CLASS 12 PHYSICS CHAPTER 12 – ATOMS
NCERT CLASS 12 PHYSICS
CHAPTER 12 – ATOMS
1. IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS
Q1. What is an atom?
Answer: An atom is the smallest unit of an element that retains its chemical properties. It consists of a small, positively charged nucleus surrounded by negatively charged electrons.
Q2. Who discovered the electron?
Answer: J. J. Thomson discovered the electron in 1897 through experiments involving the discharge of electricity through gases.
Q3. What are the main constituents of an atom?
Answer: The main constituents of an atom are:
Electrons: Negatively charged particles present outside the nucleus.
Protons: Positively charged particles present in the nucleus.
Neutrons: Electrically neutral particles present in the nucleus.
Q4. What is Thomson's atomic model?
Answer: Thomson proposed that an atom is a sphere of positive charge with negatively charged electrons embedded in it.
This model is also known as the plum pudding model because the electrons were imagined to be embedded in a positively charged sphere.
Q5. What are the limitations of Thomson's atomic model?
Answer: Thomson's atomic model could not explain the results of Rutherford's alpha-particle scattering experiment.
It could not explain:
The large deflection of some alpha particles.
The backward scattering of a very small number of alpha particles.
The concentration of positive charge and most of the mass in a very small region.
Q6. What is Rutherford's atomic model?
Answer: According to Rutherford's atomic model:
An atom contains a very small, positively charged nucleus at its centre.
Almost the entire mass of the atom is concentrated in the nucleus.
Electrons revolve around the nucleus.
Most of the volume of an atom is empty space.
The nucleus is much smaller than the overall size of the atom.
Q7. What is the alpha-particle scattering experiment?
Answer: Rutherford's alpha-particle scattering experiment was performed to investigate the structure of an atom.
In this experiment, a beam of fast-moving alpha particles was directed towards a very thin gold foil. A fluorescent screen was used to observe the scattered alpha particles.
Q8. Why did Rutherford use a thin gold foil?
Answer: Rutherford used a thin gold foil because gold is highly malleable and can be converted into an extremely thin sheet.
A very thin foil allows most alpha particles to pass through it while a small number experience scattering due to their interaction with gold nuclei.
Q9. What were the observations of Rutherford's alpha-particle scattering experiment?
Answer:
Rutherford made the following observations:
Most alpha particles passed through the gold foil without any deflection.
Some alpha particles were deflected through small angles.
Very few alpha particles were deflected through large angles.
A very small number of alpha particles returned nearly in the direction from which they came.
Q10. What conclusions were drawn from Rutherford's experiment?
Answer:
The conclusions were:
Most of the volume of an atom is empty space because most alpha particles passed through the foil without deflection.
The positive charge of an atom is concentrated in a very small central region.
Almost the entire mass of an atom is concentrated in the nucleus.
The nucleus is extremely small compared with the size of the atom.
Electrons occupy the region surrounding the nucleus.
Q11. What is the size of an atom and its nucleus?
Answer:
The approximate size of an atom is of the order of:
10⁻¹⁰ m
The approximate size of a nucleus is of the order of:
10⁻¹⁵ m
Therefore, the nucleus is extremely small compared with the atom.
Q12. What is the limitation of Rutherford's atomic model?
Answer: According to classical electromagnetic theory, an accelerating charged particle emits electromagnetic radiation.
An electron revolving around the nucleus is continuously accelerated because its direction of velocity changes.
Therefore, it should continuously lose energy and spiral towards the nucleus. This would make the atom unstable.
However, atoms are stable. Rutherford's model could not explain this stability.
Q13. What is the atomic spectrum?
Answer: The atomic spectrum is the characteristic spectrum of electromagnetic radiation emitted or absorbed by an atom when its electrons undergo transitions between different energy levels.
Q14. What is a line spectrum?
Answer: A line spectrum consists of separate, distinct spectral lines corresponding to particular wavelengths or frequencies.
It is produced because atoms have definite allowed energy levels.
Q15. What is the difference between a continuous spectrum and a line spectrum?
Answer:
Continuous spectrum:
Contains a continuous range of wavelengths.
Does not have separate gaps between its colours or wavelengths.
Is commonly produced by hot dense materials.
Line spectrum:
Contains separate, distinct spectral lines.
Each line corresponds to a particular energy transition.
Is characteristic of the atoms or elements producing it.
Q16. What is Bohr's atomic model?
Answer: Niels Bohr proposed a model of the hydrogen atom in which electrons revolve around the nucleus only in certain permitted orbits.
The electron does not radiate energy while remaining in an allowed stationary orbit.
Energy is absorbed or emitted only when an electron moves from one allowed energy level to another.
Q17. State Bohr's postulates.
Answer:
Bohr's model is based on the following postulates:
Electrons revolve around the nucleus only in certain permitted circular orbits called stationary orbits.
Electrons do not radiate energy while moving in these stationary orbits.
The angular momentum of an electron in an allowed orbit is quantised.
mvr = nh/2π
Where n = 1, 2, 3, ...
Radiation is emitted or absorbed when an electron jumps from one energy level to another.
Q18. What are stationary orbits?
Answer: Stationary orbits are the permitted orbits in which an electron can revolve around the nucleus without radiating energy.
Each stationary orbit has a definite energy.
Q19. What is the ground state of an atom?
Answer: The lowest energy state of an atom is called its ground state.
For a hydrogen atom, the ground-state energy is:
E₁ = −13.6 eV
Q20. What is an excited state?
Answer: When an electron absorbs energy and moves from a lower energy level to a higher energy level, the atom is said to be in an excited state.
Q21. What is ionisation energy?
Answer: Ionisation energy is the minimum energy required to remove an electron completely from an isolated atom in its ground state.
For a hydrogen atom, the ionisation energy is:
13.6 eV
Q22. What is the principal quantum number?
Answer: The principal quantum number, represented by n, identifies the allowed energy level or orbit of an electron in Bohr's model.
Its possible values are:
n = 1, 2, 3, 4, ...
The value n = 1 represents the ground state.
Q23. What is the relation between the radius of an orbit and the principal quantum number?
Answer: According to Bohr's model, the radius of the nth orbit of a hydrogen atom is:
rₙ = n²a₀
Where a₀ is the radius of the first Bohr orbit.
Therefore:
rₙ ∝ n²
The radius increases as the principal quantum number increases.
Q24. What is the Bohr radius?
Answer: The radius of the first allowed orbit of a hydrogen atom is called the Bohr radius.
Its value is approximately:
a₀ = 5.3 × 10⁻¹¹ m
Q25. What is the energy of an electron in the nth orbit of hydrogen?
Answer: According to Bohr's model, the energy of an electron in the nth orbit of hydrogen is:
Eₙ = −13.6/n² eV
Where n = 1, 2, 3, ...
Q26. Why is the energy of an electron in a bound state negative?
Answer: The energy of a bound electron is taken as zero when the electron is infinitely far away from the nucleus and is at rest.
A bound electron has less energy than this reference state. Therefore, its total energy is negative.
Q27. What happens when an electron moves from a higher energy level to a lower energy level?
Answer: When an electron moves from a higher energy level to a lower energy level, it releases energy in the form of electromagnetic radiation.
The energy of the emitted photon is:
hν = Eᵢ − E𝒻
Where Eᵢ is the initial higher energy and E𝒻 is the final lower energy.
Q28. What happens when an electron moves from a lower energy level to a higher energy level?
Answer: When an electron moves from a lower energy level to a higher energy level, it absorbs energy equal to the difference between the two energy levels.
hν = E𝒻 − Eᵢ
Q29. What is the relation between photon energy and the difference in atomic energy levels?
Answer:
The energy of the photon is equal to the difference between the two atomic energy levels.
hν = |E₂ − E₁|
Where:
h = Planck's constant
ν = Frequency of photon
E₁ and E₂ = Atomic energy levels
Q30. Why does hydrogen produce a line spectrum?
Answer: The electron in a hydrogen atom can occupy only certain permitted energy levels.
When the electron moves between these levels, photons of definite energies are emitted or absorbed.
Since photon energy determines frequency and wavelength, only specific spectral lines are produced.
2. NCERT EXERCISE QUESTIONS AND ANSWERS
Question 12.1
Choose the correct alternative from the clues given at the end of each statement:
(a) The size of the atom in Thomson's model is .......... the atomic size in Rutherford's model. (much greater than/no different from/much less than.)
(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson's model/Rutherford's model.)
(c) A classical atom based on .......... is doomed to collapse. (Thomson's model/Rutherford's model.)
(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson's model/Rutherford's model.)
(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford's model/both the models.)
Answer:
(a) No different from.
Both models consider the overall size of an atom to be of the same order of magnitude.
(b) Thomson's model; Rutherford's model.
In Thomson's model, electrons are in stable equilibrium in the ground state. In Rutherford's model, revolving electrons experience a net electrostatic force towards the nucleus.
(c) Rutherford's model.
According to classical electromagnetic theory, revolving electrons continuously radiate energy and would eventually spiral into the nucleus.
(d) Thomson's model; Rutherford's model.
In Thomson's model, positive charge and mass are distributed throughout the atom. In Rutherford's model, most of the mass is concentrated in the tiny nucleus.
(e) Both the models.
In both models, the positively charged part of the atom is considered to possess most of the mass.
Question 12.2
Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?
Answer:
If a thin sheet of solid hydrogen is used instead of gold foil, the results will be different from Rutherford's original experiment.
An alpha particle has a mass of approximately:
mα = 6.64 × 10⁻²⁷ kg
The hydrogen nucleus (proton) has a mass of approximately:
mp = 1.67 × 10⁻²⁷ kg
The alpha particle is heavier than the proton.
Therefore:
The alpha particles will not undergo the same large-angle scattering observed with gold nuclei.
They will generally continue moving forward after interacting with hydrogen nuclei.
The hydrogen nuclei may recoil due to the collision.
Alpha particles will not be scattered backwards in a head-on elastic collision with a stationary proton.
Final Answer:
Large-angle and backward scattering will not occur in the same manner as in the gold foil experiment because the hydrogen nucleus is much lighter than the alpha particle.
Question 12.3
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?
Solution:
Given:
Energy difference:
ΔE = 2.3 eV
Planck's constant:
h = 6.63 × 10⁻³⁴ J s
1 eV = 1.6 × 10⁻¹⁹ J
First, convert the energy difference into joules.
ΔE = 2.3 × 1.6 × 10⁻¹⁹
ΔE = 3.68 × 10⁻¹⁹ J
We know:
ΔE = hν
Therefore:
ν = ΔE/h
Substituting:
ν = (3.68 × 10⁻¹⁹)/(6.63 × 10⁻³⁴)
ν = 5.55 × 10¹⁴ Hz
Final Answer:
Frequency of emitted radiation = 5.55 × 10¹⁴ Hz.
Question 12.4
The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?
Solution:
Given:
Total energy:
E = −13.6 eV
For an electron in a circular orbit under electrostatic attraction:
Total energy:
E = K + U
Also, according to the virial relation for the Coulomb force:
U = −2K
Therefore:
E = K − 2K
E = −K
Hence:
K = −E
Substituting:
K = −(−13.6)
K = 13.6 eV
Now:
U = −2K
U = −2 × 13.6
U = −27.2 eV
Final Answer:
Kinetic energy = +13.6 eV
Potential energy = −27.2 eV
Question 12.5
A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.
Solution:
Given:
Initial energy level:
n₁ = 1
Final energy level:
n₂ = 4
Energy of the nth level of hydrogen:
Eₙ = −13.6/n² eV
Energy of ground state:
E₁ = −13.6/1²
E₁ = −13.6 eV
Energy of fourth level:
E₄ = −13.6/4²
E₄ = −13.6/16
E₄ = −0.85 eV
Energy absorbed:
ΔE = E₄ − E₁
ΔE = −0.85 − (−13.6)
ΔE = 12.75 eV
Converting into joules:
ΔE = 12.75 × 1.6 × 10⁻¹⁹
ΔE = 2.04 × 10⁻¹⁸ J
(a) Frequency of photon
We know:
ΔE = hν
Therefore:
ν = ΔE/h
ν = (2.04 × 10⁻¹⁸)/(6.63 × 10⁻³⁴)
ν = 3.08 × 10¹⁵ Hz
(b) Wavelength of photon
We know:
c = νλ
Therefore:
λ = c/ν
λ = (3 × 10⁸)/(3.08 × 10¹⁵)
λ = 9.74 × 10⁻⁸ m
Converting into nanometres:
λ = 97.4 nm
Final Answer:
Frequency = 3.08 × 10¹⁵ Hz
Wavelength = 97.4 nm.
Question 12.6
(a) Using Bohr's model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.
(b) Calculate the orbital period in each of these levels.
Solution:
(a) Speed of electron in different orbits
According to Bohr's model:
vₙ = v₁/n
Where:
v₁ = 2.18 × 10⁶ m/s
For n = 1:
v₁ = 2.18 × 10⁶/1
v₁ = 2.18 × 10⁶ m/s
For n = 2:
v₂ = 2.18 × 10⁶/2
v₂ = 1.09 × 10⁶ m/s
For n = 3:
v₃ = 2.18 × 10⁶/3
v₃ = 7.27 × 10⁵ m/s
(b) Orbital period
Orbital period is the time taken by an electron to complete one revolution around the nucleus.
Formula:
T = 2πr/v
The radius of the nth orbit is:
rₙ = n²a₀
Where:
a₀ = 5.3 × 10⁻¹¹ m
For n = 1:
r₁ = 5.3 × 10⁻¹¹ m
T₁ = 2πr₁/v₁
T₁ = (2 × 3.14 × 5.3 × 10⁻¹¹)/(2.18 × 10⁶)
T₁ = 1.53 × 10⁻¹⁶ s
For n = 2:
r₂ = 2² × 5.3 × 10⁻¹¹
r₂ = 2.12 × 10⁻¹⁰ m
T₂ = 2πr₂/v₂
T₂ = (2 × 3.14 × 2.12 × 10⁻¹⁰)/(1.09 × 10⁶)
T₂ = 1.22 × 10⁻¹⁵ s
For n = 3:
r₃ = 3² × 5.3 × 10⁻¹¹
r₃ = 4.77 × 10⁻¹⁰ m
T₃ = 2πr₃/v₃
T₃ = (2 × 3.14 × 4.77 × 10⁻¹⁰)/(7.27 × 10⁵)
T₃ = 4.12 × 10⁻¹⁵ s
Final Answer:
| Orbit | Electron speed | Orbital period |
|---|---|---|
| n = 1 | 2.18 × 10⁶ m/s | 1.53 × 10⁻¹⁶ s |
| n = 2 | 1.09 × 10⁶ m/s | 1.22 × 10⁻¹⁵ s |
| n = 3 | 7.27 × 10⁵ m/s | 4.12 × 10⁻¹⁵ s |
Question 12.7
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10⁻¹¹ m. What are the radii of the n = 2 and n = 3 orbits?
Solution:
Given:
Radius of first orbit:
r₁ = 5.3 × 10⁻¹¹ m
According to Bohr's model:
rₙ = n²r₁
For n = 2:
r₂ = 2² × r₁
r₂ = 4 × 5.3 × 10⁻¹¹
r₂ = 21.2 × 10⁻¹¹ m
r₂ = 2.12 × 10⁻¹⁰ m
For n = 3:
r₃ = 3² × r₁
r₃ = 9 × 5.3 × 10⁻¹¹
r₃ = 47.7 × 10⁻¹¹ m
r₃ = 4.77 × 10⁻¹⁰ m
Final Answer:
Radius of second orbit = 2.12 × 10⁻¹⁰ m
Radius of third orbit = 4.77 × 10⁻¹⁰ m.
Question 12.8
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Solution:
Given:
Energy of incident electron:
E = 12.5 eV
The hydrogen atom is initially in its ground state at room temperature.
Energy of the nth level:
Eₙ = −13.6/n² eV
Ground-state energy:
E₁ = −13.6 eV
The electron beam can excite hydrogen atoms if it supplies sufficient energy to raise their electrons to higher energy levels.
Step 1. Find the energy required for excitation to n = 2
E₂ = −13.6/2²
E₂ = −3.4 eV
Energy required:
ΔE₂ = E₂ − E₁
ΔE₂ = −3.4 − (−13.6)
ΔE₂ = 10.2 eV
Step 2. Find the energy required for excitation to n = 3
E₃ = −13.6/3²
E₃ = −13.6/9
E₃ = −1.51 eV
Energy required:
ΔE₃ = E₃ − E₁
ΔE₃ = −1.51 − (−13.6)
ΔE₃ ≈ 12.09 eV
Step 3. Check excitation to n = 4
E₄ = −13.6/16
E₄ = −0.85 eV
Energy required:
ΔE₄ = E₄ − E₁
ΔE₄ = −0.85 − (−13.6)
ΔE₄ = 12.75 eV
Since the available electron energy is only 12.5 eV, excitation to n = 4 is not possible.
Therefore, the hydrogen atom can be excited to n = 2 or n = 3.
Step 4. Calculate the emitted wavelengths
The possible transitions are:
n = 3 to n = 1
n = 3 to n = 2
n = 2 to n = 1
Transition 1: n = 3 to n = 1
Energy difference:
ΔE = 13.6(1 − 1/9)
ΔE = 12.09 eV
Using:
λ = 1240/ΔE nm
λ = 1240/12.09
λ ≈ 102.6 nm
Transition 2: n = 3 to n = 2
Energy difference:
ΔE = 13.6(1/4 − 1/9)
ΔE = 13.6 × 5/36
ΔE ≈ 1.89 eV
λ = 1240/1.89
λ ≈ 656 nm
Transition 3: n = 2 to n = 1
Energy difference:
ΔE = 13.6(1 − 1/4)
ΔE = 10.2 eV
λ = 1240/10.2
λ ≈ 121.6 nm
Final Answer:
The possible emitted wavelengths are approximately:
102.6 nm
121.6 nm
656 nm
These correspond to transitions between the n = 1, 2 and 3 energy levels.
Question 12.9
In accordance with Bohr's model, find the quantum number that characterises the Earth's revolution around the Sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of Earth = 6.0 × 10²⁴ kg.)
Solution:
Given:
Mass of Earth:
m = 6.0 × 10²⁴ kg
Orbital radius:
r = 1.5 × 10¹¹ m
Orbital speed:
v = 3 × 10⁴ m/s
Planck's constant:
h = 6.63 × 10⁻³⁴ J s
According to Bohr's quantisation condition:
mvr = nh/2π
Therefore:
n = 2πmvr/h
Substituting the values:
n = [2 × 3.14 × 6.0 × 10²⁴ × 1.5 × 10¹¹ × 3 × 10⁴]/(6.63 × 10⁻³⁴)
n = (1.696 × 10⁴¹)/(6.63 × 10⁻³⁴)
n ≈ 2.56 × 10⁷⁴
Final Answer:
Quantum number:
n ≈ 2.56 × 10⁷⁴
The extremely large value shows why quantisation effects are not noticeable for macroscopic objects such as the Earth.
3. IMPORTANT DERIVATIONS
Q1. Derive the expression for the radius of the nth orbit of hydrogen according to Bohr's model.
Answer:
Consider an electron of mass m revolving around a nucleus of charge +e in a circular orbit of radius r.
The electrostatic force provides the necessary centripetal force.
Therefore:
(1/4πε₀) × e²/r² = mv²/r
Simplifying:
mv² = e²/(4πε₀r) ...(1)
According to Bohr's quantisation condition:
mvr = nh/2π
Therefore:
v = nh/(2πmr)
Squaring both sides:
v² = n²h²/(4π²m²r²)
Substituting in equation (1):
m × n²h²/(4π²m²r²) = e²/(4πε₀r)
Simplifying:
n²h²/(4π²mr²) = e²/(4πε₀r)
Multiplying both sides by r²:
n²h²/(4π²m) = e²r/(4πε₀)
Therefore:
rₙ = ε₀n²h²/(πme²)
Since all quantities except n are constant:
rₙ ∝ n²
For hydrogen:
rₙ = n²a₀
Where:
a₀ = 5.3 × 10⁻¹¹ m
This is the expression for the radius of the nth Bohr orbit.
Q2. Derive the expression for the energy of an electron in the nth orbit of hydrogen.
Answer:
The total energy of an electron is the sum of its kinetic energy and potential energy.
E = K + U
From the centripetal force equation:
mv²/r = e²/(4πε₀r²)
Therefore:
mv² = e²/(4πε₀r)
Kinetic energy:
K = ½mv²
Substituting:
K = e²/(8πε₀r)
Potential energy of the electron-nucleus system:
U = −e²/(4πε₀r)
Therefore:
E = K + U
E = e²/(8πε₀r) − e²/(4πε₀r)
E = −e²/(8πε₀r)
Using the Bohr radius expression:
rₙ = ε₀n²h²/(πme²)
Substituting:
Eₙ = −e²/[8πε₀ × ε₀n²h²/(πme²)]
Simplifying:
Eₙ = −me⁴/(8ε₀²n²h²)
For hydrogen, this becomes:
Eₙ = −13.6/n² eV
This is the expression for the energy of an electron in the nth orbit of hydrogen.
Q3. Derive the expression for the speed of an electron in the nth orbit.
Answer:
According to Bohr's quantisation condition:
mvr = nh/2π
Therefore:
v = nh/(2πmr)
Using the Bohr radius expression:
rₙ = n²a₀
Substituting:
vₙ = nh/(2πmn²a₀)
Therefore:
vₙ = h/(2πma₀n)
For hydrogen:
vₙ = 2.18 × 10⁶/n m/s
Hence:
vₙ ∝ 1/n
The speed of the electron decreases as the principal quantum number increases.
Q4. Derive the expression for the frequency of radiation emitted during an electronic transition.
Answer:
Suppose an electron moves from a higher energy level Eᵢ to a lower energy level E𝒻.
According to the conservation of energy:
Energy of emitted photon = Difference in atomic energy levels
Therefore:
hν = Eᵢ − E𝒻
Dividing both sides by h:
ν = (Eᵢ − E𝒻)/h
Since:
ν = c/λ
We get:
λ = hc/(Eᵢ − E𝒻)
These equations give the frequency and wavelength of the emitted radiation.
4. IMPORTANT QUESTIONS FOR EXAMINATION
Q1. Why did most alpha particles pass through the gold foil without deflection?
Answer: Most of the volume of an atom is empty space. Therefore, most alpha particles pass through the gold foil without coming close to the small, positively charged nucleus.
Q2. Why were a few alpha particles deflected through large angles?
Answer: Some alpha particles passed very close to the positively charged nucleus. Due to strong electrostatic repulsion between the positively charged alpha particles and the nucleus, they were deflected through large angles.
Q3. Why did a very small number of alpha particles return backwards?
Answer: A very small number of alpha particles approached the nucleus almost directly. The strong electrostatic repulsion caused them to turn back nearly in the direction from which they came.
Q4. Why is Rutherford's atomic model unstable according to classical physics?
Answer: According to classical electromagnetic theory, an accelerating charged particle emits electromagnetic radiation.
An electron revolving around the nucleus is accelerated because its direction changes continuously. It should therefore lose energy and spiral towards the nucleus.
This would make the atom unstable, which contradicts the observed stability of atoms.
Q5. How did Bohr explain the stability of an atom?
Answer: Bohr proposed that electrons can revolve only in certain permitted stationary orbits.
An electron does not radiate energy while moving in one of these allowed orbits. Therefore, it does not continuously lose energy and spiral into the nucleus.
This explains the stability of the atom within Bohr's model.
Q6. Why are atomic energy levels discrete?
Answer: According to Bohr's quantisation condition, only certain angular momentum values are allowed for an electron.
As a result, only certain orbits and corresponding energy values are permitted.
Therefore, atomic energy levels are discrete.
Q7. Why is the ground-state energy of hydrogen −13.6 eV?
Answer: The ground state is the lowest energy state of the hydrogen atom.
The zero of energy is taken when the electron is infinitely far from the nucleus and is free.
The ground-state electron is bound to the nucleus and has less energy than this reference state. Its energy is therefore −13.6 eV.
Q8. What is the ionisation energy of hydrogen in the ground state?
Answer: The ionisation energy of hydrogen in the ground state is 13.6 eV.
It is the energy required to move the electron from the ground state to infinity.
Q9. What happens to the radius of an electron orbit when the principal quantum number is doubled?
Answer:
We know:
rₙ = n²a₀
If n is doubled:
r' = (2n)²a₀
r' = 4n²a₀
Therefore:
r' = 4r
The radius becomes four times its original value.
Q10. What happens to the energy of an electron when it moves to a higher orbit?
Answer: The energy of the electron increases as it moves to a higher orbit.
However, the energy remains negative for a bound electron and approaches zero as n increases towards infinity.
Q11. Why does an atom emit radiation when an electron moves from a higher energy level to a lower energy level?
Answer: The electron loses energy during the transition. This energy is released in the form of an electromagnetic photon.
The photon energy is equal to the difference between the two energy levels.
Q12. Why does an atom absorb radiation when an electron moves from a lower energy level to a higher energy level?
Answer: An electron requires additional energy to move to a higher energy level. It absorbs a photon whose energy is equal to the difference between the two energy levels.
Q13. Why is the hydrogen spectrum a line spectrum?
Answer: The electron in hydrogen can occupy only certain allowed energy levels. Transitions between these levels produce photons with definite energies.
Therefore, the emitted radiation contains specific wavelengths and appears as a line spectrum.
Q14. Why does the energy of an electron approach zero as the principal quantum number increases?
Answer:
The energy of the nth orbit is:
Eₙ = −13.6/n² eV
As n increases, n² also increases. Therefore, the magnitude of energy decreases and the energy approaches zero from the negative side.
At n = infinity, the electron is free from the atom.
5. IMPORTANT FORMULAS
Bohr's angular momentum quantisation:
mvr = nh/2π
Radius of the nth orbit of hydrogen:
rₙ = n²a₀
Bohr radius:
a₀ = 5.3 × 10⁻¹¹ m
Radius in terms of fundamental constants:
rₙ = ε₀n²h²/(πme²)
Energy of the nth orbit:
Eₙ = −13.6/n² eV
Kinetic energy of electron:
K = −E
Potential energy of electron:
U = 2E
Speed of electron in nth orbit:
vₙ = 2.18 × 10⁶/n m/s
Orbital period:
T = 2πr/v
Energy of emitted or absorbed photon:
hν = |Eᵢ − E𝒻|
Frequency of emitted radiation:
ν = ΔE/h
Wavelength of emitted radiation:
λ = hc/ΔE
Photon energy in electron volts:
E = 1240/λ nm
Ionisation energy of hydrogen in ground state:
13.6 eV
Conversion of electron volt into joule:
1 eV = 1.6 × 10⁻¹⁹ J
6. IMPORTANT CONSTANTS
| Constant | Value |
|---|---|
| Planck's constant (h) | 6.63 × 10⁻³⁴ J s |
| Speed of light (c) | 3 × 10⁸ m/s |
| Electron charge (e) | 1.6 × 10⁻¹⁹ C |
| Electron mass (m) | 9.1 × 10⁻³¹ kg |
| Bohr radius (a₀) | 5.3 × 10⁻¹¹ m |
| Permittivity of free space (ε₀) | 8.85 × 10⁻¹² C²/(N m²) |
| Ground-state energy of hydrogen | −13.6 eV |
| Ionisation energy of hydrogen | 13.6 eV |
7. IMPORTANT SI UNITS
| Physical Quantity | SI Unit |
|---|---|
| Atomic radius | Metre (m) |
| Nuclear radius | Metre (m) |
| Energy | Joule (J) |
| Frequency | Hertz (Hz) |
| Wavelength | Metre (m) |
| Speed | Metre per second (m/s) |
| Momentum | kg m/s |
| Angular momentum | kg m²/s |
| Time period | Second (s) |
| Mass | Kilogram (kg) |
| Charge | Coulomb (C) |
8. QUICK REVISION POINTS
J. J. Thomson discovered the electron in 1897.
Thomson proposed the plum pudding model of the atom.
Rutherford performed the alpha-particle scattering experiment.
Most alpha particles passed through the gold foil without deflection.
A very small number of alpha particles were scattered through large angles.
Rutherford concluded that most of an atom is empty space.
The positive charge and most of the mass of an atom are concentrated in its nucleus.
Rutherford's model could not explain atomic stability.
Bohr proposed the concept of stationary orbits.
Electrons do not radiate energy while remaining in permitted stationary orbits.
Angular momentum is quantised according to mvr = nh/2π.
The Bohr radius of hydrogen is 5.3 × 10⁻¹¹ m.
The radius of the nth orbit is proportional to n².
The energy of the nth orbit of hydrogen is −13.6/n² eV.
The ground-state energy of hydrogen is −13.6 eV.
The ionisation energy of hydrogen in its ground state is 13.6 eV.
An electron absorbs energy when it moves to a higher energy level.
An electron emits energy when it moves to a lower energy level.
The energy of an emitted photon is equal to the difference between the two energy levels.
Atomic spectra consist of distinct spectral lines because atomic energy levels are discrete.
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