CBSE Important Questions Class 12 Physics Chapter 12


With class 12 physics chapter 12 important questions, students will get elaborated and authentic solutions to their doubts regarding the important chapter ATOMS.

After preparing physics class 12 chapter 12 important questions, students will be able to solve

CBSE Sample papers, CBSE Past year question papers, NCERT Books as well as EXEMPLAR Questions.

In addition, these chapter 12 class 12 physics important questions also contain important formulas, short derivations, and CBSE extra questions that can help students to test their understanding.

Chapter 12 Atoms Class 12 Physics which is considered to be an important part of the CBSE syllabus, can easily be prepared with the help of these important questions as this also contains CBSE revision notes. 



In important questions class 12 physics chapter 12 we will consider different models of atoms such as the Plum pudding model proposed by J.J. Thomson, Planetary Model of Atom proposed by Rutherford.

We will also consider important concepts such as electron orbits, spectral series, energy levels, etc., that will help students have a clear understanding of this important chapter.

Also, as this is considered to be an important chapter of the CBSE Syllabus, it should be prepared well with the help of these class 12 physics chapter 12 important questions.


By the nineteenth century, enough evidence had accumulated in favour of the atomic hypothesis of matter. The first Model of atoms was proposed by J.J Thomson in 1898, but this was not enough to explain the structure of atoms hence various other scientists, such as Rutherford, also experimented with this important concept of the atom.


  1. Name the series of Hydrogen spectrums lying in ultraviolet and visible regions.  

Ans. The two series of hydrogen spectrum lying in the ultraviolet and visible region are –

  1. Layman series in ultraviolet region
  2. Balmer series in visible region
  1. State the meaning of ionization energy. What is its value for a hydrogen atom?

Ans. Ionization energy can be referred to as the energy required to knock out an electron from an atom.

For hydrogen atoms, it is 13.6 eV

  1. What is the ratio of radii of the orbits corresponding to the first excited state and ground state in a hydrogen atom?

Ans. The radius of Bohr’s stationary orbits, r = n²h4ℼ²mKe²

Clearly, r ∝ n² and in the ground state, n=1

For 1st excited state n = 2

Therefore Ratio of radii of the orbits = = 41 = 4:1

  1. When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why can’t it be emitted as other forms of energy?

Ans. The accelerated electron produces an electric as well as the magnetic field due to which the transition of an electron from a higher energy level to a lower energy level, the difference in the energy levels appears in the form of electromagnetic radiation.

  1. Calculate the radii of the 2nd and 3rd electron orbit of a hydrogen atom. It is given that the radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10⁻¹¹m.

Ans. The radius of the innermost orbit of a hydrogen atom r₁ =  5.3 × 10⁻¹¹m.

Let r₂ be the radius of the orbit at n=2 

By relation between r₁ and r₂ 

r₂ = n²r₁ 

    = 2²×  5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m

For n=3, we can write the corresponding electron radius as:

r₃ = (n)²r₁ 

    = 9 ×   5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m

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


  1. Does a nucleus lose mass when it suffers gamma decay?

Ans. The answer is No. This is because in a gamma decay, neither the proton number nor the neutron number changes. Only the quantum numbers of nucleons change. 

  1. 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 the quantisation of the orbits of planets around the sun?

 Ans. Since the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h), we never speak of the quantization of orbits of planets around the Sun. The angular momentum of the earth in its orbit is found to be 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 found to be relatively very well.

Hence, the quantum levels for planetary motion are always considered continuous. 

  1. In accordance with Bohr’s model, find the quantum number that characterizes 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. The radius of the orbit around the sun, r = 1.5 × 10¹¹ m

Orbital speed of the Earth, v = 3 × 10⁴ m/s

Mass of the Earth m = 6.0 × 10²⁴kg

According to Bohr’s model, angular momentum is quantized and given as :

mvr = nh2ℼ


h= Planck’s constant = 6.62 × 10⁻³⁴ Js

n = Quantum number, 

Therefore, n = mvr2ℼh

                      = 2ℼ×6×10²⁴×3×10⁴×1.5×10¹¹6.62×10⁻³⁴

                      = 25.61 × 10⁷³ = 2.6 × 10⁷⁴

Hence, the quants number that characterizes the Earth’s revolution is 2.6 

  1. (a) The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain.

(b) Plot a graph showing the variation of the potential energy of a pair of nucleons as a function of their separation.


(a) When nucleons approach each other to form a nucleus, they strongly attract each other. Their potential energy decreases and becomes negative. It is this potential energy which holds the nucleons together in the nucleus. The decrease in potential energy results in a decrease in the mass of the nucleons inside the nucleus. 


  1. A Hydrogen atom initially at the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength of the photon.

Ans. For ground level, n₁ = 1

Let E1 = -13.6(n₁)²eV

           = -13.6 = -13.6 eV

The atom is excited to a higher level, n₂ = 4.

Let E₂ be the energy of this level.

Therefore, E₂ =  -13.6(n₂)² eV

                        =  -13.6 = -13.616 eV

The amount of energy absorbed by the photon is given as :

E = E₂ –  E1

= -13.61613.61

= 13.6 × 1516 × 1.6 × 10-19 = 2.04 × 10-18 J

For a photon of wavelength 𝛌, the expression of energy is written as :

E = hc𝛌


h = Planck’s constant = 6.6 × 10-34 Js

c = Speed of light = 3 × 108 m/s

Therefore,  𝛌 = hcE

                    = 6.6×10⁻³⁴ × 3 × 10⁸2.04 × 10⁻¹⁸

                    = 9.7 × 10-18m = 97 nm

  1. Does, de Broglie hypothesis have any relevance to macroscopic matter?

Ans. De Broglie relations can be applied to both microscopic and macroscopic. Taking, for example, a macro-sized 100 Kg car moving at a speed of 100m/s, will have a-

Wavelength of

𝜆 = hmv = 6.63 × 10⁻³⁴100×100 = 10-30 m

High-energy γ-radiations have wavelengths of only 10-12 m.  

Very small wavelength corresponds to high frequencies. Waves below a certain wavelength or beyond certain frequencies undergo particle-antiparticle annihilation to create mass. So, wave nature or de Broglie wavelength is not observable in the macroscopic matter.


  1. Why is the classicalRUTHERFORDmodel for an atom of an electron orbiting around the nucleus not able to explain the atomic structure?

Ans. The classical method of Rutherford could not explain the atomic structure as the electrons revolving around the nucleus are accelerated and emit energy, as the result the radius of the circular path goes on decreasing, and ultimately, electrons fall into the nucleus, which is not practical because the stability of the nucleus will get disturbed.

  1. Define the term LASER.

Ans. The acronym LASER stands for light amplification by stimulated emission of radiation. It has found applications in Physics, Chemistry, Biology, Medicine, etc. 

  1. What is the FRANK-HERTZ experiment?

Ans. The existence of discrete energy levels in an atom was directly verified in 1914 by James Frank and Gustav Hertz. 

They studied the spectrum of mercury vapour when electrons having different kinetic energies passed through the vapour. The electron energy was varied by subjecting the electrons to electric fields of varying strength.

The electrons collide with the mercury atoms and can transfer energy to the mercury  atoms.

For e.g. the difference between an occupied energy level of Hg and a higher unoccupied  level is 4.9 eV.

If an electron having an energy of 4.9 eV or more passes through mercury, an electron in a mercury atom can absorb energy from the bombarding electron and get excited to a higher level. 

The excited electron would subsequently fall back to the ground state by the emission of radiation.

By direct measurement using the formula, 𝛌 = hcE, Frank, and Hertz calculated the wavelength.

For this experimental verification of Bohr’s Basic idea of discrete energy levels in atoms and the process of photon emission, Frank and Hertz were awarded the Nobel Prize in 1925.

  1. What are the limitations of Bohr’s Atomic Model?

Ans. The limitations of Bohr’s atomic model are mentioned below –

  1. The Bohr atomic model theory considers electrons to have both a known radius and orbit that is known position and momentum at the same time which is impossible according to the Heisenberg uncertainty principle. 
  2. The Bohr atomic model theory made correct predictions for smaller-sized atoms like Hydrogen but poor spectral predictions are obtained when large atoms are considered.
  3. It failed to explain the Zeeman Effect when the spectral line is split into several components in a magnetic field.
  4. Bohr’s Atomic Model theory failed to explain the stark effect when the spectral line gets split up into fine lines in the presence of an electric field.
  1. Briefly explain Rutherford’s Alpha Scattering experiment.

Ans. To explain the structure of the atom Rutherford conducted a light scattering experiment where he placed a gold foil and bombarded the gold sheet with the alpha particles.

After the bombardment, Rutherford studied the trajectory of the alpha particles. There was a radioactive source that emitted alpha particles which are positively charged particles that were enclosed within a Lead shield in a protective manner.

The radiation then passed in a narrow beam after it passed through a slit that was made in the Lead Screen. 

A very thin section of Gold foil is placed before the Lead screen and the L.E.D. The screen was covered with Zinc Sulfide so as to give it a fluorescent nature that served as a counter-detection to the alpha particles.                           

As soon as the alpha particles strike the fluorescent screen, it is shattered into a burst of light which is known as Scintillation. As the screen was movable, it allowed Rutherford to study whether or not alpha particles get deflected by the gold foil.

  1. Define the distance of the closest approach. What will be the distance of the closest approach for an alpha particle if kinetic energy gets doubled?

Ans. The distance of the closest approach is defined as “the distance of a charged particle from  the centre of the nucleus at which the whole of the initial kinetic energy of the charged particle gets converted into the electric potential energy of the system.

Distance of the closest approach is given by 

r= 14ℼε₀ × 2Ze²K

From this expression rc 1K

Therefore, When K is doubled, rc  becomes half.

  1. Explain the spectral series of Hydrogen atoms. 

Ans. Hydrogen is the simplest atom and therefore has the simplest spectrum. In 1885 the first spectral series was observed by a Swedish school teacher Johann Jakob Balmer in the visible region of the hydrogen spectrum. 

Fig. Emission lines in the spectrum of hydrogen.

If we group the spectral lines appearing in a hydrogen atom, they are of five series.

  1. Lyman series – In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5………) to the first energy level (p=1).

The wavelength of light emitted in the Layman series lies in the ultraviolet region of the electromagnetic spectrum.

  1. Balmer series – In this series, spectral lines are obtained by the transition of electrons from any high energy level ( n= 3,4,5,6…..) to the second energy level (p=2). The wavelength emitted in this series lies in the visible region of the electromagnetic spectrum. The first line in this series (n=3 to p=2) is called H𝞪  line and the second line ( n=4 to p = 2) is called Hᵦ line.
  2. Paschen series – In this series,  the spectral lines are obtained when an electron makes a transition from any high energy level ( n=5,6,7,8……) to the third  energy level (p=3). The wavelength of light emitted in this series lies in the infrared region of the electromagnetic spectrum.
  3. Bracket series – In this series, spectral lines are  obtained when an electron makes a transition from any high energy level (n=5,6,7,8…..) to the fourth energy level( p= 4). The wavelength of light emitted in this series also lies in the infrared region of the electromagnetic spectrum. 
  1. Pfund Series – In this series, the spectral lines are obtained when an electron makes a transition from any high energy level ( n= 6,7,8,9….) to the fifth energy level (p=5). The wavelength of light emitted in this series also lies in the infrared region of the electromagnetic spectrum. 

So this is how the spectral series of hydrogen atoms is explained.

  1. Show that Bohr’s second postulate can be explained on the basis of de Broglie’s hypothesis of the wave nature of electrons.

Ans. Bohr’s 2nd postulate tells us that electrons orbit the nucleus only in those orbits for which the angular momentum is an integral multiple of nh/2π. 

By that time, de Broglie established his wave matter duality principle, which explains matter waves and how their wavelengths are inversely proportional to the mass of the body and the velocity of the body by the relation:

(wavelength) 𝜆 = h /mv

Thus,  we come to know about why massive macro objects do not show wave nature. The wavelength is too small and negligible.

 De Broglie included his wave nature principle in Bohr’s 2nd postulate. De Broglie imagined the circumference of the orbit as a string and extended it in a straight line applying Bohr’s concept and deriving the same equation of angular momentum as Bohr. 

   2πr = 𝜆 (wavelength)

2πr = nh/mv

Where 2πr   is the circumference of the orbit = the length of the string and h/mv is the wavelength equation from de Broglie and n is the number of wavelengths. 

mvr = nh/2π

L = nh/2π

Solving the equation;

 We get  Bohr’s 2nd postulate equation.

The explanation concluded by de Broglie was that wavelengths of matter waves were quantized. This means that the electrons can exist in those orbits which have a complete set of n number of wavelengths (matter wave wavelengths depend on the mass and velocity of the electron) where n is a whole number (and not an integer like 1.5 or 2.7 etc). And since each of those orbits will have a constant angular momentum, hence the phenomenon can also be explained as the electrons will orbit the nucleus in those orbits for which the angular momentum is nh/2π), where n is again a whole number.


Atomic structure determines bonding magnetic properties, electrical conductivity, melting and boiling points, the spectrum of radiation, and chemical reaction rates. Hence we can say that atomic structure is a cornerstone of chemistry.

The greatest advancement in chemistry in the last forty years is that it recognized the Pauli-Aufbau filling order of sub-shells. 

In the most simple terms, without atoms, there would not be a functioning world because atoms make up matter, and matter makes up everything in the world, with a few exceptions.

For example – Oxygen atoms are in the air and keep up alive because we need oxygen we need oxygen in order to breathe.

Hydrogen and Oxygen atoms bond together to form water.

Q1-Name the series of hydrogen spectrum which lies in the visible region of electromagnetic spectrum? 

Ans-Balmer series lies in the visible region of electromagnetic spectrum.

Q2-Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: For the heavier nucleus, the impact parameter will be greater.
Reason: The scattering of ? rays depend on the mass of the nucleus.


Assertion is true but reason is false.

Assertion and reason both are false.

Both assertion and reason are true and the reason is the correct explanation of the assertion.

Both assertion and reason are true but reason is not the correct explanation of the assertion.

Ans-The scattering occurs due to the electrostatic field of the nucleus. However, the rays never enter in the nucleus of an atom during the scattering, so the impact parameter depends on the charge of an atom but it is independent of its mass.

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FAQs (Frequently Asked Questions)

1. How do atoms become molecules?

As atoms come together to form molecules, chemical bonds bind them together. As a consequence of sharing or exchanging electrons between the atoms, these bonds form. It is only the electrons that are ever active in bonding in the outermost shell.

2. What is a simple molecule?

Water is known to be a basic molecule consisting of a few atoms. Basic molecular substances are molecules in which strong covalent bonds bind the atoms. Nevertheless, weak forces bind the molecules together so that they have high melting and boiling points.

3. What is the structure of an atom?

The structure of an atom can be described as – 

Three fundamental particles are composed of atoms: protons,  electrons, and neutrons.

Protons which are positively charged particles. Neutrons are in the nucleus (centre) of the atom and are negatively charged particles. Electron shells are considered the outermost regions of the atom, which contain electrons, and these are the negatively charged particles.

4. 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 solid at a temperature below 14 K). What results do you expect?

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.

5. What are the basic characteristics of the main subatomic particles of atoms?

The basic characteristics of the main subatomic particles of atoms are as follows –


  1. Protons are positively charged subatomic particles. The charge of a proton is 1e, which corresponds to approximately 1.602 × 10-19.
  2. The mass of a proton is approximately 1.672 × 10-24.
  3. Protons are over 1800 times heavier than electrons.
  4. The total number of protons in the atoms of an atom of an element is always equal to the atomic number of the element.



  1. The mass of a neutron is almost the same as that of a proton i.e 1.674 × 10-24  
  2. Neutrons are electrically neutral particles and carry no charge.
  3. Different isotopes of an element have the same number of protons but vary in the number of neutrons present in their respective nuclei.



  1. The charge of an electron is -1e, which approximates to -1.602 × 10-19.
  2. The mass of an electron is approximately 9.1 × 10-31.
  3. Due to the relatively negligible mass of electrons, they are ignored when calculating the mass of an atom.

6. How can the total number of neutrons in the nucleus of a given isotope be determined?

The mass number of an isotope is given by the sum of the total number of protons and neutrons in it. The atomic number describes the total number of protons in the nucleus.

Therefore, the number can be determined by subtracting the atomic number from the mass number.

7. What are Quantum numbers? Explain the Quantum numbers in brief.

The numbers which identify the state of an electron, specify the energy associated with it, and its location around the nucleus are called quantum numbers. These numbers are used to designate and distinguish electrons in atomic orbitals.

There are four quantum numbers, the principal quantum number(n)

Azimuthal quantum number(1), the magnetic quantum number (m)

and the spin quantum number(s).

8. What is a wave function?

In quantum physics, a wave function is a mathematical description of a quantum state of a particle as a function of momentum, time, position and spin.

The symbol used for a wave function is a Greek letter called psi,𝜳.

By using a wave function, the probability of finding an electron within the matter wave can be explained. This can be obtained from an imaginary number that is squared to get a real number solution resulting in the position of an electron. The concept of wave function was introduced in 1925 with the help of the Schrodinger equation.

9. Briefly explain Dalton’s Atomic theory.

The English chemist John Dalton suggested that all matter is made up of atoms, which are indivisible and indestructible. He also stated that all the atoms of an element were exactly the same, but the atoms of different elements differ in size and mass.

Chemical reactions, according to Dalton’s atomic theory, involve a rearrangement of atoms to form products. According to the postulates proposed by Dalton, the atomic structure comprised atoms, the smallest particle responsible for the chemical reactions to occur.

The following are the postulates of Dalton’s atomic theory –


  1. Every matter is made up of atoms.
  2. Atoms are indivisible.
  3. Specific elements have only one type of atom in them.
  4. Each atom has its own constant mass that varies from element to element.
  5. Atoms undergo rearrangement during a chemical reaction.
  6. Atoms can neither be created nor destroyed but can be transformed from one form to another.


Dalton’s atomic theory successfully explained the Laws of chemical reactions, namely, the law of conservation of mass, the Law of constant properties, the Law of multiple proportions and the Law of reciprocal proportions.

10. What are the limitations of Thomson’s atomic model?

The limitations of Thomson’s atomic model are –

  1. It failed to explain the stability of an atom because his model of the atom failed to explain how a positive charge holds the negatively charged electrons in an atom. Therefore, this theory failed to account for the position of the nucleus in an atom.
  2. Thomson’s model failed to explain the scattering of alpha particles by thin metal foils.
  3. No experimental evidence in its support.


Thomson’s model was not an accurate model to account for the atomic structure; it proved to be the base for the development of other atomic models.