12th Class Physics Notes 2024 All Punjab Board

12th Class Physics Notes 2024 All Punjab Board.Physics is an important subject for FSC students, so it is very important to prepare it in the best way. Exam Don’t worry about taking bundles of different Class 12 Physics notes in different formats. Here we are providing you all in one. This is the selection of our toppers and our highly qualified, capable teachers. So don’t forget to download these helpful notes. We care about you and your future and provide you with the best content.This is the selection of our toppers and our highly qualified, capable teachers. So don’t forget to download these helpful notes.12th Class Physics Notes 2024 All Punjab Board.

Physics Notes 2024 All Punjab Board

12th Class Physics.PDF
ChaptersChapter NameMedium
1ElectrostaticsEnglish Medium
2Current ElectricityEnglish Medium
3ElectromagnetismEnglish Medium
4Electromagnetic InductionEnglish Medium
5Alternating CurrentEnglish Medium
6Physics of SolidsEnglish Medium
7ElectronicsEnglish Medium
8Dawn of Modern PhysicsEnglish Medium
9Atomic SpectraEnglish Medium
10Nuclear PhysicsEnglish Medium

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Physics Notes All Chapters 12th Class

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Second Year Physics Notes PDF

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Related: Physics Book Punjab Board

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12th Class Physics Notes 2024

Rutherford’s Atomic Model

The correct description of the distribution of positive and negative charges within an atom was made in 1911 by a New Zealander when working at Manchester University in England. This was Ernest Rutherford, who was later made Lord Rutherford for his many scientific achievements. He entered into physics during that crucial period of its development when the phenomenon of natural radioactivity had just been discovered, and he was first to realize that radioactivity represents a spontaneous disintegration of heavy unstable atoms.

Rutherford realized that important information about the inner structure of atoms could be obtained by the study of collisions between on rushing α particles and the atoms of various materials forming the target.


The basic idea of the experimental arrangement used by Rutherford in his studies was explained as follows:

a speck of α – emitting radioactive material; a lead shield with a hole that allowed a narrow beam of the α − particles to pass through; a thin metal foil to deflect or scatter them; and a pivoted flourescent screen with a magnifier through which the tiny flashes of light were observed whenever an α-particle struck the screen. The whole apparatus was evacuated, so that the particles would not collide with air molecules.


Rutherford, knowing the kinetic energy of the α-particles, calculated that they would be within about 10-12cm from its centre if α – particles were to be turned back in the direction from which they came.

Because there would be a Coulomb force of attraction between the positive nucleus and the negative electrons, the two would be down together and the atom would vanish unless some provisions were made to prevent it. It was suggested that the electrons might be orbiting rapidly around the nucleus, so that the electrostatic attraction would merely provide the necessary centripetal force.


(i) Rutherford’s atomic model was unable to make any predictions about the light that an atom would emit 

(ii) More serious than this was its conflict with the accepted laws of electromagnetic theory. An electron revolving rapidly around a nucleus must have a continual centripetal acceleration, and this acceleration would cause a continuous loss of energy by radiation. Bohr calculated that this emission of radiation would cause the electrons in an atom to lose all their energy and fall into the nucleus within a hundred – millionth of a second. Since matter composed of atoms exists permanently, as far as we know, there was obviously something wrong here. Bohr’s conclusion was that the conventional classical laws of physics must be wrong, at least when applied to the motion of electrons within an atom.

Bohr’s Theory

3. Electrons may jump from one orbit to another, in which case the difference in energy between the two states of motion is radiated as a photon whose frequency is determined by the quantum rule ΔE = hf.

Bohr’s Orbits

For an electron orbiting in a hydrogen atom, the necessary centripetal force is the electrostatic attraction between the negative electron and the massive, positively-charged proton, that is the nucleus.

Thus,   ke2/r2 = mv2/r or r = ke2/mv2(1)

According to Bohr’s theory

mvr = nh/2π where, n = 1, 2, 3,….

r = nh/2πmv (2)

From (1) and (2)

v = 2πke2 /nh

structure-of-atoms  where  h = 6.63 × 10-34 J –s (16.1)

Electron energies

Kinetic energy K = 1/2mv2 = structure-of-atoms     (16.2)Potential energy U = −ke2/r= −ke2structure-of-atoms(16.3)Total energy E = K + U = − structure-of-atoms (16.4)Putting the values of k = 9 × 109 Nm2/C2e = 1.6 × 10-19 C and h = 6.63 × 10-34 Js, we get E = −1/n2 (2.18 × 10-18) J = -13.6/n2 eV (16.5)structure-of-atoms

Radiation and Energy Levels

  ΔE = hf

  ΔE = E2 − E1

Using equation (16.5)

ΔE = 2.18 × 10-18 structure-of-atoms

Applying Planck’s Law,

f = ΔE/h = 3.29 × 1015structure-of-atomsHz (16.6)

Dividing the above equation by c = 3 × 108 m/s, we get 


orstructure-of-atoms  m-1 (16.7)

where  R = Rydberg constant = 1.097 × 107 m-1 .


Successes and Limitations

Bohr showed that Planck’s quantum idea were a necessary part of the atom and its inner mechanism; he introduced the idea of quantized energy levels and explained the emission or absorption of radiation as being due to the transition of an electron from one level to another. As a model for even multielectron atoms, the Bohr picture is still useful. It leads to a good, simple, rational ordering of the electrons in larger atoms and qualitatively helps to predict a great deal about chemical behaviour and spectral details.

Bohr’s theory is unable to explain the following facts

1. The spectral lines of hydrogen atom is not a single line but a collection of several lines very close together.

2. The structure of multielectron atoms is not explained.

3. No explanation for using the principle of quantisation of angular momentum.

4. No explanation for Zeeman effect

If a substance which gives a line emission spectrum is placed in a magnetic field, the lines of the spectrum get split up into a number of closely spaced lines.

This phenomenon is known as Zeeman effect.


The atom consists of a heavy positively charged nucleus and negatively charged electrons moving around it. The electron is an elementary particle having a mass
me ≈ 9.1 × 10−31 kg and a charge −e, e being an elementary charge approximately equal to
1.60 × 10-19C.

The nuclear charge is equal to +Ze, where Z is the atomic number. The atom contains Z electrons, their total charge being −Ze. Consequently, the atom is an electrically neutral system. The size of the nucleus varies depending on Z from 10-13 cm to 10-12 cm. The size of the atom is a quantity of the order of  10-8 cm.

The energy of the atom is quantized. This means that it can assume only discrete (i.e. separated by finite gaps) values: E1, E2, E3,…, which are called the energy levels of the atom
(E1 < E2 < E3 < …). Atoms with different Z’s have different sets of energy levels.

12th Class Physics Notes 2024 All Chapter

In a normal (unexcited) state, the atom is on the lowest possible energy level. In such a state, the atom may stay for an infinitely long time. By imparting an energy to the atom, it is possible to transfer it to an excited state with an energy higher than the energy of the ground state. A transition of the atom to a higher energy level may occur as a result of absorption of a photon or as a result of a collision with another atom or a particle, say, an electron.

Excited states of the atom are unstable. The atom can stay in an excited state for about 10-8 s. After that the atom spontaneously (by its own) goes over to a lower energy level, emitting in this process a photon with an energy

Eik = hf ik (i > k),

where i is the number of the energy level in the initial state and k is the number of the level to which a spontaneous transition of the atom occurred. For example, an atom which is in an excited state with the energy E3 can return to the ground state either directly, by emitting a photon of frequency f31 = (E3 − E1)/h, or through an intermediate state with the energy E2, as a result of which two photons with frequencies f32 = (E3 − E2)/h and f21 = (E2 − E1)/h are emitted.

Important Formulae

1. Radius of nth orbit

rn = 0.53 n2/Z Å   where Z = atomic number (16.8)

2. Velocity of the electron in the nth orbit 

vn = Z/n(c/137) where c = 3 × 108 m/s (16.9)

3. Energy of the electron in the nth orbit 

En = -13.6 Z2/n2 (eV) (16.10 a)

En = -(2.18 × 10-18)Z2/n2  (J) (16.10 b)

E = K + U (16.11 a )

K = -E = -U/2 (16.11 b)

U = 2E = -2 K (16.11 c)

4. Wavelength of photon emitted for a transition from n2 to n1


where R = 1.096 × 107 m-1 (for a stationary nucleus)

If nucleus is not considered to be stationary 

R =  R/1+m/M(16.13)

where m is the mass of electron and M is the mass of nucleus.

5. Wavelength (Å) of a photon of energy E (eV) is given by 

λ = 12400/E(eV)Å (16.14)

6. Momentum of a photon of energy E 

p = E/c (16.15)

Example 16.1

A single electron orbits around a stationary nucleus of charge +Ze, where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from second Bohr orbit to the third Bohr orbit. 

(a) Find the value of Z 

(b) Find the energy required to excite the electron from n = 3 to n = 4 

(c) Find the wavelength of radiation required to remove electron from first Bohr’s Orbit to infinity. 

(d) Find the kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.


(a) Given ΔE23 = 47.2 eV 

Since ΔE = 13.6Z2 structure-of-atoms eV

∴ 47.2 = 13.6 Z2 (1/22 – 1/32) ⇒ Z = 5 

(b) To find ΔE34;   n1 = 3;     n= 4 

ΔE = 13.6 Z2 structure-of-atoms eV

∴ ΔE = 13.6 × 52 (1/32 – 1/42) = 16.53 eV 

(c) Ionization energy is the energy required to excite the electron from n = 1 to n = ∞ 

Thus, ΔE = 13.6 × 52 (1/12– 1/∞2) = 340 eV 

The respective wavelength is 

λ =  hc/ΔE = 12400/ΔE = 12400/340 = 36.47 Å

(d) K = -E = +340 eV 

U = 2E = -680 eV 

L = h/2π=  6.63 x 10-34/2π= 1.056 × 10-34 J-s

Example 16.2

Find the quantum number n corresponding to excited state of He+ ion if on transition to the ground state, the ion emits two photons in succession with wavelengths 108.5 nm and
30.4 nm. The ionization energy of H atom is 13.6 eV. 


The energy transitions for the given wavelengths are 

ΔE1 = 12400/λ1 = 12400/1085 = 11.34eV

ΔE2 = 12400/λ2 = 12400/304 = 40.79 eV

Total energy emitted ΔE = ΔE1 + ΔE2 = 52.22 eV 

Now ΔE = 13.6 Z2structure-of-atoms  eV ΔE = energy emitted 

or 52.34 = 13.6 × 22 (1/12 – 1/n2)

Thus, n = 5

Example 16.3

An isolated hydrogen atom emits a photon of 10.2 eV.

(a) Determine the momentum of photon emitted 

(b) Calculate the recoil momentum of the atom 

(c) Find the kinetic energy of the recoil atom.

[Mass of proton, mP = 1.67 × 10-27 kg]


(a) Momentum of the photon is  

p1 =  E/c= structure-of-atoms = 5.44 × 10-27 kg m/s 

(b) Applying the momentum conservation 


p2 = p1 = 5.44 × 10-27 kg m/s 

(c) K = 1/2 mv2  (v = recoil speed of atom, m = mass of hydrogen atom)

or K = structure-of-atoms

Substituting the value of the momentum of atom, we get 

K = structure-of-atoms = 8.86 × 10-27 

Example 16.4

A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV. Find the energy and wavelength of photon emitted.


Since the binding energy is always negative, therefore,

Ei = -0.85 eV 

Let ni be the initial binding state of the electron, then 

En = structure-of-atoms

or -0.85 = -13.6 structure-of-atoms

or ni = 4 

Binding energy = En = -13.6 Z2/n2  

⇒ −0.85 eV = -13.6(1)2/n22    ⇒ n2 = 4 

Let the electron now goes to an energy level n whose excitation energy is 10.2 eV. Since the excitation energy ΔE is defined with respect to ground state, therefore 

ΔE = 13.6 Z2 structure-of-atoms eV 

or 10.2 = 13.6 × 12 structure-of-atoms

thus nf = 2 

So the electron makes a transition from energy level ni = 4 to nf = 2.

Thus, the energy released is ΔE = E4 – E2

or ΔE = 13.6 [1/22 – 1/42]= 2.55 eV  

Since λ = hc/ΔE = 12400/2.25eV = 5511 Å

Example 16.5

A particle of charge equal to that of an electron, -e and mass 208 times the mass of electron (called a μ-meson) moves in a circular orbit around a nucleus of charge +3e (take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system:

(i) Calculate the radius of nth Bohr orbit 

(ii) Find the value of n, for which the radius of orbit is approximately the same as that of first Bohr orbit for the hydrogen atom;

(iii) Find the wavelength of radiation emitted when the μ-meson jumps from the third orbit to first orbit 

(Rydberg’s constant = 1.097 × 107/m)


(i) Radius of the nth Bohr orbit for hydrogen atom is 

rn = 0.53 n2/Z

Since r ∝ 1/m

∴ Radius of nth orbit for μ-meson is 

rn = -0.53n2/(208)Z

or rn = (8.5 × 10-4)n2

(ii) (8.5 × 10-4)n2 = 0.53 

∴ n2 = 623 

or n ≈ 25 

(iii) In case of hydrogen like atom, 

ΔE = E– E1 = 13.6 Z2 (1 – 1/32) = 12.08 eV 

since ΔE ∝ m 

∴ μ-meson, ΔE = (12.8)(208) = 22.6 keV

Thus λ = 12400/ΔE = 12400/22.3 x 103= 0.548 Å


Electromagnetic Induction of Class 12

The term electromagnetic induction constitutes two phenomena. The first involves a current that is induced in a conductor moving relative to magnetic field lines. The second involves the generation of an electric field associated with a time-varying magnetic field.

The observations of Michael Faraday are illustrated in the following cases:

(i)Change in Field Strength.

When a bar magnet moves relative to a loop of wire, there is an induced current in the loop (see Fig.4. 1 ) when the magnet and the loop are stationary, nothing happens.


Current Electricity of Class 12

Flow of electric charge constitutes electric current. For a given conductor AB, if ‘δQ’ charge flows through a cross-section of area A in time ‘δt’, then the electric current through the conductor AB, is given as

I = δQ/δt 

Electric Current

The current so defined above, is the average current over the period δt. The instantaneous current is given as I = dQ/dt

Direction of electric current as defined above will be taken along the direction of flow of positive charge (although in majority of conductors the charge carrier is electron which is negatively charged and hence electric current would be in a direction opposite to that of flow of electrons).

Despite the direction that we associate with electric current, it is not a vector quantity. Instead, we choose current density (j), that is current flowing through unit area of the cross−section, as a vector quantity.

Unit of Electric Current

The SI unit of electric current is ampere. It is denoted by A.

1 ampere (A)  = 1 coulomb / second

Current Density (J)

To describe the flow of charge through a cross section of the conductor at a particular point, we use the term current density J. The direction of flow of positive charge is same as that of the direction of the electric field E, and the direction of flow of –ve charge is opposite to the direction of the electric field. The direction of current density J is same as that of the electric field.

For each element of cross−section, the magnitude of current density J is equal to the current per unit area through that element.

The total current through the conductor is given as

i = Electric Current . . . (i)

where dA is the area vector of the element, perpendicular to it

If the current is uniform across the surface and parallel to dA, then current density J is also uniform and parallel to area vector dA. Hence, from equation (i)

i = Electric Current = JA

or J = i/A

where A is the total cross−sectional area of the surface. S.I. unit for current density is the ampere per square meter (A/m2).

Illustration 1.

 How many electrons pass through a wire in 1 minute if the current passing through the wire is 200 mA ?

Solution: I = q/t = ne/t

or, n = It/e

or, n = 200 x 10-3 x 60/1.6 x 10-19 = 7.5 × 1019.

Illustration 2. One billion electrons pass through a conductor AB from end A to end B in 1 ms. What is the direction and magnitude of current?

Solution : i =  ne/t = (109)(1.6 x 10-19)/10-3 = 1.6 × 10-7 Amp.

⇒ i = 0.16 μ A. The current flows from B to A.

Illustration 3. A particle having charge q coulomb describes a circular orbit. If radius of the orbit is R and frequency of the orbital motion of particle is f, then find the current in the orbit.Solution: Through any section of the orbit, the charge passes f times in one second. Therefore, through that section total charge passing in one second is fq. By definition i = fq.

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