# NCERT Solutions for Class 12 Physics Chapter 12 Atoms

** NCERT solutions for class 12 physics chapter 12 Atoms: ** ** ** Do you know that an atom is electrically neutral, which tells that atoms contain an equal number of positive and negative charges. The solutions of NCERT class 12 physics chapter 12 atoms explain questions related to different models of the atom, their drawbacks, hydrogen spectra, and Debroglie hypothesis. CBSE NCERT solutions for class 12 physics chapter 12 atoms will help you in board exam preparation. The solutions of NCERT plays a major role in making the concepts easy and hence will help in competitive exams also. The questions are divided in to exercise and additional exercise

Some of the important formulas of the chapter atoms which will help in NCERT solutions for class 12 physics are given below.

The various series in line spectrum of atomic hydrogen are:

- According to Bohr’s postulates for stationary orbit angular momentum

Where n is the quantum number

The total energy of the electron in the stationary states of the hydrogen atom in electronvolt is given by

Another important formula in NCERT is the

De Broglie wavelength

Where ‘h’ is Planck's constant and mv is the momentum. This relation can be modified in terms of the kinetic energy of the particle. Try to derive it yourself.

** NCERT solutions for class 12 physics chapter 12 atoms exercise: **

** Q 12.1 ** Choose the correct alternative from the clues given at the end of the each statement:

** Answer: **

(a) The size of the atom in Thomson’s model is __ no different from __ the atomic size in Rutherford’s model.

(b) In the ground state of __ Thomson’s model __ electrons are in stable equilibrium, while in __ Rutherford’s model __ electrons always experience a net force.

(c) A classical atom based on __ Rutherford’s model __ is doomed to collapse.

(d) An atom has a nearly continuous mass distribution in a __ Thomson’s model __ but has a highly non-uniform mass distribution in __ Rutherford’s model __ .

(e) The positively charged part of the atom possesses most of the mass in __ both the models __ .

** Answer: **

On repeating the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil we would have different observations than Rutherford, as the alpha particles won't be scattered much because of being heavier than the nucleus of the Hydrogen atom. Therefore we would not be able to confirm the presence of almost the entire mass of the atom at its centre.

** Q 12.3 ** What is the shortest wavelength present in the Paschen series of spectral lines?

** Answer: **

The Rydberg's formula for the hydrogen atom is

Where R is Rydberg constant for the Hydrogen atom and equals to 1.1 10 ^{ 7 } m ^{ -1 }

For shortest wavelength in Paschen Series n _{ 1 } =2 and n _{ 2 } =

The shortest wavelength in Paschen Series is therefore 818 nm.

** Answer: **

Frequency of radiation consisting of photons of energy E is given by

E=2.3 eV

Plank's constant(h)=6.62 10 ^{ -34 } Js

** Answer: **

The ground state energy E=-13.6 eV.

The kinetic energy= -E=13.6 eV

Also ground state energy = Kinetic energy+Potential energy

E=K+U

U=E-K

U=-13.6-13.6

U=-27.2 eV

The kinetic and potential energies are 13.6 eV and -27.2 eV respectively.

** Answer: **

The initial energy of the electron is E _{ 1 }

E _{ 1 } =-13.6 eV

The energy of the electron when it is excited to level n=4 is E _{ 2 }

E _{ 2 } =-0.85 eV

The difference between these two energy levels is equal to the energy of the photon absorbed by the electron.

The energy of the photon E = E _{ 2 } - E _{ 1 }

E = -0.85 -(-13.6)

E = 12.75 eV

The wavelength of the photon can be calculated using relation

hc=1240 eV

The wavelength and frequency of the photon absorbed by the hydrogen atom are 98.6 nm and 3.04 10 ^{ 15 } Hz respectively.

** Answer: **

As per Bohr's model the angular momentum of electrons in each orbit is constant and a multiple of

(i)

The electrostatic force of attraction between the electron and the nucleus provides the required centripetal force for the circular motion of the electron.

(ii)

Using equation (i) and (ii) we get

v _{ 1 } =2.18 10 ^{ 6 } ms ^{ -1 }

v _{ 2 } =1.09 10 ^{ 6 } ms ^{ -1 }

v _{ 3 } =7.28 10 ^{ 5 } ms ^{ -1 }

** (b) ** calculate the orbital period in each of these levels.

** Answer: **

Orbital period (T _{ n } ) is defined as time taken by the electron to complete one revolution around the nucleus and is given by

T _{ 1 } =1.53 10 ^{ -16 } s

T _{ 2 } =1.22 10 ^{ -15 } s

T3=4.12 10 ^{ -15 } s

** Answer: **

The radius of the orbit is proportional to the square of n.

For n=2 the radius of the orbit is

For n=3 the radius of the orbit is

** Answer: **

Since the energy of the electron beam is 12.5 eV the Hydrogen atoms will get excited to all requiring energy equal to or less than 12.5 eV

E _{ 1 } =-13.6 eV

E _{ 3 } = -1.5 eV

E _{ 3 } -E _{ 1 } = 12.1 eV

E _{ 4 } = -0.85 eV

E _{ 4 } -E _{ 1 } =12.75 eV

Therefore the electron can reach maximum upto the level n=3.

During de-excitations, the electron can jump directly from n=3 to n=1 or it can first jump from n=3 to n=2 and then from n=2 to n=1

Therefore two wavelengths from the Lyman series and one from the Balmer series will be emitted

To find the wavelengths emitted we will use the Rydberg's Formula

where R is the Rydberg's constant and equals 1.097 10 ^{ 7 } m ^{ -1 }

For n _{ 1 } =1 and n _{ 2 } =3

Emitted wavelength is 102.5 nm

For n _{ 1 } =1 and n _{ 2 } =2

Emitted wavelength is 121.54 nm

For n _{ 1 } =2 and n _{ 2 } =3

Emitted wavelength is 656.3 nm

** Answer: **

As per the Bohr's model, the angular of the Earth will be quantized and will be a multiple of

n = 2.56 10 ^{ 74 }

Therefore the quantum number that characterises the earth’s revolution around the sun in an orbit of radius m with an orbital speed

is 2.56 10 ^{ 74 }

## ** NCERT solutions for class 12 physics chapter 12 atoms additional exercise: **

** Answer: **

The average angle of deflection of -particles by a thin gold foil predicted by both the models is about the same.

** Answer: **

The probability of backward scattering predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

** Answer: **

Scattering at moderate angles requires head-on collision the probability of which increases with the number of target atoms in the path of -particles which increases linearly with the thickness of the gold foil and therefore the linear dependence between the number of -particles scattered at a moderate angle and the thickness t of the gold foil.

** Answer: **

It is completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of -particles by a thin foil in Thomson's model as the deflection caused by a single collision in this model is very small.

** Answer: **

As per the bohrs model

(i)

If the proton and the electron were bound only by the gravity the gravitational force between them will provide the centripetal force required for circular motion

(ii)

From equation (i) and (ii) we can calculate that the radius of the ground state (for n=1) will be

The above value is larger in order than the diameter of the observable universe. This shows how much weak the gravitational forces of attraction as compared to electrostatic forces.

** Answer: **

Using Bohr's model we have.

Since n is very large 2n-1 can be taken as 2n and n-1 as n

The frequency of the emission caused by de-excitation from n to n-1 would be

The classical frequency of revolution of the electron in the nth orbit is given by

The above is the same as the frequency of the emission during de-excitation from n to n-1.

** Answer: **

Using dimensional analysis we can see that the quantity to be constructed and consisting of m _{ e } , e and c will also have and will be equal to

and has numerical value 3.5 10 ^{ -14 } which is much smaller than the order of atomic radii.

** Answer: **

Using dimensional analysis we can see that the quantity to be constructed and consisting of m _{ e } , e and h will also have and will be equal to

and has a numerical value of approximately 6.657 10 ^{ -10 } which is about the order of atomic radii.

** Q 12.15 ** ** (a) ** The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.

What is the kinetic energy of the electron in this state?

** Answer: **

Since we know that kinetic energy is equal to the negative of the total energy

K=-E

K=-(-3.4)

K=3.4 eV

** Q 12.15 ** ** (b) ** The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.

What is the potential energy of the electron in this state?

** Answer: **

Total Energy= Potential energy + Kinetic Energy

E=U+K

U=E-K

U=-3.4-3.4

U=-6.8 eV

** Q 12.15 ** ** (c) ** The total energy of an electron in the first excited state of the hydrogen atom is about - 3.4eV.

Which of the answers above would change if the choice of the zero of potential energy is changed?

** Answer: **

The total energy would change if the choice of the zero of potential energy is changed.

** Answer: **

We never speak of Bohr's quantization postulate while studying planetary motion or even motion of other macroscopic objects because they have angular momentum very large relative to the value of h. In fact, their angular momentum is so large as compared to the value of h that the angular momentum of the earth has a quantum number of order 10 ^{ 70 } . Therefore the angular momentum of such large objects is taken to be continuous rather than quantized.

** Answer: **

As per Bohr's quantization postulate

Similarly, like the case in a simple hydrogen atom, the electrostatic force acts centripetally

From the above relations, we can see that in Bohr's model the Radius is inversely proportional to the mass of the orbiting body and Energy is directly proportional to the mass of the orbiting body.

In case of hydrogen, atom r _{ 1 } is 5.3 10 ^{ -11 } m

Therefore in case of a muonic hydrogen atom

r _{ 1 } = 2.56 10 ^{ -13 } m

In case of the hydrogen atom, E _{ 1 } is -13.6 eV

Therefore in case of a muonic hydrogen atom

E _{ 1 } =207 (-13.6)

E _{ 1 } =2.81 keV

## **NCERT solutions for class 12 physics chapter wise **

** NCERT solutions subject wise **

## ** Importance of NCERT solutions for class 12 physics chapter 12 atoms: **

For CBSE board exam on an average 3 to 4 marks questions are asked from the chapter. The solutions of NCERT class 12 physics chapter 12 atoms will help in securing full marks in board exam for this chapter. As far as the NEET exam is considered up to 2 questions are expected from the chapter. The NCERT solutions for class 12 physics chapter 12 atoms will also help to perform better in exams like JEE Main and other competitive exams.

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Physics Chapter 12 Atoms

**Question: **How many questions from the chapter atoms comes in CBSE board exams?

**Answer: **

one or two questions are usually asked in the board exams from class 12 chapter atoms. It can be a 2 or 3 mark question

**Question: **What is the importance of the chapter for engineering and medical entrance exams?

**Answer: **

The class 12 physics chapter atoms are important for both engineering and medical exams. At least one question can be expected from atoms for both JEE Main and NEET

**Question: **What are the important topics of the class 12 NCERT chapter atoms

**Answer: **

The hydrogen spectrum, energy of atoms, Bhor model of atoms, Bhor postulates etc are important topics of the chapter

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