NCERT Solutions for Class 12 Physics Chapter 12: Atoms

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NCERT Solutions for Class 12 Physics Chapter 12 Atoms are given in this article. Atoms are the smallest unit into which matter can be divided without the release of electrically charged particles. It also is the smallest unit of matter that has the characteristic properties of a chemical element. Atom is the basic building block of chemistry.

Unit 8 Atoms and Nuclei along with Unit 7 Dual Nature of Radiation and Matter has a weightage of 12 marks in the CBSE Board examinations. The Class 12 Physics Chapter 12 NCERT Solutions cover the concepts of Atomic Spectra, Magnetic Quantum Number, and energy levels.

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NCERT Solutions for Class 12 Physics Chapter 12

The NCERT solutions for Class 12 Physics Chapter 12: Atoms is given below.

NCERT Solutions

CBSE Class 12 Physics Chapter 12 – Topics Covered

Thomson’s Atomic Model: Every atom is a uniformly positive charged sphere of radius of the order of 10^-10 m, in which the entire mass is distributed uniformly and negatively charged electrons are randomly embedded randomly.

Limitations of Thomson’s Atomic Model

1. It couldn’t explain the origin of spectral series of hydrogen and other atoms.

2. It couldn’t explain the large angle scattering of α-particles.

Bohr Model of the Hydrogen Atom: An electron can revolve in certain non-radiating orbits for which the angular momentum of electron is an integer multiple of (h/2π).

$r_n = n^2({h^2 \epsilon_o \over \pi mZe^2})$

Hydrogen Spectrum Series: Each element emits a radiation spectrum, which is characteristic of the element itself. The spectrum is known as the line spectrum.

Hydrogen spectrum consists of 5 series – Lyman Series, Balmer Series, Paschen Series, Brackett Series, and Pfund Series.

Wave Model is based on wave mechanics. Quantum numbers are the numbers that are required to completely specify the state of the electrons.

In the presence of a strong magnetic field, the 4 quantum number are Principle quantum number, Orbital angular momentum quantum number, Magnetic quantum number, and Magnetic spin angular momentum quantum number.

Also Read:

Chapter 12 Related Articles
Structure of atom	Atomic Number	Mass Number
Subatomic Particles of Atom	Atomic Radii	Atomic Energy Level
NCERT Solution for class 12 Physics	Dalton’s Atomic Theory	Atomic Orbitals
Thomson's Atomic Model	Alpha-Particle scattering and Rutherford's Nuclear Model of Atom	Avogadro's Number
Particle Physics	Laser Light	Alpha Particle Mass

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CBSE CLASS XII Related Questions

1.
A wire of length $L$ is bent round into (i) a square coil having $N$ turns and (ii) a circular coil having $N$ turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.
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2.
A current of $4.0\,\text{A}$ flows through a wire of length $1\,\text{m}$ and cross-sectional area $1.0\,\text{mm}^2$, when a potential difference of $2\,\text{V}$ is applied across its ends. Calculate the resistivity of the material of the wire.
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3.
Consider a cylindrical conductor of length $l$ and area of cross-section $A$. Current $I$ is maintained in the conductor and electrons drift with velocity $\vec{v}_d$ ($|\vec{v}_d| = \frac{e|\vec{E}|}{m}\tau$), (where symbols have their usual meanings). Show that the conductivity $\sigma$ of the material of the conductor is given by
$$\sigma = \frac{ne^2}{m}\tau.$$
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4.
A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:
- 80 Wb
- 16 Wb
- $8 \times 10^{-2}$ Wb
- $8 \times 10^{-3}$ Wb
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5.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:
- 5 V
- 3.5 V
- 2.5 V
- Zero V
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6.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:
- $\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})$
- $\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})$
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