CBSE Class 12 Physics NCERT Exemplar Solutions: Chapter 12 – Atoms
NCERT Exemplar Solutions for all the questions of Class 12 Physics Chapter 12 – Atoms are available here. These questions are important for CBSE Class 12 Physics board exams and other competitive exams like NEET, JEE Main, JEE Advanced, WBJEE, UPSEE etc.
NCERT textbooks contain study material besides example and questions, whereas NCERT Exemplar book contains only exercises and are mainly used for assessment purpose. The level of questions given in Class 12 Physics NCERT Exemplar book varies from medium to very high.
Important topics of Chapter 12 – Atoms:
• Alpha-particle Scattering
• Rutherford’s Nuclear Model of Atom
• Atomic Spectra
• Bohr Model of the Hydrogen Atom
• The Line Spectra of the Hydrogen Atom
• De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation
Most of the questions given in this chapter of Class 12 Physics NCERT Exemplar are based on topics given above.
In order to solve the problems of this chapter, one needs a crystal clear understanding of the topics given above.
Types and number of questions in this chapter:
In this chapter of Class 12 Physics NCERT Exemplar, there are –
- 7 MCQ I (Multiple choice questions with single correct option)
- 6 MCQ II (Multiple choice questions with multiple correct options)
- 5 VSA (Very short answer type questions)
- 5 SA (Short answer type questions)
- 6 LA (Long answer type questions)
Few problems and their solutions from this chapter are as follows:
Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr's model, will be about
(a) 53 pm
(b) 27 pm
(c) 18 pm
(d) 13 pm
Solution 12.1: (c)
In Hydrogen-like atom, radii of the orbits (r) are inversely proportional to atomic number (Z), i.e., r ∝ 1/Z.
As the atomic number of lithium is 3; therefore, the radius of Li++ ion in its ground state, on the basis of Bohr's model, will be about 1/3 times to that of Bohr radius.
So, the radius of lithium is nearly 53/3 ≈ 18 pm.
Solution 12.2: (c)
In Hydrogen-like atom, electron revolved around the nucleus (containing proton) has centripetal acceleration. So, its frame of reference is not inertia.
The electron revolves uniformly around nucleus have certain centripetal acceleration associated with it.
When one decides to work in a frame of reference where the electron is at rest, the given expression is not true as it forms the non- inertial frame of reference.
The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because
(a) of the electrons not being subject to a central force
(b) of the electrons colliding with each other
(c) of screening effects
(d) the force between the nucleus and an electron will no longer be given by Coulomb's law
Solution 12.3: (a)
The electrostatic force of attraction between electron and nucleus is a central force which provides necessary centripetal force for the circular motion of the electron.
The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because of the electrons not being subject to a central force.
For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,
(a) because Bohr model gives incorrect values of angular momentum
(b) because only one of these would have a minimum energy
(c) angular momentum must be in the direction of spin of electron
(d) because electrons go around only in horizontal orbits
Solution 12.4: (a)
Using simple Bohr model, we get the only magnitude of the angular momentum of revolving electrons. But, angular momentum is vector quantity, so, statement (a) is not correct.
O2 molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms
(a) is not important because nuclear forces are short-ranged
(b) is as important as electrostatic force for binding the two atoms
(c) cancels the repulsive electrostatic force between the nuclei
(d) is not important because oxygen nucleus have equal number of neutrons and protons
Solution 12.5: (a)
Nuclear forces are very short range forces. So, in the molecule, nuclear force between the nuclei of the two atoms is not important.
Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is
(b) 20.40 eV
(d) 27.2 eV
When total energy associated with the two H-atoms in the ground state collide in elastically = 2 × (13.6 eV) = 27.2 eV.
Here collision is inelastic, so linear momentum is conserved, but some kinetic energy is lost in the process. If one of them goes into first excited state after the inelastic collision, then the total energy associated with the two H-atoms after collision= (13.6/22) + 13.6 = 17.0 eV.
So, maximum loss of their combined kinetic energy = 27.2 -17.0 = 10.2 eV.
A set of atoms in an excited state decays
(a) in general to any of the states with lower energy
(b) into a lower state only when excited by an external electric field
(c) all together simultaneously into a lower state
(d) to emit photons only when they collide
Solution 12.7: (a)
Option (a) is correct. A set of atoms can in excited state decays in general to any of the states with lower energy.
Link to Download NCERT Exemplar Solutions for Class 12 Physics: Chapter 12 – Atoms
Types of Question
Link to Download PDF
7 MCQ I (Multiple choice questions with single correct option)
6 MCQ II (Multiple choice questions with multiple correct options)
5 VSA (Very short answer type questions)
5 SA (Short answer type questions)
6 LA (Long answer type questions)