# 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 plays a major role in making the concepts easy and hence will help in competitive exams also. Some of the important formulas of the chapter atoms which will help in NCERT solutions for class 12 physics chapter 12 atoms are given below.

• The various series in line spectrum of atomic hydrogen are:

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

$L=\frac{nh}{2\pi}$

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

$E_n=-\frac{13.6}{n^2}eV$

• Another important formula which is used in NCERT solutions for class 12 physics chapter 12 atoms is the

De Broglie wavelength

$\lambda=\frac{h}{mv}$

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:

(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)

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

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

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

(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.)

(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.

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.

The Rydberg's formula for the hydrogen atom is

$\frac{1}{\lambda }=R\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]$

Where R is Rydberg constant for the Hydrogen atom and equals to 1.1$\times$107 m-1

For shortest wavelength in Paschen Series n1=2 and n2=$\infty$

$\frac{1}{\lambda }=1.1\times 10^{7}\left [ \frac{1}{3^{2}}-\frac{1}{\infty^{2}} \right ]$

$\lambda =8.18\times 10^{-7}\ m$

The shortest wavelength in Paschen Series is therefore 818 nm.

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

$\nu =\frac{E}{h}$

E=2.3 eV

Plank's constant(h)=6.62$\times$10-34 Js

$\nu =\frac{2.3\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}$

$\nu =5.55\times 10^{14}\ Hz$

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.

## Q 12.6 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

The initial energy of the electron is E1

$E_{1}=-\frac{13.6}{1^{2}}$

E=-13.6 eV

The energy of the electron when it is excited to level n=4 is E2

$E_{1}=-\frac{13.6}{4^{2}}$

E2=-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 $\Delta$E = E- E1

$\Delta$E = -0.85 -(-13.6)

$\Delta$E = 12.75 eV

The wavelength of the photon can be calculated using relation

$\Delta E=\frac{hc}{\lambda }$

hc=1240 eV

$\\\lambda =\frac{hc}{\Delta E}\\ \lambda=\frac{1240}{12.57}\\ \lambda=98.6\ nm$

$\\\nu =\frac{c}{\lambda }\\ \nu =\frac{3\times 10^{8}}{98.6\times 10^{-9}}\\\nu =3.04\times 10^{15}\ Hz$

The wavelength and frequency of the photon absorbed by the hydrogen atom are 98.6 nm and 3.04$\times$1015 Hz respectively.

As per Bohr's model the angular momentum of electrons in each orbit is constant and a multiple of $\frac{nh}{2\pi }$

$m_{e}v_{n}r_{n}=\frac{nh}{2\pi }$           (i)

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

$\frac{mv_{n}^{2}}{r_{n}}=\frac{e^{2}}{4\pi \varepsilon _{0}r_{n}^{2} }$       (ii)

Using equation (i) and (ii) we get

$v_{n}=\frac{e^{2}}{2nh\varepsilon _{_{0}}}$

$r_{n}=\frac{n^{2}h^{2}\varepsilon _{_{0}}}{m_{e}\pi e^{2}}$

$\\v_{1}=\frac{e^{2}}{2h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{2\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}$

v1=2.18$\times$106 ms-1

$\\v_{2}=\frac{e^{2}}{4h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{4\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}$

v2=1.09$\times$106 ms-1

$\\v_{3}=\frac{e^{2}}{6h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{6\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}$

v3=7.28$\times$105 ms-1

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

Orbital period (Tn ) is defined as time taken by the electron to complete one revolution around the nucleus and is given by

$\\T_{n}=\frac{2\pi r_{n}}{v_{n}}\\ T_{n}=\frac{4n^{3}h^{3}\varepsilon _{0}^{2}}{m_{e}e^{4}}$

$T_{1}=\frac{4\times 1^{3}\times (6.62\times 10^{-34})^{2} \times (8.85\times 10^{-12})^{2}}{9.1\times 10^{-31}\times (1.6\times 10^{-19})^{4}}$

T1=1.53$\times$10-16 s

$T_{2}=\frac{4\times 2^{3}\times (6.62\times 10^{-34})^{2} \times (8.85\times 10^{-12})^{2}}{9.1\times 10^{-31}\times (1.6\times 10^{-19})^{4}}$

T2=1.22$\times$10-15 s

$T_{3}=\frac{4\times 3^{3}\times (6.62\times 10^{-34})^{2} \times (8.85\times 10^{-12})^{2}}{9.1\times 10^{-31}\times (1.6\times 10^{-19})^{4}}$

T3=4.12$\times$10-15 s

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

For n=2 the radius of the orbit is

$\\r_{2}=r_{1}\times 2^{2}\\ =5.3\times 10^{-11}\times 4\\ =2.12\times 10^{-10}\ m$

For n=3 the radius of the orbit is

$\\r_{3}=r_{1}\times 3^{2}\\ =5.3\times 10^{-11}\times 9\\ =4.77\times 10^{-10}\ m$

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=-13.6 eV

E= -1.5 eV

E3 -E1 = 12.1 eV

E4= -0.85 eV

E4-E1=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

$\frac{1}{\lambda }=R(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$     where R is the Rydberg's constant and equals 1.097$\times$107 m-1

For n1=1 and n2=3

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{1^{2}}-\frac{1}{3^{2}})$

Emitted wavelength is 102.5 nm

For n1=1 and n2=2

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{1^{2}}-\frac{1}{2^{2}})$

Emitted wavelength is 121.54 nm

For n1=2 and n2=3

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{2^{2}}-\frac{1}{3^{2}})$

Emitted wavelength is 656.3 nm

### As per the Bohr's model, the angular of the Earth will be quantized and will be a multiple of $\frac{h}{2 \pi}$

$\\mvr=\frac{nh}{2 \pi}\\ n=\frac{2 \pi mvr}{h}\\ n=\frac{2\pi\times 6\times 10^{24}\times 3\times 10^{4}\times 1.5\times 10^{11}}{6.62\times 10^{-34}}$

n = 2.56$\times$1074

Therefore the quantum number that characterises the earth’s revolution around the sun in an orbit of radius   $1.5 \times 10^{11}m$ m with an orbital speed $3 \times 10^{4}m/s$

is 2.56$\times$1074

Is the average angle of deflection of $\alpha$-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

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

Keeping other factors fixed, it is found experimentally that for small thickness t, the number of $\alpha$-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?

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

## Q 12.11 (d)  Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.

In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of $\alpha$-particles by a thin foil?

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

As per the bohrs model

$m_{e}v_{n}r_{n}=\frac{nh}{2\pi}$                        (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

$\frac{m_{e}v_{n}^{2}}{r_{n}}=\frac{Gm_{e}m_{p}}{r_{n}^{2}}$                   (ii)

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

$\\r_{1}=\frac{h^{2}}{4\pi Gm_{p}m_{e}^{2}}\\ r_{1}=\frac{(6.62\times 10^{-34})^{2}}{4\pi \times 6.67\times 10^{-11}\times 1.67\times 10^{-27}\times (9.1\times 10^{-31})^{2}}$

$r_{1}\approx 1.2\times 10^{29}\ m$

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.

Using Bohr's model we have.

$v_{n}=\frac{e^{2}}{2nh\varepsilon _{_{0}}}$

$r_{n}=\frac{n^{2}h^{2}\varepsilon _{_{0}}}{m_{e}\pi e^{2}}$

$\\E_{n}=\frac{1}{2}mv_{n}^{2}-\frac{e^{2}}{4\pi \epsilon _{0}r_{n}^{2}}\\ \\E_{n}=-\frac{me^{4}}{8n^{2}h^{2} \epsilon_{0}^{2} }$

$\\E_{n}-E_{n-1}=-\frac{me^{4}}{8n^{2}h^{2} \epsilon_{0}^{2} }-(-\frac{me^{4}}{8(n-1)^{2}h^{2} \epsilon_{0}^{2} })\\ \\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{1}{n^{2}}-\frac{1}{(n-1)^{2}}]\\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{-2n+1}{n^{2}(n-1)^{2}}]$

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

$\\E_{n}-E_{n-1}=-\frac{me^{4}}{8h^{2} \epsilon_{0}^{2} }[\frac{-2n}{n^{2}(n)^{2}}]\\ \\E_{n}-E_{n-1}=\frac{me^{4}}{4n^{3}h^{2} \epsilon_{0}^{2} }$

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

$\\\nu =\frac{E_{n}-E_{n-1}}{h}\\ \nu =\frac{me^{4}}{4n^{3}h^{3} \epsilon_{0}^{2} }$

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

$\nu =\frac{v_{n}}{2\pi r_{n}}$

$\nu =\frac{e^{2}}{2nh\epsilon _{_{0}}}\times \frac{m_{e}\pi e^{2}}{2\pi n^{2}h^{2}\epsilon _{_{0}}}$

$\nu =\frac{me^{4}}{4n^{3}h^{3} \epsilon_{0}^{2} }$

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

(a)    construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.

Using dimensional analysis we can see that the quantity to be constructed and consisting of me, e and c will also have $\epsilon _{0}$ and will be equal to

$\frac{e^{2}}{\epsilon _{0}m_{e}c^{2}}$ and has numerical value 3.5$\times$10-14 which is much smaller than the order of atomic radii.

(b)  You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.

Using dimensional analysis we can see that the quantity to be constructed and consisting of me, e and h will also have $\epsilon _{0}$ and will be equal to

$\frac{\epsilon _{0}h^{2}}{m_{e}e^{2}}$  and has a numerical value of approximately 6.657$\times$10-10 which is about the order of atomic radii.

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

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

K=-E

K=-(-3.4)

K=3.4 eV

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

Total Energy= Potential energy + Kinetic Energy

E=U+K

U=E-K

U=-3.4-3.4

U=-6.8 eV

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

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

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 1070 . Therefore the angular momentum of such large objects is taken to be continuous rather than quantized.

As per  Bohr's quantization postulate

$m_{\mu ^{-}}v_{n}r_{n}=\frac{nh}{2\pi }$

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

$\frac{m_{\mu ^{-}}v_{n}^{2}}{r_{n}}=\frac{e^{2}}{4\pi\epsilon _{0}r_{n}^{2} }$

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 r1 is 5.3$\times$10-11 m

Therefore in case of a muonic hydrogen atom

$r_{1}=\frac{5.3\times 10^{-11}}{207}$

r1 = 2.56$\times$10-13 m

In case of the hydrogen atom, E1 is -13.6 eV

Therefore in case of a muonic hydrogen atom

E1=207$\times$(-13.6)

E1=2.81 keV

## NCERT solutions for class 12 physics chapter wise:

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

## 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.