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NCERT Solutions Class 12 Physics Chapter 12 - Atoms

NCERT QUESTIONS:
(Atoms)
Q12.1: Choose the correct alternative from the clues given at the end of the 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.)
Ans: (a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.
(b) In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the electrons always experience a net force.
(c) A classical atom based on Rutherford’s model is doomed to collapse.
(d) An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.
(e) The positively charged part of the atom possesses most of the mass in both the models.

Q12.2: 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? 
Ans: In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 × 10−27 kg) is less than the mass of incident α−particles (6.64 × 10−27 kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α−particles would not bounce back if solid hydrogen is used in the α-particle scattering experiment.

Q12.3: 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?
Ans: Separation of two energy levels in an atom,
E = 2.3 eV
= 2.3 × 1.6 × 10−19
= 3.68 × 10−19 J
Let ν be the frequency of radiation emitted when the atom transits from the upper level to the lower level.
We have the relation for energy as:
E = hv
Where,
h = Planck’s constant
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the frequency of the radiation is 5.6 × 1014 Hz.

Q12.4: The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?
Ans: Ground state energy of hydrogen atom, E = − 13.6 eV
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.
Kinetic energy = − E = − (− 13.6) = 13.6 eV
Potential energy is equal to the negative of two times of kinetic energy.
Potential energy = − 2 × (13.6) = − 27 .2 eV

Q12.5: 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.
Ans: For ground level, n1 = 1
Let E1 be the energy of this level. It is known that E1 is related with n1 as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
The atom is excited to a higher level, n2 = 4.
Let E2 be the energy of this level.
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
The amount of energy absorbed by the photon is given as:
E = E2E1
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
For a photon of wavelengthλ, the expression of energy is written as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
And, frequency of a photon is given by the relation,
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 × 1015 Hz.

Q12.6: (a) Using the 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.
Ans: (a) Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation,
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
e = 1.6 × 10−19 C
0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
h = Planck’s constant = 6.62 × 10−34 Js
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
For level n2 = 2, we can write the relation for the corresponding orbital speed as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
And, for n3 = 3, we can write the relation for the corresponding orbital speed as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 106 m/s, 1.09 × 106 m/s, 7.27 × 105 m/s respectively.
(b) Let T1 be the orbital period of the electron when it is in level n1 = 1.
Orbital period is related to orbital speed as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
r1 = Radius of the orbit
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
h = Planck’s constant = 6.62 × 10−34 Js
e = Charge on an electron = 1.6 × 10−19 C
0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
m = Mass of an electron = 9.1 × 10−31 kg
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
For level n2 = 2, we can write the period as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
r2 = Radius of the electron in n2 = 2
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
And, for level n3 = 3, we can write the period as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
r3 = Radius of the electron in n3 = 3
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the orbital period in each of these levels is 1.52 × 10−16 s, 1.22 × 10−15 s, and 4.12 × 10−15 s respectively.

Q12.7: 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?
Ans: The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 × 10−11 m.
Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
For n = 3, we can write the corresponding electron radius as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the radii of an electron for n = 2 and n = 3 orbits are 2.12 × 10−10 m and 4.77 × 10−10 m respectively.

Q12.8: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Ans: It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 12.5 eV i.e., −1.1 eV.
Orbital energy is related to orbit level (n) as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
For n = 3,
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
Ry = Rydberg constant = 1.097 × 107 m−1
λ= Wavelength of radiation emitted by the transition of the electron
For n = 3, we can obtain λ as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.

Q12.9: In accordance with the 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 × 104m/s. (Mass of earth = 6.0 × 1024 kg.)
Ans: Radius of the orbit of the Earth around the Sun, r = 1.5 × 1011 m
Orbital speed of the Earth, ν = 3 × 104 m/s
Mass of the Earth, m = 6.0 × 1024 kg
According to Bohr’s model, angular momentum is quantized and given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
h = Planck’s constant = 6.62 × 10−34 Js
n = Quantum number
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × 1074.


Old NCERT Solutions

Q1: Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

(a) Is the average angle of deflection of α­-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil?
Ans: (a) about the same
The average angle of deflection of α­-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.
(b) much less
The probability of scattering of α-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.
(c) Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.
(d) Thomson’s model
It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of α­−particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

Q2: The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10−40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Ans: Radius of the first Bohr orbit is given by the relation,
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
0 = Permittivity of free space
h = Planck’s constant = 6.63 × 10−34 Js
me = Mass of an electron = 9.1 × 10−31 kg
e = Charge of an electron = 1.9 × 10−19 C
mp = Mass of a proton = 1.67 × 10−27 kg
r = Distance between the electron and the proton
Coulomb attraction between an electron and a proton is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Gravitational force of attraction between an electron and a proton is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
G = Gravitational constant = 6.67 × 10−11 N m2/kg2
If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:
FG = FC
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Putting the value of equation (4) in equation (1), we get:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
It is known that the universe is 156 billion light years wide or 1.5 × 1027 m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Q3: Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n−1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Ans: It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n−1).
We have the relation for energy (E1) of radiation at level n as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
v1 = Frequency of radiation at level n
h = Planck's constant
m = Mass of hydrogen atom
e = Charge on an electron
= Permittivity of free space
Now, the relation for energy (E2) of radiation at level (n − 1) is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms    ......(ii)
Where,
v2 = Frequency of radiation at level (n-1)
Energy (E) released as a result of de-excitation:
E = E2E1
= E2E1 … (iii)
Where,
ν = Frequency of radiation emitted
Putting values from equations (i) and (ii) in equation (iii), we get:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
For large n, we can write NCERT Solutions Class 12 Physics Chapter 12 - Atoms
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Classical relation of frequency of revolution of an electron is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
Velocity of the electron in the nth orbit is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms   ....(i)
And, radius of the nth orbit is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms   ....(ii)
Putting the values of equations (vi) and (vii) in equation (v), we get:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms   ....(iii)
Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

Q4: Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10−10 m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
Ans: (a) Charge on an electron, e = 1.6 × 10−19 C
Mass of an electron, me = 9.1 × 10−31 kg
Speed of light, c = 3 ×108 m/s
Let us take a quantity involving the given quantities as
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
0 = Permittivity of free space
And,
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
The numerical value of the taken quantity will be:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.
(b) Charge on an electron, e = 1.6 × 10−19 C
Mass of an electron, me = 9.1 × 10−31 kg
Planck’s constant, h = 6.63 ×10−34 Js
Let us take a quantity involving the given quantities as
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
0 = Permittivity of free space
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
And,
The numerical value of the taken quantity will be:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the value of the quantity taken is of the order of the atomic size.

Q5: The total energy of an electron in the first excited state of the hydrogen atom is about −3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?
Ans: (a) Total energy of the electron, E = −3.4 eV
Kinetic energy of the electron is equal to the negative of the total energy.
K = −E
= − (− 3.4) = 3.4 eV
Hence, the kinetic energy of the electron in the given state is 3.4 eV.
(b) Potential energy (U) of the electron is equal to the negative of twice of its kinetic energy.
U = −2 K
= − 2 × 3.4 = − 6.8 eV
Hence, the potential energy of the electron in the given state is − 6.8 eV.
(c) The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic and potential energies, total energy of the system will also change.

Q6: If Bohr’s quantization postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantization of orbits of planets around the sun?
Ans: We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). The angular momentum of the Earth in its orbit is of the order of 1070h. This leads to a very high value of quantum levels n of the order of 1070. For large values of n, successive energies and angular momenta are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.

Q7: Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ) of mass about 207me orbits around a proton].
Ans: Mass of a negatively charged muon,
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
According to Bohr’s model,
Bohr radius,
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
And, energy of a ground state electronic hydrogen atom,
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
We have the value of the first Bohr orbit,
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Let rμ be the radius of muonic hydrogen atom.
At equilibrium, we can write the relation as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the value of the first Bohr radius of a muonic hydrogen atom is
2.56 × 10−13 m.
We have,
Ee= − 13.6 eV
Take the ratio of these energies as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Hence, the ground state energy of a muonic hydrogen atom is −2.81 keV.

Q8: What is the shortest wavelength present in the Paschen series of spectral lines?
Ans: Rydberg’s formula is given as:
NCERT Solutions Class 12 Physics Chapter 12 - Atoms
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
(n1 and n2 are integers)
The shortest wavelength present in the Paschen series of the spectral lines is given for values n1 = 3 and n2 = ∞.
NCERT Solutions Class 12 Physics Chapter 12 - Atoms

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FAQs on NCERT Solutions Class 12 Physics Chapter 12 - Atoms

1. What are atoms?
Ans. Atoms are the basic building blocks of matter. They are the smallest unit of an element that retains the properties of that element.
2. How are atoms structured?
Ans. Atoms are structured with a nucleus containing protons and neutrons, surrounded by electrons in orbitals.
3. Can atoms be divided further?
Ans. Atoms can be divided further into subatomic particles such as protons, neutrons, and electrons, but they lose their chemical properties in the process.
4. How do atoms combine to form molecules?
Ans. Atoms combine through chemical bonds to form molecules. This can happen through sharing electrons (covalent bonds) or transferring electrons (ionic bonds).
5. What is the significance of studying atoms?
Ans. Studying atoms is crucial for understanding the properties of matter, chemical reactions, and various fields of science such as chemistry and physics.
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