove electrons from the first energy level than to remove electrons from any of the other higher levels.
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where e = charge (Coulomb "C")
V = p.d (Volts "V")
Atoms can be given energy by heating (application of heat) or passing electricity through the atom.
Excitation is the process whereby electrons move from a lower energy level to a higher energy level.
De-excitation is the process whereby electrons lose energy as photon and returns to a lower energy level
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Atoms exist in specific energy states, and transitions between these states lead to the emission or absorption of energy.
Electron enter higher energy orbital by absorbing light and as they return from the excited state to the ground, they emit energy in the form of light.
I. Ground State:
- The ground state of an atom is the lowest energy state it can occupy.- Electrons in the ground state are in their lowest energy orbitals or energy levels.
- The ground state is considered the most stable state for an atom.
- Transition to a lower energy level or orbital in the ground state is not possible.
II. Excited State:
- When an atom absorbs energy, its electrons move to higher energy levels.- This results in the atom being in an excited state.
- Electrons in the excited state have more energy than in the ground state.
- Excited states are temporary, and electrons tend to quickly return to lower energy levels.
III. Emission of Energy on Light Return to Ground State (Atomic Spectra):
- When electrons in an excited state return to lower energy levels, they release energy in the form of electromagnetic radiation, typically in the visible range.- This emission of light is what creates the atomic spectra.
- Atomic spectra are unique patterns of lines or bands of light corresponding to different elements.
- Atomic spectra can be of two main types:
- Emission Spectra: These are produced when electrons return to the ground state, emitting specific wavelengths of light.
- Absorption Spectra: These are produced when electrons absorb energy and move to higher energy levels. They are characterized by missing lines in the spectrum.
Notable Points:
Ionization potential
Ionization potential is the minimum energy required to remove the most loosely bond electron from an atom in a gaseous form.Excited energy levels:
When electrons absorb enough energy they jump to higher energy levels. The higher energy levels are called excited energy levels. The first excited energy level (E1) corresponds to the second principal quantum number (n = 2). The Third excited energy level (E2) corresponds to n = 3 and so on.Electrons occupying the excited energy levels are said to be excited. Excited energy levels are unstable and tend to become stable by losing excess energy. They lose energy in the form of electromagnetic radiation when they jump or fall to lower energy levels, the energy radiated is given by.
Ei – Ef = hf
The difference between the two energy levels determine the size of the energy released i.e the frequency.
Excitation and ionization energies.
Excitation energy is the minimum energy absorbed by an electron to move it to higher energy levels. Ionization energy is the energy absorbed by an electron to move it completely out of the atom.Ionization energy = E∞ - En = 0 - En
An electron in the ground state needs 13.6eV to free it from the attractive force of the nucleus.
- Energy levels in atoms are quantized, meaning electrons can only occupy specific energy levels.
- The energy levels are represented by quantum numbers, with the principal quantum number (n) indicating the main energy level.
- Electrons cannot exist between energy levels; they can only transition between these quantized levels.
- The energy of a photon of light emitted during a transition is directly related to the energy difference between the energy levels involved.
- One electron volt (1eV) is the energy acquired by an electron in falling freely through a p.d of 1 Volt = 1.6 x 10-19J.
- During the excitation from lower energy level to higher energy level, the potential energy is converted into Kinetic energy so that the electrons eventually acquire a velocity given by:
K.E = ½ MV2 = eV.
- The energy gained by electron = charge x p.d = eV.
Therefore, the electron moves from one level to the other according to the relation. En – Eo = hf/λ= eV
3. Worked Problems on Energy Level in an atom
Example 1.If the p.d by which an electron moves is 1.5kv.
Calculate
(a) the velocity with which the electron moves if the ratio of its charge to mass is 1.8 x 1011 Ckg-1
(b) the kinetic energy.
SOLUTION
V = 1.5kv = 1.5 x 103V
m = 1kg
e = 1.8 x 1011C
(a) K.E = ½ mv2 = ev
2ev = mv2
v2 = 2ev/m
But ratio of charge to mass i.e e/m = 1.8 x 1011
V2 = 2 x 1.8 x 1011 x 1.5 x 103 ÷ 1
V2 = 5.4 x 1014
V = √5.4 x 1014
V = 2.3 x 107 m/s
(b) Ke = eV
Ke = 1.8 x 1011 x 1.5 x 103
Ke = 2.7 x 1014J.
Example 2
The diagram above illustrate an electron transition from level n=3 to n=1. Calculate the; (i) energy of the photon.
(ii) frequency of the photon.
(iii) wavelength of the photon.
[h= 6.6 x 10-38 Js; c = 3.0 x 108 m/s; 1eV = 1.6 x 10-19 J]
SOLUTION
(i) Energy of photon = ΔE
ΔE = E3 - E0
E3 = -1.51 eV
E0 = -13.6 eV
ΔE = [-1.51 - (-13.6)] eV
ΔE = 12.09 eV
But 1eV = 1.6 x 10-19J
:. 12.09 eV = 12.09 x 1.6 x 10-19J
= 1.93 x 10-18J
(ii) ΔE = hf
f = ΔE/h
f = 1.93 x 10-18 ÷ 6.6 x 10-34
f = 2.92 x 1015
(iii) ΔE = hf
ΔE = hc/λ since f = c/λ
λ =hc/ΔE
λ = 6.6 x 10-18 x 3 x 108 ÷ 1.93 x 10-18
λ = 1.03 x 107m
4. Explain Photoelectric effect
Photoelectric effect is the emission of electrons from a metal surface when radiation of enough frequency or energy falls on its surface.It can also be said as the phenomenon in which electrons are emitted from a metal when light shines on it.
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5. Explain the characteristics of photoelectric effects
1. Existence of threshold frequency: A minimum frequency called threshold frequency is required to emit electrons from a surface. if the frequency of incident light is below the threshold frequency, no electron is removed from the surface no matter the intensity of the incident radiation.2. The maximum kinetic energy of the photoelectrons depends on the frequency of the incident radiation. It is proportional to the frequency of the incident radiation.
3. The number of photoelectrons emitted is proportional to the intensity of the incident radiation. Once the incident energy is enough to remove electrons from the surface, the number of photo electrons removed from the surface increases as intensity of incident light increases.
6. Explain the application of photoelectric effects
Photoelectric emissions is used in the following:i. Burglary alarm
ii. Television camera
iii. Automatic devices for switching light at dusk and street light.
iv. Sound production of film track
v. industrial controls and counting operations.
7. Explain EINSTEN PHOTOELECTRIC EQUATION
According to Einstein, electrons are bound to the material. The work done to remove electrons from the surface is called work function. Work function is the minimum energy an electron should absorb before it is removed from the surface of a material. The work function of a material is given by:Einstein photoelectric equation is given by
E = W – Wo
E = hf - hfo
But E = ½mv2
:. ½mv2 = hf – Wo
½mv2 = hf - hc/𝜆0
where
E or ½mv2 = maximum kinetic energy that can be given to a photo electrons
𝑊 = work function
fo = Threshold frequency
𝜆0 = threshold wavelength
h = Planck’s constant
Wo = maximum energy of the liberated photoelectrons.
THRESHOLD FREQUENCY (fo)
This is the lowest frequency that can cause photo emission of electrons from a metallic surface. Below threshold frequency, emission will not occur.The Threshold frequency (f0) can also be defined as the frequency of light which falling on the surface is just enough to liberate electrons without given additional kinetic energy
The threshold wavelength is the longest wavelength that will produce photo electrons when the surface is illuminated.
WORK FUNCTION (W = hfo)
This is the minimum energy required to liberate electrons from a metallic surface.8. Solve problems on PHOTO-ELECTRIC EFFECT.
EXAMPLE 1.The maximum kinetic energy of the photo-electrons emitted from a metal surface is 0.34 eV if the work function of the metal surface is 1.83 eV. Find the stopping potential
[h= 6.6 x 10-38Js; c = 3.0 x 108m/s; 1eV = 1.6 x 10-19J]
SOLUTION
E = ½mv2 = maximum kinetic energy that can be given to a photo electrons, while
W = hf = maximum energy of the liberated photoelectrons.
W = 0.34 eV
E = 1.83 eV
Maximum K.E= 0.34 eV = 0.34 x 1.6 x 10-19 J
K.E = 5.44 x 10-20
Stopping Potential Vs = K.E/1eV
Vs = 5.44 x 10-20 ÷ 1.6 x 10-19
Vs = 0.34 m/s
EXAMPLE 2
Light of wavelength 5 x 10-7m is incident on a metal of work function 1.9eV. Calculate
(1) Photon energy
(ii) Kinetic energy of the most energetic photon
(iii) The stopping Potential Vs
[h= 6.6 x 10-38Js; c = 3.0 x 108m/s; 1eV = 1.6 x 10-19J]
SOLUTION
λ = 5 x 10-7 m
W = 1.9 eV = 1.9 x 1.6 x 10-19 J
W = 3.o4 x 10-19 J
(1) Photon energy hf
hfo or 𝑊 = hc/𝜆0
hf = 6.6 x 10-38 x 3.0 x 108 ÷ 5 x 10-7
hf = 3.96 x 10-19 J
(ii) Kinetic energy of the most energetic photon E or K.E
E = hf – W
E = 3.96 x 10-19 J - 3.o4 x 10-19 J
E = 9.2 x 10-20 J
(iii) Stopping Potential Vs = K.E/1eV
Vs = 9.2 x 10-20 ÷ 1.6 x 10-19
Vs = 0.575 V
Example 3
The diagram bove illustrate energy level in the hydrogen atom. E is he energy of the ground state.
(i) When an electron makes a transition from level n=3 to level n-=1, it emits a photon of wavelength 1.02 x 10-7m. Calculate Eo.
(ii) Calculate the ionization potential of the hydrogen atom.
(iii) Explain the statement, the work function of sodium is 2.0 eV.
(iv) Light of wavelength 160 nm is shore on the surface of a sodium metal of work function 2.0 eV. Determine whether photo electrons will be emitted.
[h= 6.6 x 10-38 Js; c = 3.0 x 108 m/s; 1eV = 1.6 x 10-19 J]
SOLUTION
(i) from n=3 to n=0
ΔE = E3 - E0
But ΔE = hf
ΔE = hc/λ since f = c/λ
h= 6.6 x 10-38Js;
c = 3.0 x 108m/s;
λ = 1.02 x 10-7m.
:. ΔE = 6.6 x 10-38 x 3.0 x 108 ÷ 1.02 x 10-7
ΔE = 1.941 x 10-18J
But 1 eV = 1eV = 1.6 x 10-19J
:. 1.941 x 10-18 = 1eV x 1.941 x 10-18 ÷ 1.6 x 10-19
ΔE = 12.13 eV
Hence ΔE = E3 - E0
12.1 eV = -1.50 eV - E0
E0 = -1.5 - 12.13
E0 = -13.63 eV
(ii) Ionization Energy = E∞ - E0
Ionization Energy = 0 - (-13.6)
Ionization Energy = 13.6 eV
(iii) The statement "the work function of sodium is 2.0 eV" means that the minimum energy needed to remove an electron from the surface of sodium metal is 2.0 eV.
(iv) Given λ = 160 nm = 160 x 10-9
Wo = hfo
N/B: For photoelectron emission
W > Wo
i.e hf > hfo
But E = hc/λ
E = 6.6 x 10-38 x 3.0 x 108 ÷ 160 x 10-9
ΔE = 1.2375 x 10-18 J
In electron volt
1eV = 1.6 x 10-19J
:. 1.2375 x 10-18 J = 1.2375 x 10-18 J x 1eV ÷ 1.6 x 10-19J
1.2375 x 10-18 J = 7.73 eV
Since E = hf
hf = 7.73 eV is greater than hfo = 2.0 eV = work function
:. Photoelectrons would be emitted when light of 160 nm shone on the metal surface.
9. Definition of X-ray:
x-rays are electromagnetic radiation of high penetrating power of short wavelength of the order of 10-10 m.X-rays are produced when a fast moving, high energy electron beam is suddenly slowed down by the atoms of the target. The kinetic energy of these moving electrons is converted to heat energy with less than 1% converted to x-rays.
X-ray was discovered in 1895 by Williams Rontgen. X–rays are produced when thermally generated electrons from a hot filament are accelerated through a high potential difference and focused on to a tungsten target, where the electrons are suddenly stopped.
X-ray have similar properties to gamma rays. The difference between the two is that gamma rays originate from the nucleus during radioactive decay, whereas electrons moving from one energy level to another lower energy level produce X-rays.
The x-ray tube consists of the following parts:
1. Electron gun to shoot out electrons. Electron gun consists of a filament heated by a low voltage. Electrons are shot out at high speed from the electron gun towards the anode.
2. A target to stop fast moving electron beams. This is usually very hot when the x-ray tube is working and therefore must have a means to cool it. The target is made of materials like tungsten and molybdenum which can withstand high temperatures.
3. An enclosed container to stop air from hindering the motion of the electron beam.
4. High voltage of between 40kv and 100kv is applied across the target and the electron gun. The high voltage accelerates the electrons and gives them enough energy before they strike the target. Less than 1% of the electron beam is converted to x-ray. The rest are transformed into heat.
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HARDNESS OF X-RAY: This is a measure of the strength or penetrating ability of the x–ray.
INTENSITY OF X-RAY: This is the energy radiated per unit time per unit area by the x–ray. It depends on the current of the filament.
Heisenberg Uncertainty Principle: Heisenberg Uncertainty Principle states that the momentum and position of a particle cannot be determined or measured simultaneously
10. List and explain TYPES OF X–RAY
There are two types of x-rays(i) Hard x–rays:
Have high penetrating power or ability and Shorter wavelength.(ii) Soft x- rays:
Have low penetrating power and longer wavelength11. List Characteristics of X-ray
1. They are electromagnetic waves of very high frequency2. X–rays are electromagnetic waves of very short wavelength. (2x10-10 m)
3. They move in a straight line
4. They have the speed of light of 3x108 ms-1
5. They are not diffracted by electric or magnetic field.
6. They are not deflected by alpha particles
7. They are not diffracted by crystals.
8. They have high penetrating power, X-Rays penetrate most materials that are opaque to light.
9. They ionize gases
10. They cause zinc sulphide to fluorescence (make certain material to glow).
11. X-Rays affect photographic film like light.
12. X-rays produce interference, diffraction and polarisation like other waves
12. Explain MODE OF OPERATION of the X-ray:
In the X- ray tube, a high potential difference is applied between the hot cathode and the anode. Electrons are emitted from the cathode and are accelerated to an extremely high speed. They are abruptly decelerated when they strike the anode causing the emission of high energy radiation of short wavelength i.e X-rays. The anode becomes very hot in the process and requires cooling gins on the outside of the tube.13. Explain ENERGY CONVERSION (transformation) DURING X–RAY PRODUCTION
During X – ray production, electrical energy is converted to thermal energy. The thermal energy is converted into mechanical energy (kinetic energy) to accelerate the electron. The mechanical energy is converted into electromagnetic energy of the x-rayElectrical → Heat → Mechanical → Electromagnetic
15. List Application and uses of X–ray
(A) APPLICATIONS1. For examining body to locate broken bones
2. To detect metals and contra band in a baggage
3. They are used to detect cracks n welded joints
4. For investigating crystal structure
5. Treatment of tumors and malignant growth
6. It is used in agriculture to kill germs.
7. Medical uses of x-ray
8. Industral uses of x-ray
9. Scientific uses of x-ray
(B) USES OF X-RAYS
(a) Medical application of x-rays
i. X-rays are used to reveal broken bones, bullets and other dense objects hidden in the body
ii. X-Ray photographs can reveal obstruction in digestive tract and internal growth
iii. Right dose of har x-rays is used to kill cancerous cells and malignant growth in the body.
(b) Industrial and scientific application of x rays.
i. Hard x-ray are used in the industries to reveal hidden crack in welded joints and castings
ii. X-ray diffraction of crystal reveal how the atom are arranged in different crystal.
iii. X-rays diffraction is used to study the structure of complex organic molecules.
14. List Hazards and Precautions of exposure to X-ray
1. It causes genetic mutation2. It can destroy body cells
3. it causes leukemia, by damaging body tissues
4. it causes skin burns and cancer.
Precautions: Those who work with x-rays should put on lead coat and they should always go for regular medical check-up.
Rationale:
Experiments performed by Franck-Hertz gave one of the most direct proofs of the existence of energy levels or electron shells within an atom. In these experiments it was possible to measure the energy necessary to raise an electron from the ground state in an atom to an outer orbit or higher state excitation or to remove it from the atom entirely ionisation. Excitation energy is the minimum energy required to change a system from its ground state.Prerequisite/ Previous knowledge:
Atomic structure.Learning Resources:
Flash cards, an audio and video youtube examples, Available useful objects.Reference Materials:
1. New system physics for secondary school by Dr. Charles chew etal2. New school physics by M. W Anyakoha
3. Internet facility
Lesson Development:
STAGE |
TEACHER'S ACTIVITY |
LEARNER'S ACTIVITY |
LEARNING POINTS |
|---|---|---|---|
| STEP 1: INTRODUCTION Individual Student |
The teacher asks the students the following questions: 1. What is the Concept of the Atom? 2. List and explain the four models of an atom. 3. Explain the limitation of each model. |
The students expected answers:1. Concept of the AtomThe concept of the atom is the fundamental idea that matter is composed of tiny, indivisible particles known as atoms.2. The models of the atomi. Sir J. J. Thompson model: Thompson proposed an atomic model which visualized the atom as a homogenous sphere of positive charge inside of which are embedded negatively charged electrons.He also determined the ratio of the charged to mass, e/m, of electrons, and found e/m to be identical for all cathode rays particles, irrespective of the kind of gas in the tube or the metal the electrons are made of.
Thompson’s model explains that: (a) A normal atom is electrically neutral. (b) Ions are formed by atom gaining or loosing electron. ii. Rutherford model: He proposed a planetary model of the atom which suggested that the atom consists of positively charged heave core called the nucleus where most of the mass of the atom was concentrated around this nucleus, negatively charged electrons circle in orbits much as planets move around the sun. Each nucleus must be surrounded by a number of electrons necessary to produce an electrically neutral atom.
The model was created to explain the back-scattering of alpha particles from thin metal foils. The postulates are statedas: (a) The diameter of the atom is about 10−10 m. (b) The diameter of the nucleus of an atom is 10−14 m. (c) The region around the nucleus is large compare to the space occupied by th enucleus. (d) The electrons are located around the nucleus. (e) The mass of an atom is concentrated at the nucleus iii. Bohr Model: Niels Bohr introduced the atomic Hydrogen model in 1913. He described it as a positively charged nucleus, comprised of protons and neutrons, surrounded by a negatively charged electron cloud.
Niels Bohr suggested a model of hydrogen atom in which: (a) The electrons in such stationary statements no radiation (b) If an electron jumps to a lower state, it emits a photon whose energy equals the difference in energy between the two states. (c) The angular momentum ‘L’ of the atomic electron is quantized by the rule (Where n=1, 2, 3, etc). (d) The chemical and physical properties of an element depend on the number of electrons in the outermost shell. (e) The chemical properties of element depend on the number of electrons in the outermost shell. iv. The electron cloud Model: This model visualizes the atom as consisting of a tiny nucleus of radius of the order of 10-10 m - 10-15 m. The electron is visualized as being in rapid motion within a relatively large region around the nucleus, but spending most of its time in certain high probability regions. Thus, the electron is not considered as a ball revolving around the nucleus but as a particle or wave with a specified energy having only a certain probability of being in a given region in the space outside the nucleus. The electron is visualized as spread out around the nucleus in a sort of electron–cloud.
Chemists prefer to consider the electron in terms of a cloud of negative charges (electron cloud), with a cloud being dense in regions of high electron probability and more diffuse in region of low probability. The probability of finding the electron inside the spherical boundary is high. The probability then decreases rapidly as the distance of the thin shell from the nucleus increases. 3. LIMITATIONS OF THE ATOMIC MODELSi. Limitation of Thompson’s model(a) The scattering of alpha-particle by gold foil in Geiger and Marsden experiment. (b) The existence of line spectra of hydrogen atom and other complex gases. ii. Limitations of Rutherford model (a) The assumed nature of the orbit not could be said to be fixed due to the electro static attraction of the electrons towards the nucleus by the positively charged protons. (b) The electrostatic force that is supposed to maintain the electrons in the irrespective orbits are weak due to emission of photon when accelerating electrons fall into the nucleus or from one orbit to the other. (c) The model was unable to explain some observations behind the line of spectrum emitted by incandescent objects. iii. Limitations of Bohr’s model (a) It does not explain how fixed orbits for electrons are chosen when they are not radiating energy. (b) The model is not applicable to the atom with more than one electron in the outermost shell. (c) The model cannot be used to predict energy level for complex atoms with many electrons. iv. Limitation of electron cloud atomic model (a) The model does not give the exact position of an electron in the orbital at a given time. (b) Electrons couldn't orbit the nucleus like miniature planets (c) The electron cloud model, based on quantum mechanics, is the most accurate representation of the atom to date, but it can be conceptually challenging as it involves probability distributions rather than fixed orbits. |
Identification of Prior Ideas and revising previous lesson |
| STEP 2: EXPLORATION Entire Class |
The teacher’s leads the students to understand that Energy quantization is the process by which a packet of radiation energy known as photons or quanta is emitted when heat is applied or when temperature of a substance is high. The energy E of a photon or quantum is given by: E = hf En - Eo = hf ΔE = hf But v = λ/t ≡ c = λ/t also, f = 1/t :. v = λf ≡ c = λf :. f = c/λ Hence ΔE = hc/λ where: ΔE or E = energy of a photon or quantum En - Eo is the difference in the energy of the photon or quantum between the initial and final state of the atom h = plank's constant = 6.6 x 10-34 Js c = velocity of electromagnetic waves in free space = 3 x 108 m/s λ = wavelenght of electromagnetic wave (m) f = frequency of electromagnetic wave (Hz)
where e = charge (Coulomb "C") V = p.d (Volts "V") 1. THE GROUND STATE: The ground state describes the lowest possible energy that an atom can have. This is the energy state that would be considered normal for the atom. 2. THE EXCITED STATE: An excited state is an energy level of an atom, ion, or molecule in which an electron is at a higher energy level than its ground state. An electron is normally in its ground state, the lowest energy state available. After absorbing energy, it may jump from the ground state to a higher energy level, called an excited state. 3. THE EMISSION STATE: Atomic emission spectra are produced when excited electrons return to the ground state. When electrons return to a lower energy level, they emit energy in the form of light. When an atom is in emission state the electron can drop all the way to the ground state in one go, or stop on the way in an intermediate level. Electrons do not stay in excited States for very long. They soon return to their ground States emitting a photon with the same energy as the one that was absorbed. Emission lines are seen when electrons return to the ground state releasing light. Light is emitted by an electron when it loses energy and returns to the ground state. |
The students write down what the teacher explains and listen attentively. | Definition of Energy quantization and the energy level in an atom |
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| STEP 3: DISCUSSION Entire class |
The teacher worked some Problems on Energy Level in an atom with the students. Example 1. The change in energy level of an electron in an atom is 6.2 x 10-21J. Calculate: (a) the frequency of the photon (b) the wavelength [c = 3.0 x 108 ms-1, h = 6.625 x 10-34Js] SOLUTION (a) ΔE = En -Eo = 6.2 x 10-21J ΔE = hf f = ΔE/h f = 6.2 x 10-21 ÷ 6.625 x 10-34 f = 9.358 x 1012Hz (b) But c = fλ :. λ = c/f λ = 3.0 x 108 ÷ 9.358 x 1012 λ = 3.206 x 10-5 m Example 2 An atom excited to an energy level E2 = -12.42 x 10-19 J falls to a ground level of energy Eo = -30.3 x 10-19 J. Calculate the frequency and the wavelength of the emitted photon. SOLUTION ΔE = E2 - Eo ΔE = -12.42 x 10-19 – (-30.3 x 10-19) ΔE = 1.788 x 10-18J (a) f = ΔE/h f = 1.788 x 10-18 ÷ 6.625 x 10-34 f = 2.699 x 1015 Hz. (b) λ = c/f λ = 3.0 x 108 ÷ 2.699 x 1015 λ = 1.1115 m Example 3 The ground state of hydrogen is -26.3eV and the second state is -10.3eV. Calculate the wavelength of the radiation if the electron returns to the ground state. (c = 3.0 x 108 ms-1, h = 6.625 x 10-34 Js) SOLUTION ΔE = E2 – Eo ΔE = -10.3eV – (-26.3eV) ΔE = 16eV But 1eV = 1.6 x 10-19J :. 16eV = 16 x 1.6 x 10-19J ΔE = hf ΔE = hc/λ since f = c/λ λ = hc/ΔE :. λ = 6.625 x 10-34 x 3.0 x 108 ÷ 16 x 1.6 x 10-19 λ = 1.9875 x 10-25 ÷ 25.6 x 10-18 λ = 7.76 x 10-8m |
The students copied the worked examples. | Better understanding of of Energy quantization. |
| The teacher explains Photoelectric effect with worked examples. Photoelectric effect is a phenomenon in which electrons are emitted from a metal when light shines on it. Einstein photoelectric equation is given by E = W – Wo E = hf - hfo But E = ½mv2 :. ½mv2 = hf – Wo ½mv2 = hf - hc/𝜆0 where E or ½mv2 = maximum kinetic energy that can be given to a photo electrons 𝑊 = work function fo = Threshold frequency 𝜆0 = threshold wavelength h = Planck’s constant Wo = maximum energy of the liberated photoelectrons. EXAMPLE 1. Compute the frequency of the photon whose energy is required to eject a surface electron with a kinetic energy of 3.5 x 10-16 eV if the work function of the metal is 3.0 x 10-16 eV (h = 6.6 x 10-34 JS, 1eV = 1.6 x 10-19 J) SOLUTION E = 3.5 x 10-16 eV W = 3.0 x 10-16 eV h = 6.6 x 10-34 JS 1eV = 1.6 x 10-19 J E = hf – w E + w = hf hf = (3.5 + 3.0) x 10-16 eV hf = 6.5 x 10-16 eV To convert hf from eV to J But 1eV = 1.6 x 10-19 J :. hf = 6.5 x 10-16 x 1.6 x 10-19 J hf = 10.4 x 10-35 J f = 10.4 x 10-35 ÷ h f = 10.4 x 10-35 ÷ 6.6 x 10-34 f = 1.58 x 10-1 Hz EXAMPLE 2 The work frequency of Lithium is 2.30 eV, calculate (i) the maximum energy in Joules of photoelectrons liberated by light of wavelength 3.3 x 10-17 m (ii) the threshold wavelength of the metal. [h= 6.6 x 10-38 Js; c = 3.0 x 108 m/s; 1eV = 1.6 x 10-19 J] SOLUTION W = 2.30 eV = 2.3 x 1.6 x 10-19 J h = 6.6 x 10-34 JS c = 3.0 x 108m/s λ = 3.3 x 10-17 m (i) E = hf – W E = hc/λ - W E = (6.6 x 10-34 x 3.0 x 108 ÷ 3.3 x 10-17) – (2.3 x 1.6 x 10-19) E = 2.208 x 10-27 J (ii) W = hc/λo λo =hc/W λo = 6.6 x 1034 x 3.0 x 108 ÷ (2.3 x 1.6 x 10-19) λo = 8.61 x 10-7 m EXAMPLE 3 A 500Kv is applied across an X-ray tube, calculate the maximum velocity of the electrons produced. m = 9.1 x 10-31 kg, 1eV = 1.6 x 10-19 V SOLUTION p.d = 500Kv = 500 x 103 p.d is the work done which is converted to K.E :. K.E = 500 x 103 x 1.6 x 10-19 J K.E = 8.0 x 10-14 J But K.E = ½ mv2 ½ mv2 = 8.0 x 10-14 v2 = 2 x 8.0 x 10-14 ÷ 9.1 x 10-31 v2 = 1.75 x 1017 V = 4.19 x 108 m/s |
The students write down what the teacher explains and listen attentively. | Definition of Photoelecric effect with worked examples | |
| The teacher explains X-ray tube to the students X–rays are electromagnetic waves of short wavelength of the order of 10-10 m.
1. They are electromagnetic waves of very high frequency 2. X–rays are electromagnetic waves of very short wavelength. (2x10-10 m) 3. They move in a straight line 4. They have the speed of light of 3x108 ms-1 5. They are not diffracted by electric or magnetic field. 6. They are not deflected by alpha particles 7. They are not diffracted by crystals. 8. They have high penetrating power, X-Rays penetrate most materials that are opaque to light. 9. They ionize gases 10. They cause zinc sulphide to fluorescence (make certain material to glow). 11. X-Rays affect photographic film like light. 12. X-rays produce interference, diffraction and polarisation like other waves |
The students write down what the teacher explains and listen attentively. | The modern x-ray tube | |
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| STEP 4: APPLICATION Entire class |
The teacher ask the students to read through all they have copied and give more discussion/explanation as directed by the teacher. They take corrections where they are wrong. | The students did what the teacher ask them to do. | Better understanding of Energy quantization. |
| STEP 5: EVALUATION Individual students |
The teacher asks the students questions to test them. 1. An electron jumps from one energy level to another in an atom radiating 4.5 × 10-19 J. if planck’s constant is 6.6×10-34 Js. What is the wavelength of this radiation? (C = 3.0 × 108) 2. An atom excited to an energy level E2 (-2.42 × 10-19 J) falls to the ground state E0 (-21.8 × 10-19 J). Calculate the frequency and the wavelength of the emitted photon (h = 6.6 x 10-34 JS) 3. An electron make a transition from a certain energy level Ek to the ground state E0. If the frequency of emission is 8 x 1014 Hz, what is the energy emitted? (h = 6.6 x 10-34 Js) 4. Mention the application of the Photo cells. |
The students respond to the questions correctly. SOLUTION 1. ΔE = 4.5 × 10-19 J h = 6.6 × 10-34 JS ΔE = hc/λ λ = hc/ΔE λ = 6.6 × 10-34 x 3.0 × 108 ÷ 4.5 × 10-19 λ = 4.4 × 10-7 m 2. h = 6.6 × 10-34 JS E2 = -2.42 × 10-19 J E0 = -21.8 × 10-19 J ΔE = E2 -E0 = -(2.42 × 10-19 )J - (-21.8 × 10-19) J ΔE = -2.42 × 10-19 J + 21.8 × 10-19 J ΔE = 19.38 × 10-19 J ΔE = hf f = ΔE/h f = 19.38 × 10-19 ÷ 6.6 × 10-34 f = 2.94 X 1015 Hz f = c/λ λ = c/f λ = 3.0 × 108 ÷ 2.94 X 1015 λ = 1.02 x 10-7 m 3. h = 6.6 x 10-34 Js f = 8 x 1014 Hz E = hf E = 6.6 x 10-34 x 8 x 1014 E = 5.28 x 10-19 J 4. (a) In TVs and cameras as light metres. (b) In electronic igniting circuit to operate or cause doors to open or close when light falling on the phocells is cut off. (c) To operate loudspeakers and burglary alarms. (d) In automatic street lighting and automatic digital counters. |
Asking the learners questions to assess the achievement of the set objectives. |
| CONCLUSION | The teacher concluded the lesson Understanding the energy levels in atoms, transitions between ground and excited states, and the emission of energy as atomic spectra is crucial in the field of physics. It has practical applications in various scientific and technological domains. Applications: 1. Spectroscopy: Atomic spectra are widely used in spectroscopy to identify elements based on their unique emission or absorption patterns. 2. Lasers: The principles of energy quantization and atomic transitions are employed in laser technology, producing highly focused and coherent light. |
The students write down the conclusion of the lesson and listen attentively. | Better understanding of Energy quantization, photoelectric effect and x-ray. |
| ASSIGNMENT | The teacher gives learners take home. 1. Draw the energy level in the atom. 2. An electron jumps from one energy level to another in an atom radiating 6.5×10-19 J. If the planck constant = 6.6×10-34 Js. what is the wavelength of the radiation? Take velocity of light = 3×108 m/s. 3. What is the amount of energy released when 0.5kg of uranium is burnt completely? 4. Explain the function of x-ray 5. State two medical uses of x-ray 6. List three properties of x-ray 7. Draw a modern x-ray tube. |
The learners copy the assignment | Better understanding of Energy quantization, photoelectric effect and x-ray. |