CBSE Important Questions Class 12 Physics Chapter 12

IMPORTANT  QUESTIONS FOR CBSE CLASS 12 PHYSICS CHAPTER 12- ATOMS

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. 

CBSE CLASS 12 PHYSICS CHAPTER- 12 IMPORTANT QUESTIONS

STUDY IMPORTANT QUESTIONS FOR CLASS 12 PHYSICS CHAPTER 12 – ATOMS

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.

INTRODUCTION TO ATOMS 

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.

IMPORTANT QUESTIONS FOR CLASS 12 PHYSICS CHAPTER 12

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 = 2²1² = 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.

5. 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

respectively.

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

Where, 

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.

Ans. 

(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. 

(b)

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

           = -13.61² = -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

                        =  -13.64² = -13.616 eV

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

E = E₂ –  E1

= -13.616 – –13.61

= 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

Where, 

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.

ATOMS AND NUCLEI CLASS 12 IMPORTANT QUESTIONS 

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 

rc  = 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.

CONCLUSION

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.