Chapter 12 Atoms

NCERT Solutions Class 12 Physics Chapter 12 - Atoms

Atoms are the twelfth chapter in the Physics Course within Class 12. The chapter will provide the basis for learning the rest of the modern physics course. The chapter discusses the internal structure of atoms, shifting from the traditional "solid sphere" model of atoms to a quantum perspective on atomic structure. It begins with the first important work on alpha-particle scattering that allowed us to discover the nucleus of the atom and proceeds to examine both the success and failures of Rutherford's and Bohr's atomic models. The study of atomic structure is fundamental to understanding the nature of spectral lines and the stability of matter. For this reason, this chapter will also play an important role in CBSE Board Exams, Joint Entrance Examinations (JEE), and National Eligibility Assessment Test (NEET) by providing both key historical experiments and precise mathematical descriptions of energy levels.

NCERT Solutions Of Class 12 Physics Chapter 12 (Atoms) to visualize the subatomic world of subatomic particles. It shows the journey from the trajectory of alpha particles through to the quantization of electron orbits, in a simple and orderly manner, and explains it in an enjoyable and informative manner.

1.0Class 12 Physics Chapter 12 Atoms: Key Concepts

Class 12 Physics Chapter 12, Atoms, explains the structure of an atom, the development of atomic models, and the quantisation of energy. The chapter mainly focuses on Rutherford’s atomic model, Bohr’s model of the hydrogen atom, energy levels, and the origin of hydrogen spectral lines. The main concepts covered in this chapter include:

Alpha-Particle Scattering and Rutherford’s Nuclear Model: Explains the Geiger-Marsden experiment, alpha-particle scattering, the distance of closest approach, and the impact parameter. It also describes how Rutherford’s experiment led to the nuclear model of the atom.
Bohr’s Model of the Hydrogen Atom: Covers Bohr’s postulates, including stationary orbits, quantisation of angular momentum, and the emission or absorption of energy when an electron moves between energy levels.
Energy Levels and Electron Orbits: Derives the expressions for the radius, velocity, kinetic energy, potential energy, and total energy of an electron in the $n^{t} h$ orbit of a hydrogen atom.
Hydrogen Atom and Line Spectra: Explains how spectral lines are produced when electrons move between different energy levels. The chapter covers the Lyman, Balmer, Paschen, Brackett, and Pfund series of the hydrogen spectrum.
de Broglie’s Explanation of Bohr’s Second Postulate: Connects the wave nature of matter with Bohr’s quantisation condition. The concept of standing matter waves explains why only certain electron orbits are allowed.
Limitations of Atomic Models: Discusses the limitations of Rutherford’s and Bohr’s atomic models, including their inability to fully explain the spectra and behaviour of atoms with more than one electron.

2.0NCERT Solutions Class 12 Physics Chapter 12 Atoms : Detailed Solutions

SOLVED EXAMPLES

In the Rutherford's nuclear model of the atom, the nucleus (radius about $1 0^{- 15} m$ ) is analogous to the sun about which the electron move in orbit (radius $\approx 1 0^{- 10} m$ ) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the Sun than actually it is? The radius of earth's orbit is about $1.5 \times 1 0^{11} m$ . The radius of Sun is taken as $7 \times 1 0^{8} m$ . Sol. The ratio of the radius of electron's orbit to the radius of nucleus is $(1 0^{- 10} m) / (1 0^{- 15} m) = 1 0^{5}$ , that is, the radius of the electron's orbit is $1 0^{5}$ times larger than the radius of nucleus. If the radius of the earth's orbit around the Sun were $1 0^{5}$ times larger than the radius of the Sun, the radius of the earth's orbit would be $1 0^{5} \times 7 \times 1 0^{8} m = 7 \times 1 0^{13} m$ . This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the Sun. It implies that an atom contains a much greater fraction of empty space than our solar system does.
In a Geiger-Marsden experiment, what is the distance of closest approach to nucleus of a 7.7 MeV $α$ -particle before it comes momentarily to rest and reverses its direction? Sol. Let d be the centre-to-centre distance between the $α$ -particle and the gold nucleus when the $α$ -particle is at its stopping point. Then we can write the conservation of energy $E_{i} = E_{f}$ as

$K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Ze )}{d} = \frac{2 Ze ^{2}}{4 π ε _{0} d}$

Thus the distance of closest approach d is given by

$d = \frac{2 Ze ^{2}}{4 π ε _{0} K}$

The maximum kinetic energy found in $α$ -particles of natural origin is 7.7 MeV or $1.2 \times 1 0^{- 12} J$ . Since $1/4 π ε_{0} = 9.0 \times 1 0^{9} Nm^{2} / C^{2}$ . Therefore with $e = 1.6 \times 1 0^{- 19} C$ , We have

$d = \frac{( 2 ) ( 9.0 \times 1 0 ^{9} ) ( 1.6 \times 1 0 ^{- 19} ) ^{2} Z}{1.2 \times 1 0 ^{- 12}} d = 3.84 \times 1 0^{- 16} Z m$

The atomic number of foil material gold is $Z = 79$ , so that

$d (Au) = 3.0 \times 1 0^{- 14} m d = 30 fm . {1 fm (i.e. fermi) = 1 0^{- 15} m}$

The radius of gold nucleus is, therefore, less than $3.0 \times 1 0^{- 14} m$ . This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the $α$ -particle. Thus, the $α$ -particle reverses its motion without even actually touching the gold nucleus.

It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom. Sol. Total energy of the electron in hydrogen atom is $- 13.6 eV = - 13.6 \times 1.6 \times 1 0^{- 19} J$

$E = - 2.2 \times 1 0^{- 18} J .$

Thus We have,

$E = - \frac{e ^{2}}{8 π ε _{0} r} = - 2.2 \times 1 0^{- 18} J$

This gives the orbital radius

$r = - \frac{e ^{2}}{8 π ε _{0} E} r = - \frac{( 9 \times 1 0 ^{9} Nm ^{2} / C ^{2} ) ( 1.6 \times 1 0 ^{- 19} C ) ^{2}}{( 2 ) ( - 2.2 \times 1 0 ^{- 18} J )} r = 5.3 \times 1 0^{- 11} m$

The velocity of the revolving electron can be computed from

$\frac{mv ^{2}}{r} = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}} with m = 9.1 \times 1 0^{- 31} kg . v = \frac{e}{4 π ε _{0} mr} = 2.2 \times 1 0^{6} m / s$

According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom. Sol. We know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius $5.3 \times 1 0^{- 11} m$ is $2.2 \times 1 0^{- 6} m / s$ . Thus, the frequency of the electron moving around the proton is

$v = \frac{v}{2 π r} = \frac{2.2 \times 1 0 ^{6} ms ^{- 1}}{2 π ( 5.3 \times 1 0 ^{- 11} m )} \approx 6.6 \times 1 0^{15} Hz$

According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is $6.6 \times 1 0^{15} Hz$ .

EXERCISE QUESTIONS WITH SOLUTIONS

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)

Sol. (a) No different from

(b) Thomson's model, Rutherford's model (c) Rutherford's model (d) Thomson's model, Rutherford's model (e) Both the models

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? Sol. Hydrogen nuclei (or protons) are much lighter than $α$ -particle are not scattered by solid hydrogen. They pass through solid hydrogen almost undeflected from their paths.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level? Sol. Frequency $v = \frac{Δ E}{h}$

$= \frac{2.3 \times 1.6 \times 1 0 ^{- 19}}{6.62 \times 1 0 ^{- 34}} sec^{- 1} = 5.6 \times 1 0^{14} Hz$

The ground state energy of hydrogen atom is -13.6eV. What are the kinetic and potential energies of the electron in this state? Sol. Given total energy in ground state $= - 13.6 eV$ Now, (a) Kinetic energy for ground state

$= - (total energy in ground state) = 13.6 eV$

(b) Potential energy of electron in ground state

$= 2 (T.E. in ground state) = - 27.2 eV$

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 photon. Sol.

$\frac{hc}{λ} + E_{1} = E_{4} (∵ E_{n} = \frac{- 13.6}{n ^{2}} eV) \frac{hc}{λ} = E_{4} - E_{1} = (\frac{- 13.6}{4 ^{2}} + \frac{13.6}{1 ^{2}}) eV \frac{12400 eV - A ˚}{λ} = 12.75 eV λ = \frac{12400 eV - A ˚}{12.75 eV} ≃ 972.5 \overset{˚}{A} v = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{9.7 \times 1 0 ^{- 8}} ≃ 3.1 \times 1 0^{15} Hz$

(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. Sol. (a) Speed of electron in the $n^{th}$ orbit

$v_{n} = \frac{1}{n} \frac{e ^{2}}{4 π ϵ _{0}} \times \frac{2 π}{h} v_{n} = \frac{21.871 \times 1 0 ^{5}}{n} m / s$

for

$n v_{1} = 1, = \frac{21.871}{1} \times 1 0^{5} = 2.187 \times 1 0^{6} m / s$

for

$n v_{2} = 2 = \frac{21.871}{2} \times 1 0^{5} = 1.093 \times 1 0^{6} m / s$

for

$n = 3 v_{3} = \frac{21.871}{3} \times 1 0^{5} = 7.29 \times 1 0^{5} m / s$

(b) Time period in $n^{th}$ orbit is

$T_{n} = \frac{2 π r _{n}}{v _{n}} = \frac{n ^{3} h ^{3}}{4 π ^{2} k ^{2} me ^{4}}$

by putting values

$= (1.52 \times 1 0^{- 16}) n^{3} s$

for

$n = 1, T_{1} = 1.52 \times 1 0^{- 16} s$

for

$n T_{2} = 2, = 1.52 \times 1 0^{- 16} \times (2)^{3} = 1.216 \times 1 0^{- 15} s$

for

$n T_{3} = 3, = 1.52 \times 1 0^{- 16} \times (3)^{3} = 4.104 \times 1 0^{- 15} s$

The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 1 0^{- 11} m$ . What are the radii of the $n = 2$ and $n = 3$ orbits? Sol. Given radius of the innermost orbit of electron in hydrogen atom is :

$r_{1} = 5.3 \times 1 0^{- 11} m$

$∵ r_{n} = n^{2} . r_{1}$ Now for

$n r_{2} = 2, r_{2} = (2)^{2} r_{1} = 4 \times 5.3 \times 1 0^{- 11} m = 2.12 \times 1 0^{- 10} m$

for $n = 3$ ,

$r_{3} = (3)^{2} r_{1} = 9 \times 5.3 \times 1 0^{- 11} m r_{3} = 4.77 \times 1 0^{- 10} m$

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? Sol. Here,

$Δ E = 12.75 eV$

Energy of an electron in $n^{th}$ orbit of hydrogen atom is

$E_{n} = \frac{13.6}{n ^{2}} eV$

In ground state, $n = 1$

$E_{1} = - \frac{13.6}{1 ^{2}} = - 13.6 eV$

Energy of an electron in the excited state after absorbing a photon of 12.75 eV energy becomes

$∴ n^{2} or n E_{n} = - 13.6 + 12.75 = - 0.85 eV = - \frac{13.6}{E _{n}} = - \frac{13.6}{- 0.85} = 16 = 4$

Thus the electron gets excited to $n = 4$ state. $∴$ Total number of wavelength in emission spectrum.

$= \frac{n ( n - 1 )}{2} = \frac{4 \times 3}{2} = 6$

The possible emission lines are shown in fig.

Emitted wavelength,

$λ_{if} λ_{43} = \frac{hc}{E _{i} - E _{f}} = \frac{6.6 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{( E _{i} - E _{f} )} = \frac{19.8 \times 1 0 ^{- 26}}{E _{i} - E _{f}} m = \frac{19.8 \times 1 0 ^{- 26}}{( - 0.85 + 1.51 ) \times 1.6 \times 1 0 ^{- 19}} = \frac{19.8 \times 1 0 ^{- 7}}{0.66 \times 1.6} = 18.75 \times 1 0^{- 7} m = 18750 \overset{˚}{A}$

$λ_{42} = \frac{19.8 \times 1 0 ^{- 26}}{( - 0.85 + 3.4 ) \times 1.6 \times 1 0 ^{- 19}} = \frac{19.8 \times 1 0 ^{- 7}}{2.55 \times 1.6} = 4.8529 \times 1 0^{- 7} m = 4852.9 \overset{˚}{A}$

$λ_{41} = \frac{19.8 \times 1 0 ^{- 26}}{( - 0.85 + 13.6 ) \times 1.6 \times 1 0 ^{- 19}} = \frac{19.8 \times 1 0 ^{- 7}}{12.75 \times 1.6} = 0.9706 \times 1 0^{- 7} m = 970.6 \overset{˚}{A}$

$λ_{31} = \frac{19.8 \times 1 0 ^{- 26}}{( - 1.51 + 13.6 ) \times 1.6 \times 1 0 ^{- 19}} = \frac{19.8 \times 1 0 ^{- 7}}{12.09 \times 1.6} = 1.0236 \times 1 0^{- 7} m = 1023.6 \overset{˚}{A}$

$λ_{21} = \frac{19.8 \times 1 0 ^{- 26}}{( 1.2132 \times 1 0 ^{- 7} ) \times 1.6 \times 1 0 ^{- 19}} = \frac{19.8 \times 1 0 ^{- 7}}{10.2 \times 1.6} = 1.2132 \times 1 0^{- 7} m = 1213.2 \overset{˚}{A}$

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 $1.5 \times 1 0^{11} m$ with orbital speed $3 \times 1 0^{4} m / s$ . (Mass of earth $= 6.0 \times 1 0^{24} kg$ ) Sol. From Bohr's second postulate

$mvr = \frac{nh}{2 π}; n = \frac{mvr 2 π}{h} n = \frac{6 \times 1 0 ^{24} \times 3 \times 1 0 ^{4} \times 1.5 \times 1 0 ^{11} \times 2 \times 3.14}{6.62 \times 1 0 ^{- 34}} n ≃ 2.6 \times 1 0^{74}$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Find NCERT Solutions for Class 12 Physics chapter-wise, with exercise answers, formulas, and solved questions to help students study each topic and practise Physics.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 12

Step-by-Step Energy Calculations: The solutions explain how to calculate the energy of an electron in the nth orbit. The derivation also shows why the total energy of an electron in a hydrogen atom is negative, which indicates that the electron is in a bound state.
Clear Explanation of Hydrogen Spectral Series: The solutions explain the Lyman, Balmer, Paschen, Brackett, and Pfund series with clear diagrams and information about their regions of the electromagnetic spectrum. This helps students understand which series falls in the ultraviolet, visible, or infrared region.
Easy Understanding of Impact Parameter: The relationship between the impact parameter and the scattering angle is explained with simple examples. This helps students understand alpha-particle scattering and solve related questions more easily.
Step-by-Step Numerical Solutions: Numerical problems based on the Rydberg formula, energy levels, and electronic transitions are solved step by step. The calculations use simple algebra and show how to reach the final answer.
Clear Understanding of Bohr’s Model: The solutions explain why electrons in Bohr’s model remain in allowed stationary orbits instead of continuously losing energy and falling into the nucleus. This also helps students understand the limitations of Rutherford’s atomic model.

Frequently Asked Questions

NCERT Solutions Class 12 Physics Chapter 12 - Atoms

1.0Class 12 Physics Chapter 12 Atoms: Key Concepts

Alpha-Particle Scattering and Rutherford’s Nuclear Model: Explains the Geiger-Marsden experiment, alpha-particle scattering, the distance of closest approach, and the impact parameter. It also describes how Rutherford’s experiment led to the nuclear model of the atom.
Bohr’s Model of the Hydrogen Atom: Covers Bohr’s postulates, including stationary orbits, quantisation of angular momentum, and the emission or absorption of energy when an electron moves between energy levels.
Energy Levels and Electron Orbits: Derives the expressions for the radius, velocity, kinetic energy, potential energy, and total energy of an electron in the $n^{t} h$ orbit of a hydrogen atom.
Hydrogen Atom and Line Spectra: Explains how spectral lines are produced when electrons move between different energy levels. The chapter covers the Lyman, Balmer, Paschen, Brackett, and Pfund series of the hydrogen spectrum.
de Broglie’s Explanation of Bohr’s Second Postulate: Connects the wave nature of matter with Bohr’s quantisation condition. The concept of standing matter waves explains why only certain electron orbits are allowed.
Limitations of Atomic Models: Discusses the limitations of Rutherford’s and Bohr’s atomic models, including their inability to fully explain the spectra and behaviour of atoms with more than one electron.

2.0NCERT Solutions Class 12 Physics Chapter 12 Atoms : Detailed Solutions

SOLVED EXAMPLES

In the Rutherford's nuclear model of the atom, the nucleus (radius about $1 0^{- 15} m$ ) is analogous to the sun about which the electron move in orbit (radius $\approx 1 0^{- 10} m$ ) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the Sun than actually it is? The radius of earth's orbit is about $1.5 \times 1 0^{11} m$ . The radius of Sun is taken as $7 \times 1 0^{8} m$ . Sol. The ratio of the radius of electron's orbit to the radius of nucleus is $(1 0^{- 10} m) / (1 0^{- 15} m) = 1 0^{5}$ , that is, the radius of the electron's orbit is $1 0^{5}$ times larger than the radius of nucleus. If the radius of the earth's orbit around the Sun were $1 0^{5}$ times larger than the radius of the Sun, the radius of the earth's orbit would be $1 0^{5} \times 7 \times 1 0^{8} m = 7 \times 1 0^{13} m$ . This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the Sun. It implies that an atom contains a much greater fraction of empty space than our solar system does.
In a Geiger-Marsden experiment, what is the distance of closest approach to nucleus of a 7.7 MeV $α$ -particle before it comes momentarily to rest and reverses its direction? Sol. Let d be the centre-to-centre distance between the $α$ -particle and the gold nucleus when the $α$ -particle is at its stopping point. Then we can write the conservation of energy $E_{i} = E_{f}$ as

$K = \frac{1}{4 π ε _{0}} \frac{( 2 e ) ( Ze )}{d} = \frac{2 Ze ^{2}}{4 π ε _{0} d}$

Thus the distance of closest approach d is given by

$d = \frac{2 Ze ^{2}}{4 π ε _{0} K}$

$d = \frac{( 2 ) ( 9.0 \times 1 0 ^{9} ) ( 1.6 \times 1 0 ^{- 19} ) ^{2} Z}{1.2 \times 1 0 ^{- 12}} d = 3.84 \times 1 0^{- 16} Z m$

The atomic number of foil material gold is $Z = 79$ , so that

$d (Au) = 3.0 \times 1 0^{- 14} m d = 30 fm . {1 fm (i.e. fermi) = 1 0^{- 15} m}$

It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom. Sol. Total energy of the electron in hydrogen atom is $- 13.6 eV = - 13.6 \times 1.6 \times 1 0^{- 19} J$

$E = - 2.2 \times 1 0^{- 18} J .$

Thus We have,

$E = - \frac{e ^{2}}{8 π ε _{0} r} = - 2.2 \times 1 0^{- 18} J$

This gives the orbital radius

$r = - \frac{e ^{2}}{8 π ε _{0} E} r = - \frac{( 9 \times 1 0 ^{9} Nm ^{2} / C ^{2} ) ( 1.6 \times 1 0 ^{- 19} C ) ^{2}}{( 2 ) ( - 2.2 \times 1 0 ^{- 18} J )} r = 5.3 \times 1 0^{- 11} m$

The velocity of the revolving electron can be computed from

$\frac{mv ^{2}}{r} = \frac{1}{4 π ε _{0}} \frac{e ^{2}}{r ^{2}} with m = 9.1 \times 1 0^{- 31} kg . v = \frac{e}{4 π ε _{0} mr} = 2.2 \times 1 0^{6} m / s$

According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom. Sol. We know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius $5.3 \times 1 0^{- 11} m$ is $2.2 \times 1 0^{- 6} m / s$ . Thus, the frequency of the electron moving around the proton is

$v = \frac{v}{2 π r} = \frac{2.2 \times 1 0 ^{6} ms ^{- 1}}{2 π ( 5.3 \times 1 0 ^{- 11} m )} \approx 6.6 \times 1 0^{15} Hz$

EXERCISE QUESTIONS WITH SOLUTIONS

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)

Sol. (a) No different from

(b) Thomson's model, Rutherford's model (c) Rutherford's model (d) Thomson's model, Rutherford's model (e) Both the models

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? Sol. Hydrogen nuclei (or protons) are much lighter than $α$ -particle are not scattered by solid hydrogen. They pass through solid hydrogen almost undeflected from their paths.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level? Sol. Frequency $v = \frac{Δ E}{h}$

$= \frac{2.3 \times 1.6 \times 1 0 ^{- 19}}{6.62 \times 1 0 ^{- 34}} sec^{- 1} = 5.6 \times 1 0^{14} Hz$

The ground state energy of hydrogen atom is -13.6eV. What are the kinetic and potential energies of the electron in this state? Sol. Given total energy in ground state $= - 13.6 eV$ Now, (a) Kinetic energy for ground state

$= - (total energy in ground state) = 13.6 eV$

(b) Potential energy of electron in ground state

$= 2 (T.E. in ground state) = - 27.2 eV$

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 photon. Sol.

(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. Sol. (a) Speed of electron in the $n^{th}$ orbit

$v_{n} = \frac{1}{n} \frac{e ^{2}}{4 π ϵ _{0}} \times \frac{2 π}{h} v_{n} = \frac{21.871 \times 1 0 ^{5}}{n} m / s$

for

$n v_{1} = 1, = \frac{21.871}{1} \times 1 0^{5} = 2.187 \times 1 0^{6} m / s$

for

$n v_{2} = 2 = \frac{21.871}{2} \times 1 0^{5} = 1.093 \times 1 0^{6} m / s$

for

$n = 3 v_{3} = \frac{21.871}{3} \times 1 0^{5} = 7.29 \times 1 0^{5} m / s$

(b) Time period in $n^{th}$ orbit is

$T_{n} = \frac{2 π r _{n}}{v _{n}} = \frac{n ^{3} h ^{3}}{4 π ^{2} k ^{2} me ^{4}}$

by putting values

$= (1.52 \times 1 0^{- 16}) n^{3} s$

for

$n = 1, T_{1} = 1.52 \times 1 0^{- 16} s$

for

$n T_{2} = 2, = 1.52 \times 1 0^{- 16} \times (2)^{3} = 1.216 \times 1 0^{- 15} s$

for

$n T_{3} = 3, = 1.52 \times 1 0^{- 16} \times (3)^{3} = 4.104 \times 1 0^{- 15} s$

The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 1 0^{- 11} m$ . What are the radii of the $n = 2$ and $n = 3$ orbits? Sol. Given radius of the innermost orbit of electron in hydrogen atom is :

$r_{1} = 5.3 \times 1 0^{- 11} m$

$∵ r_{n} = n^{2} . r_{1}$ Now for

$n r_{2} = 2, r_{2} = (2)^{2} r_{1} = 4 \times 5.3 \times 1 0^{- 11} m = 2.12 \times 1 0^{- 10} m$

for $n = 3$ ,

$r_{3} = (3)^{2} r_{1} = 9 \times 5.3 \times 1 0^{- 11} m r_{3} = 4.77 \times 1 0^{- 10} m$

A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? Sol. Here,

$Δ E = 12.75 eV$

Energy of an electron in $n^{th}$ orbit of hydrogen atom is

$E_{n} = \frac{13.6}{n ^{2}} eV$

In ground state, $n = 1$

$E_{1} = - \frac{13.6}{1 ^{2}} = - 13.6 eV$

Energy of an electron in the excited state after absorbing a photon of 12.75 eV energy becomes

$∴ n^{2} or n E_{n} = - 13.6 + 12.75 = - 0.85 eV = - \frac{13.6}{E _{n}} = - \frac{13.6}{- 0.85} = 16 = 4$

Thus the electron gets excited to $n = 4$ state. $∴$ Total number of wavelength in emission spectrum.

$= \frac{n ( n - 1 )}{2} = \frac{4 \times 3}{2} = 6$

The possible emission lines are shown in fig.

Emitted wavelength,

$λ_{42} = \frac{19.8 \times 1 0 ^{- 26}}{( - 0.85 + 3.4 ) \times 1.6 \times 1 0 ^{- 19}} = \frac{19.8 \times 1 0 ^{- 7}}{2.55 \times 1.6} = 4.8529 \times 1 0^{- 7} m = 4852.9 \overset{˚}{A}$

$λ_{41} = \frac{19.8 \times 1 0 ^{- 26}}{( - 0.85 + 13.6 ) \times 1.6 \times 1 0 ^{- 19}} = \frac{19.8 \times 1 0 ^{- 7}}{12.75 \times 1.6} = 0.9706 \times 1 0^{- 7} m = 970.6 \overset{˚}{A}$

$λ_{31} = \frac{19.8 \times 1 0 ^{- 26}}{( - 1.51 + 13.6 ) \times 1.6 \times 1 0 ^{- 19}} = \frac{19.8 \times 1 0 ^{- 7}}{12.09 \times 1.6} = 1.0236 \times 1 0^{- 7} m = 1023.6 \overset{˚}{A}$

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 $1.5 \times 1 0^{11} m$ with orbital speed $3 \times 1 0^{4} m / s$ . (Mass of earth $= 6.0 \times 1 0^{24} kg$ ) Sol. From Bohr's second postulate

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Find NCERT Solutions for Class 12 Physics chapter-wise, with exercise answers, formulas, and solved questions to help students study each topic and practise Physics.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 12

Step-by-Step Energy Calculations: The solutions explain how to calculate the energy of an electron in the nth orbit. The derivation also shows why the total energy of an electron in a hydrogen atom is negative, which indicates that the electron is in a bound state.
Clear Explanation of Hydrogen Spectral Series: The solutions explain the Lyman, Balmer, Paschen, Brackett, and Pfund series with clear diagrams and information about their regions of the electromagnetic spectrum. This helps students understand which series falls in the ultraviolet, visible, or infrared region.
Easy Understanding of Impact Parameter: The relationship between the impact parameter and the scattering angle is explained with simple examples. This helps students understand alpha-particle scattering and solve related questions more easily.
Step-by-Step Numerical Solutions: Numerical problems based on the Rydberg formula, energy levels, and electronic transitions are solved step by step. The calculations use simple algebra and show how to reach the final answer.
Clear Understanding of Bohr’s Model: The solutions explain why electrons in Bohr’s model remain in allowed stationary orbits instead of continuously losing energy and falling into the nucleus. This also helps students understand the limitations of Rutherford’s atomic model.

Frequently Asked Questions

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Lyman, Balmer, Paschen, Brackett and Pfund — how are these hydrogen spectral series explained in NCERT Solutions Class 12 Physics Chapter 12?

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NCERT Solutions Class 12 Physics Chapter 12 - Atoms

1.0Class 12 Physics Chapter 12 Atoms: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 12 Atoms : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 12

Table of Contents

Frequently Asked Questions

Where can I get NCERT Solutions Class 12 Physics Chapter 12 Atoms?

What topics are covered in NCERT Solutions Class 12 Physics Chapter 12 Atoms?

What did Rutherford’s alpha-particle scattering experiment reveal according to NCERT Solutions Class 12 Physics Chapter 12?

How does NCERT Solutions Class 12 Physics Chapter 12 describe the allowed orbits in Bohr’s model?

Why does a hydrogen atom produce different spectral lines in NCERT Solutions Class 12 Physics Chapter 12?

Lyman, Balmer, Paschen, Brackett and Pfund — how are these hydrogen spectral series explained in NCERT Solutions Class 12 Physics Chapter 12?

What is the connection between de Broglie’s matter waves and Bohr’s quantisation condition in NCERT Solutions Class 12 Physics Chapter 12?

Can NCERT Solutions Class 12 Physics Chapter 12 help with JEE and NEET preparation?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 12 - Atoms

1.0Class 12 Physics Chapter 12 Atoms: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 12 Atoms : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 12

Table of Contents