Class 12 Physics Chapter 12 Atoms NCERT Solutions: Topics, Question with Solutions PDF

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

(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.)

Ans.12.1 The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.

In the ground state of Thomson’s model, electrons are in stable equilibrium. While in Rutherford’s model, electrons always experience a net force.

A classical atom based on Rutherford’s model, is doomed to collapse.

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.

The positively charged part of the atom possesses most of the mass in both the models.

 

Q.12.2 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 a solid at temperatures below 14 K.) What results do you expect?

Ans.12.2 In the alpha-particle scattering experiment, if a thin sheet of hydrogen is used in place of a gold film, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 $\times {10}^{-27}$ kg) is less than the mass of incident  $\alpha -$ particles (6.64 $\times {10}^{-27}$ kg). Thus, the mass of the scattering particles is more than the target nucleus (hydrogen). As a result, the $\alpha -$ particles would not bounce back if solid hydrogen is used in the $\alpha -$ particle scattering experiment.

 

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

Ans.12.3 Rydberg’s formula is given as:

$\frac{hc}{\lambda }$ = 21.76 $\times {10}^{-19}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

Where, h = Planck’s constant = 6.6 $\times {10}^{-34}$ Js

c = speed of light = 3 $\times {10}^8$ m/s

$\lambda$ = Wavelength

$n_1$ and $n_2$ are integers.

The shortest wavelength present in the Paschen series of the spectral lines is given for the values $n_1=3$ and $n_2=\infty$

Therefore, $\frac{hc}{\lambda }$ = 21.76 $\times {10}^{-19}\left[\frac{1}{3^2}-\frac{1}{{\infty }^2}\right]$

$\frac{hc}{\lambda }=$ 2.42 $\times {10}^{-19}$

$\lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2.42\times {10}^{-19}}$ = 8.189 $\times {10}^{-7}$ m= 818.9 nm

 

Q.12.4 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Ans.12.4 Separation of two energy level of atom, E = 2.3 eV = 2.3 $\times 1.6\times {10}^{-19}$ J = 3.68 $\times {10}^{-19}$

Let $\nu$ be the frequency of radiation emitted when the atom transits from upper level to lower level.

We have the relation for energy as $E=h\nu$  , where

h = Planck’s constant = 6.626 $\times {10}^{-34}$ Js

Then $\nu =\frac{E}{h}$ = $\frac{3.68\times {10}^{-19}}{6.626\times {10}^{-34}}$ Hz = 5.55 $\times {10}^{14}$ Hz

Hence the frequency is 5.55 $\times {10}^{14}$ Hz

 

Q.12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Ans.12.5 Ground state energy of hydrogen atom, E = -13.6 eV

Kinetic energy is equal to the negative of the total energy (ground state energy) = 13.6 eV

Potential energy is equal to the negative two times of kinetic energy = -13.6 $\times 2=-27.2eV$

 

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.

Ans.12.6 For ground level, $n_1$ = 1

Let $E_1$ be the energy level at $n_1$ . From the relation

$E_1=\frac{E}{n_1^2}$ where E = -13.6 eV, we get

$E_1=\frac{E}{n_1^2}$ eV = -13.6 V

For higher level, $n_2$ = 4

Let $E_2$ be the energy level at $n_2$ . From the relation

$E_2=\frac{E}{n_2^2}$ where E = -13.6 eV, we get

$E_2=\frac{-13.6}{4^2}$ eV = -0.85 V

The amount of energy absorbed by proton is given as $E_p$ = $E_2-E_1$ = -0.85 + 13.6 eV = 12.75 eV = 12.75 $\times 1.6\times {10}^{-19}$ J = 2.04 $\times {10}^{-18}$ J

For a photon of wavelength $\lambda ,$ the expression of energy is written as

$E_p$ = $\frac{hc}{\lambda }$ , where

c = speed of light = 3 $\times {10}^8$ m/s

h = Planck’s constant = 6.626 $\times {10}^{-34}$ Js

We get $\lambda =\frac{hc}{E_p}$ = $\frac{6.626\times {10}^{-34}\times 3\times {10}^8\mathrm{}}{2.04\times {10}^{-18}}$ = 9.744 $\times {10}^{-8}$ m = 97.44 nm

The frequency of the proton is given by,

$\nu =\frac{c}{\lambda }$ = $\frac{3\times {10}^8}{9.744\times {10}^{-8}}$ Hz= 3.1 $\times {10}^{15}$ Hz

 

Q.12.7 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

Ans.12.7 Let $v_1$ be the speed of the electron of the hydrogen atom in the ground state level, $n_1=1.$ For charge e of an electron, $v_1$ is given by the relation

$v_1=\frac{e^2}{n_{1}4\pi {ϵ}_0\left(\frac{h}{2\pi }\right)}$ = $\frac{e^2}{2n_1{ϵ}_{0}h}$

 

$where$ , e = 1.6 $\times {10}^{-19}$ C 

${ϵ}_0=$ Permittivity of free space = 8.85 $\times {10}^{-12}$ $N^{-1}{C}^2{m}^{-2}$

$h$ = Planck’s constant = 6.626 Js $\times {10}^{-34}$

$Hence$ $v_1=\frac{e^2}{2n_1{ϵ}_{0}h}$ $\frac{{(1.6\times {10}^{-19})}^2}{2\times 1\times 8.85\times {10}^{-12}\times 6.626\times {10}^{-34}}$ = 2.182 $\times {10}^6$ m/s

 

For level 2, $n_2=2$ ,

$v_2=\frac{e^2}{2n_2{ϵ}_{0}h}$ = $\frac{{(1.6\times {10}^{-19})}^2}{2\times 2\times 8.85\times {10}^{-12}\times 6.626\times {10}^{-34}}$ = 1.091 $\times {10}^6$ m/s

 

For level 3, $n_3=3$ ,

$v_3=\frac{e^2}{2n_3{ϵ}_{0}h}$ = $\frac{{(1.6\times {10}^{-19})}^2}{2\times 3\times 8.85\times {10}^{-12}\times 6.626\times {10}^{-34}}$ = 0.727 $\times {10}^6$ m/s

 

Let be the orbital period of the electron when it is in level . Orbital period is given by the expression

$T_1$ = $\frac{2\pi r_1}{v_1}$ where $r_1$ = radius of the orbit = $\frac{n_1^2{h}^2{ϵ}_0}{\pi m{e}^2}$ , where

$m=$ mass of an electron = 9.1 $\times {10}^{-31}$ kg

e = 1.6 $\times {10}^{-19}$ C

${ϵ}_0=$ Permittivity of free space = 8.85 $\times {10}^{-12}$ $N^{-1}{C}^2{m}^{-2}$

$h$ = Planck’s constant = 6.626 $\times {10}^{-34}$ Js

 

$T_1$ = $\frac{2\pi }{v_1}\times \frac{n_1^2{h}^2{ϵ}_0}{\pi m{e}^2}$ = $\frac{2\times n_1^2\times h^2\times {ϵ}_0}{v_1\times m\times e^2}$ = $\frac{2\times 1\times {(6.626\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{2.182\times {10}^6\times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$ = $\frac{7.77\times {10}^{-78}}{5.083\times {10}^{-62}}$

=1.53 $\times {10}^{-16}s$

For level $n_2$ =2, $T_2$ = $\frac{2\times n_2^2\times h^2\times {ϵ}_0}{v_2\times m\times e^2}$ = $\frac{2\times 4\times {(6.626\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{1.091\times {10}^6\times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$ = $\frac{3.108\times {10}^{-77}}{2.542\times {10}^{-62}}$ =

1.22 $\times {10}^{-15}$ s

 

For level $n_3$ = 3, $T_3$ = $\frac{2\times n_3^2\times h^2\times {ϵ}_0}{v_3\times m\times e^2}$ = $\frac{2\times 9\times {(6.626\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{0.727\times {10}^6\times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$ = $\frac{6.993\times {10}^{-77}}{1.694\times {10}^{-62}}$ = 4.13 $\times {10}^{-15}$ s

 

Q.12.8 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits?

Ans.12.8 The radius of the innermost orbit of a hydrogen atom, $r_1$ = 5.3 $\times {10}^{-11}$ m

Let $r_2$ be the radius of the orbit at n = 2. The relation between the radius of the orbit is

$r_2$ = $n^2\times r_1$ =4 $\times 5.3\times {10}^{-11}\mathrm{}\mathrm{m}$ =21.2 $\times {10}^{-11}\mathrm{}\mathrm{m}$

Let $r_3$ be the radius of the orbit at n = 3. The relation between the radius of the orbit is

$r_3$ = $n^2\times r_1$ =9 $\times 5.3\times {10}^{-11}\mathrm{}\mathrm{m}$ =47.7 $\times {10}^{-11}\mathrm{}\mathrm{m}$

 

(b) Is the probability of backward scattering (i.e., scattering of $\mathit{\alpha }$ -particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?

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

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

Ans.12.11 About the same - The average angle of deflection $\alpha$ -particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.

Much less - The probability of scattering of $\alpha$ -particles at angles greater than 90 $°$ predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.

Thomson’s model - It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of $\alpha$ -particles by a thin foil. This is because a single-collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

 

Q.12.12 The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10–40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Ans.12.12 Radius of the first Bohr orbit is given by the relation:

$r_1$ = $\frac{4\pi {ϵ}_0{\left(\frac{h}{2\pi }\right)}^2}{m_e{e}^2}$ ………………(1)

Where,

${ϵ}_0=$ Permittivity of free space

h = Planck’s constant = 6.626 $\times {10}^{-34}$ Js

$m_e$ = mass of electron = 9.1 $\times {10}^{-31}$ kg

e = Charge of electron = 1.9 $\times {10}^{-19}$ C

$m_p$ = mass of a proton = 1.67 $\times {10}^{-27}$ kg

r = distance between the electron and proton

Coulomb attraction between an electron and a proton is given as:

$F_c$ = $\frac{e^2}{4\pi {ϵ}_0{r}^2}$ ……………….(2)

Gravitational force of attraction between an electron and a proton is given as:

$F_G$ = $\frac{G{m}_p{m}_e}{r^2}$ ……………….(3)

Where, G = Gravitational constant = 6.67 $\times {10}^{-11}$ N $m^2$ / ${kg}^2$

If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write

$F_c=F_G$

$\frac{e^2}{4\pi {ϵ}_0{r}^2}=\frac{G{m}_p{m}_e}{r^2}$

or

$G{m}_p{m}_e=\frac{e^2}{4\pi {ϵ}_0}$

By putting the above value in equation (1), we can write

$r_1$ = $\frac{4\pi {ϵ}_0}{e^2}\times \frac{{\left(\frac{h}{2\pi }\right)}^2}{m_e}$ = $\frac{1}{G{m}_p{m}_e}$ $\times \frac{{\left(\frac{h}{2\pi }\right)}^2}{m_e}$ = $\frac{{\left(\frac{h}{2\pi }\right)}^2}{G{m}_p{m}_e^2}$

= $\frac{{\left(\frac{6.626\mathrm{}\times {10}^{-34}}{2\pi }\right)}^2}{6.67\mathrm{}\times {10}^{-11}\times 1.67\mathrm{}\times {10}^{-27}\times {9.1\mathrm{}\times {10}^{-31}}^2}$ = $\frac{1.112\times {10}^{-68}}{9.224\times {10}^{-98}}$ = 1.21 $\times {10}^{29}$ m

It is known that the universe is 156 billion light years wide or 1.5 $\times {10}^{27}$ m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

 

Q.12.13 Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Ans.12.13 It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n-1)

We have the relation for energy ( $E_1$ ) of radiation at level n as:

$E_1$ = h ${\nu }_1$ = $\frac{hm{e}^2}{{\left(4\pi \right)}^2{ϵ}_0^2({\frac{h}{2\pi })}^2}$ $\times \frac{1}{{\left(n\right)}^2}$ ………………(i)

Where,

${\nu }_1$ = Frequency of radiation at level n

h = Planck’s constant

m = mass of hydrogen atom

e = charge of an electron

${ϵ}_0$ = Permittivity of free space

Now, the relation for energy ( $E_2$ ) of radiation at level (n-1) is given as:

$E_2$ = h ${\nu }_2$ = $\frac{hm{e}^2}{{\left(4\pi \right)}^2{ϵ}_0^2({\frac{h}{2\pi })}^2}$ $\times \frac{1}{{(n-1)}^2}$ ………………(ii)

Where,

${\nu }_2$ = Frequency of radiation at level (n-1)

Energy (E) released as a result of de-excitation:

E = $E_2-E_1$

$h\nu$ = $E_2-E_1$ …………………………(iii)

where,

$\nu$ = Frequency of radiation emitted

Putting values of equation (i) and (ii) in equation (iii), we get,

$\nu$ = $\frac{m{e}^2}{{\left(4\pi \right)}^2{ϵ}_0^2({\frac{h}{2\pi })}^2}$ $\left[\frac{1}{{(n-1)}^2}-\frac{1}{{\left(n\right)}^2}\right]$

= $\frac{m{e}^2(2n-1)}{{\left(4\pi \right)}^2{ϵ}_0^2({\frac{h}{2\pi })}^2{n}^2{(n-1)}^2}$

For large n, we can write (2n-1) = 2n and (n-1) = n

Therefore, $\nu$ = $\frac{m{e}^2}{32{\pi }^3{ϵ}_0^2({\frac{h}{2\pi })}^3{n}^3}$ ………..(iv)

Classical relation of frequency of revolution of an electron is given as:

${\nu }_c$ = $\frac{v}{2\pi r}$ …………………..(v)

Velocity of the electron in the $n^{th}$ orbit is given as:

v = $\frac{e^2}{4\pi {ϵ}_0\left(\frac{h}{2\pi }\right)n}$ ……………...(vi)

And, radius of the $n^{th}$ orbit is given as:

r = $\frac{4\pi {ϵ}_0({\frac{h}{2\pi })}^2}{m{e}^2}{n}^2$ ……………….(vii)

By putting the values of (vi) and (vii) in equation (v), we get

${\nu }_c$ = $\frac{m{e}^2}{32{\pi }^3{ϵ}_0^2({\frac{h}{2\pi })}^3{n}^3}$

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

 

Q.12.14 Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10–10m).

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

(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 recognizing 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.

Ans. 12.14 Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Mass of an electron, $m_e$ = 9.1 $\times {10}^{-31}$ kg

Speed of light, c = 3 $\times {10}^8$ m/s

Let us take a quantity involving the given quantities as $\left(\frac{e^2}{4\pi {ϵ}_0{m}_e{c}^2}\right)$

Where

${ϵ}_0$ = permittivity of free space and

$\frac{1}{4\pi {ϵ}_0}$ = 9.1 $\times {10}^9$ N $m^2{C}^{-1}$

Hence, $\left(\frac{e^2}{4\pi {ϵ}_0{m}_e{c}^2}\right)$ = 9.1 $\times {10}^9\times \frac{{(1.6\mathrm{}\times {10}^{-19})}^2}{9.1\mathrm{}\times {10}^{-31}\times {(3\mathrm{}\times {10}^8)}^2}$ = 2.844 $\times {10}^{-15}$ m

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.

Charge of an electron, e = 1.6 $\times {10}^{-19}$ C

Mass of an electron, $m_e$ = 9.1 $\times {10}^{-31}$ kg

Planck’s constant, h = 6.623 $\times {10}^{-34}$ Js

Hence, the value of the quantity taken is in the order of the atomic size.

 

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

(a) What is the kinetic energy of the electron in this state?

(b) What is the potential energy of the electron in this state?

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

Ans.12.15 Total energy of the electron, E = -3.4 eV

Kinetic energy of the electron is equal to the negative of the total energy

K = -E = 3.4 eV

Potential energy (U) of the electron is equal to twice the negative of kinetic energy

U = -2K = -6.8 eV

The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic and potential energies, total energy of the system will also change.

 

Q.12.16 If Bohr’s quantization postulate (angular momentum = nh/2n) 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 quantization of orbits of planets around the sun?

Ans.12.16 We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). The angular momentum of the Earth in its orbit is of the order of ${10}^{70}$ h. This leads to a very high value of quantum levels n of the order of ${10}^{70}$ . For large values of n, successive energies and angular momentum are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.

 

Q.12.17 Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ–) of mass about 207me orbits around a proton].

Ans.12.17 Mass of a negatively charged muon, $m_{\mu }$ = 207 $m_e$

According to Bohr’s model

Bohr radius, $r_e$ $\propto$ $\frac{1}{m_e}$

And, energy of a ground state electronic hydrogen atom $E_e$ $\propto$ $m_e$

Also, the energy of a ground state muonic hydrogen atom, $E_u$ $\propto$ $m_u$

We have the value of the first Bohr orbit, $r_e$ = 0.53 Å = 0.53 $\times {10}^{-10}$ m

Let $r_o$ be the radius of muonic hydrogen atom

At equilibrium, we can write the relation as:

$m_e{r}_{\mu }$ = $m_e{r}_e$

207 $m_e{r}_{\mu }$ = $m_e{r}_e$

$r_{\mu }=\frac{r_e}{207}$ = $\frac{0.53\mathrm{}\times {10}^{-10}}{207}$ =2.56 $\times {10}^{-13}$ m

Hence, the value of the first Bohr radius of a muonic hydrogen atom is 2.56 $\times {10}^{-13}$ m

We have $E_e$ = -13.6 eV