NCERT Solutions for Class 12 Physics Chapter 12 Atoms

1. Atomic Structure and it’s History

1.1 Thomson’s Atom Model

The positive charge is uniformly distributed over the entire sphere and the electrons are embedded in the sphere of positive charges just like seeds in a watermelon or plums in the pudding.  For this reason, Thomson’s atom model is also known as plum-pudding model.  The total positive charge inside the atom is equal to the total negative charge carried by electrons, so that every atom is electrically neutral.

Failure of Thomson’s Atom Model

It had to be discarded, because of the following reasons:

• It could not explain the origin of the spectral lines in the form of series as in case of hydrogen atom.

• It could not account for the scattering of α-particles through large angles as in case of Rutherford’s α-scattering experiment.

1.2. Rutherford’s Alpha Scattering Experiment Observations

• Most of α-particles were found to pass through the gold foil without any appreciable deflection.

• The different α-particles in passing through the gold foil undergo different amounts of deflections.  A large number of α-particles suffer fairly large deflections.

• A very small number of α-particles (about 1 in 8000) practically retracted their paths or suffered deflection of nearly 180º.

• The graph between the total number of α–particles N(θ) scattered through angle θ and the scattering angle θ was found to be as shown in fig.

The experimental observations led Rutherford to the following conclusions:-

• Since most of the α-particles passed undeviated, the atom has a lot of empty space in it.

• Since fast and the heavy α-particles could be deflected even through 180º, the whole of the positive charge and practically the entire mass of the atom was confined to an extremely small central core.  It was called nucleus.  Since 1 in about 8000 α-particles is deflected through 180º, the size of the nucleus is about 1/10000th of the size of the atom.

1.1.1 Rutherford’s Atom Model

On the basis of the results of α-scattering experiment, Rutherford suggested the following picture of the atom:

• Atom may be regarded as a sphere of diameter 10–10 m but whole of the positive charge and almost the entire mass of the atom is concentrated in a small central core called nucleus having diameter of about 10–14 m.

• The nucleus is surrounded by electrons. In other words, the electrons are spread over the remaining part of the atom leaving plenty of empty space in the atom.

1.1.2 Drawbacks of Rutherford’s Atom Model

• When the electrons revolve round the nucleus, they are continuously accelerated towards the centre of the nucleus.  According to Lorentz, an accelerated charged particle should radiate energy continuously.  Therefore, in the atom, a revolving electron should continuously emit energy and hence the radius of its path should go on decreasing and ultimately it should fall into the nucleus.  However, electrons revolve round the nucleus without falling into it.  Therefore, Rutherford’s atom model cannot explain the stability of the atom.

• If the Rutherford’s atom model is true, the electron can revolve in orbits of all possible radii and hence it should emit continuous energy spectrum.  However, the atoms like hydrogen possess line spectrum.

1.1.3 Distance of Closest Approach

Consider an α-particle of mass m possesses initial velocity u, when it is at a large distance from the nucleus of an atom having atomic number Z.  At the distance of closest approach, the kinetic energy of α-particle is completely converted into potential energy. Mathematically.

$1/2m{u}^2=\frac{1}{4\pi {\epsilon }_0}.\frac{(2e)(Ze)}{r_0}\therefore r_0=\frac{1}{4\pi {\epsilon }_0}.\frac{2Z{e}^2}{1/2m{u}^2}$

2. Bohr Atomic Model

Bohr adopted Rutherford model of the atom & added some arbitrary conditions. These conditions are known as his postulates:

• The electron in a stable orbit does not radiate energy. i.e. $\frac{m{v}^2}{r}=\frac{kz{e}^2}{r^2}$

• A stable orbit is that in which the angular momentum of the electron about nucleus is an integral (n) multiple of $\frac{h}{2\pi }$ . i.e. mvr  = n$\frac{h}{2\pi }$;  n = 1 , 2 ,  3 , .......(n ≠ 0).

• The electron can absorb or radiate energy only if the electron jumps from a lower to a higher orbit or falls from a higher to a lower orbit.

• The energy emitted or absorbed is a light photon of frequency ν  and of energy E  =  hν.

2.1 For Hydrogen Atom: (Z = Atomic Number = 1)

• Ln = angular momentum in the nth orbit = n $\frac{h}{2\pi }$.

• rn = radius of nth  circular orbit  = (0.529 Aº) n2 ;  

(1Aº = 10-10 m) ;  rn   n2.

• En  Energy  of the electron in the nth  orbit  $=\frac{-13.6eV}{n^2}$ i.e. $E_n\propto \frac{1}{n^2}.$

NOTE:

Binding  Energy (BE)n  =  – En  =.$\frac{-13.6V}{n^2}$

• $E_{n_2}-E_{n_1}$ = Energy emitted when an electron jumps from n2th orbit to n1th orbit (n2 > n1) .

ΔE = (13.6 ev) $[\frac{1}{n_1^2}-\frac{1}{n_2^2}].$    

ΔE = hν ; ν =  frequency of spectral line emitted  .

$\frac{1}{\lambda }=\overline{v}=$ wave no. ( no. of waves in unit length (1m)) 

$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}].$

Where R = Rydberg's constant for hydrogen = 1.097 × 107  m-1  .

• For hydrogen like atom/species of atomic number Z: 

$r_{nz}={\displaystyle \frac{Bohrradius}{Z}}{n}^2=(0.529A^{{}^{\circ }}){\displaystyle \frac{n^2}{Z}}$

$E_{nz}=(-13.6)\frac{Z^2}{n^2}ev$ 

Rz  =  RZ2 –  Rydberg's constant for element of atomic no. Z

2.2 Spectral Series

• Lyman Series: (Landing orbit n = 1).

Ultraviolet region $\overline{v}=R[\frac{1}{1^2}-\frac{1}{n_2^2}];$ n2 > 1

• Balmer Series: (Landing orbit n = 2) 

Visible region $\overline{v}=R[\frac{1}{2^2}-\frac{1}{n_2^2}];$ n2 > 2

• Paschan Series: (Landing orbit n = 3) 

In the near infrared region $\overline{v}=R[\frac{1}{3^2}-\frac{1}{n_2^2}];$ n2 > 3

• Bracket Series: (Landing orbit n = 4)

In the mid infrared region $\overline{v}=R[\frac{1}{4^2}-\frac{1}{n_2^2}];$ n2 > 4

• Pfund Series: (Landing orbit n = 5)

In far infrared region $\overline{v}=R[\frac{1}{5^2}-\frac{1}{n_2^2}];$ n2 > 5

In  all these  series  n2 

= n1 + 1 is the α line 

= n1 + 2 is the β line 

= n1 + 3 is the γ line .... etc.

where n1 = Landing orbit

Mastering Class 12 Physics Chapter 12: Atoms - MCQs, Question and Answers, and Tips for Success

1. Fill in the blanks using the given options:

1. The size of the atoms in Thomson's model are _______ the atomic size in Rutherford's model (much greater than/no different from/much lesser than).

Ans: The sizes of the atoms in Thomson's model are no different from the atomic size in Rutherford's model.

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

Ans: In the ground state of Thomson's model, the electrons are in stable equilibrium, while in Rutherford's model, electrons always experience a net force.

3. A classical atom based on _______ is doomed to collapse. (Thomson's model/ Rutherford's model.)

Ans: A classical atom based on Rutherford's model is doomed to collapse.

4. An atom features a nearly continuous mass distribution in _______ but features a highly non- uniform mass distribution in _______ (Thomson's model/ Rutherford's model.)

Ans: An atom features a nearly continuous mass distribution in Thomson's model, but features a highly non-uniform mass distribution in Rutherford's model.

5. The positively charged part of the atom possesses most of the mass in _______ (Rutherford's model/ Thomson's model /both the models.)

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

2. If you’re given a chance to repeat the $\alpha -$particle scattering experiment employing a thin sheet of solid hydrogen instead of the gold foil (Hydrogen is a solid at temperatures below 14 K). What results would you expect?

Ans: In the $\alpha -$particle scattering experiment, when a thin sheet of solid hydrogen is replaced with the gold foil, the scattering angle would not turn out to be large enough.

This is because the mass of hydrogen is smaller than the mass of incident $\alpha -$particles. Also, the mass of the scattering particle is more than the target nucleus (hydrogen).

As a consequence, the $\alpha -$particles would not bounce back when solid hydrogen is utilized in the $\alpha -$particle scattering experiment and hence, we cannot evaluate the size of the hydrogen nucleus.

3. What’s the shortest wavelength present within the Paschen series of spectral lines?

Ans: We know the Rydberg’s formula is given as;

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

Here,

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

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

($n_1$ and $n_2$ are integers)

The shortest wavelength present within the Paschen series of the spectral lines is for values $n_1=3$ and $n_2=\mathrm{\infty }$.

$\Rightarrow$ $\frac{hc}{\lambda }=21.76\times {10}^{-19}[\frac{1}{3^2}-\frac{1}{{\mathrm{\infty }}^2}]$

$\Rightarrow$ $\lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8\times 9}{21.76\times {10}^{-19}}$

$\Rightarrow$ $\lambda =8.189\times {10}^{-7}m$

$\Rightarrow$ $\lambda =818.9nm$

4. The two energy levels in an atom are separated by a difference of $2.3eV$. What is the frequency of radiation emitted when the atom makes a transition from the higher level to the lower level?

Ans: Given that the distance between the two energy levels in an atom is $E=2.3eV$.

$\Rightarrow E=2.3\times 1.6\times {10}^{-19}$

$\Rightarrow E=3.68\times {10}^{-19}J$

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

The relation for energy is given as;

$E=h\nu$

Here,

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

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

$\Rightarrow \nu =\frac{3.38\times {10}^{-19}}{6.62\times {10}^{-32}}$

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

Clearly, the frequency of the radiation is $5.6\times {10}^{14}Hz$.

5. For a hydrogen atom, the ground state energy is $-13.6eV$ . What are the kinetic and potential energies of the electron during this state?

Ans: Provided that the ground state energy of hydrogen atom, $E=-13.6eV$which is the total energy of a hydrogen atom. 

Here, kinetic energy is equal to the negative of the total energy.

Kinetic energy $=-E=-(-13.6)=13.6eV$

The potential energy is the same as the negative of two times kinetic energy.

Potential energy $=-2\times \left(13.6\right)=-27.2eV$

$\therefore$ The kinetic energy of the electron is $13.6eV$ and the potential energy is $-27.2eV$ .

6. A hydrogen atom absorbs a photon when it is in the ground level, this excites it to the $n=4$ level. Find out the wavelength and frequency of the photon.

Ans: It is known that for ground level absorption, $n_1=1$

Let $E_1$ be the energy of this level. It is known that $E_1$ is related with $n_1$ as;

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

$\Rightarrow E_1=\frac{-13.6}{1^2}=-13.6eV$

When the atom jumps to a higher level, $n_2=4$.

Let $E_2$ be the energy of this level.

$\Rightarrow E_2=\frac{-13.6}{n_2^2}eV$

$\Rightarrow E_2=\frac{-13.6}{4^2}=\frac{-13.6}{16}eV$

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

$$E=E_1-E_2$$

$$\Rightarrow E=\left(\frac{-13.6}{16}\right)-\left(\frac{-13.6}{1}\right)$$

$$\Rightarrow E=\frac{13.6\times 15}{16}eV$$

$$\Rightarrow E=\frac{13.6\times 16}{16}\times 1.6\times {10}^{-19}$$

$$\Rightarrow E=2.04\times {10}^{-18}J$$

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

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

Here,

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

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

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

$\Rightarrow \lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2.04\times {10}^{-18}}$

$\Rightarrow \lambda =9.7\times {10}^{-8}m$

$\Rightarrow \lambda =97nm$

Also, frequency of a photon is given by the relation,

$\nu =\frac{c}{\lambda }$

$\Rightarrow \nu =\frac{3\times {10}^8}{9.7\times {10}^{-8}}\approx 3.1\times {10}^{15}Hz$

Clearly, the wavelength of the photon is 97nm whereas the frequency is $3.1\times {10}^{15}Hz$.

7. Answer the following questions.

1. Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the $n=1,2$ and $3$ levels.

Ans: Consider ${\nu }_1$ to be the orbital speed of the electron in a hydrogen atom in the ground state level $n_1=1$ . For charge $\left(e\right)$ of an electron, ${\nu }_1$ is given by the relation,

${\nu }_1=\frac{e^2}{n_{1}4\pi {\in }_0\left(\frac{h}{2\pi }\right)}$

$\Rightarrow {\nu }_1=\frac{e^2}{2{\in }_{0}h}$

Here,

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

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

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

$\Rightarrow {\nu }_1=\frac{{(1.6\times {10}^{-19})}^2}{2\times 8.85\times {10}^{-12}\times 6.62\times {10}^{-34}}$

$\Rightarrow {\nu }_1=0.0218\times {10}^8$

$\Rightarrow {\nu }_1=2.18\times {10}^{6}m/s$

For level $n_2=2$ , we can write the relation for the corresponding orbital speed as;

${\nu }_2=\frac{e^2}{n_{2}2{\in }_{0}h}$

$\Rightarrow {\nu }_2=\frac{{(1.16\times {10}^{-19})}^2}{2\times 2\times 8.85\times {10}^{-12}\times 6.62\times {10}^{-34}}$

$\Rightarrow {\nu }_2=1.09\times {10}^{6}m/s$

And, for $n_3=3$ , we can write the relation for the corresponding orbital speed as;

${\nu }_3=\frac{e^2}{n_{3}2{\in }_{0}h}$

$\Rightarrow {\nu }_3=\frac{{(1.16\times {10}^{-19})}^2}{3\times 2\times 8.85\times {10}^{-12}\times 6.62\times {10}^{-34}}$

$\Rightarrow {\nu }_3=7.27\times {10}^{5}m/s$

Clearly, the speeds of the electron in a hydrogen atom in the levels $n=1,2$ and $3$ are $2.18\times {10}^{6}m/s$ , $1.09\times {10}^{6}m/s$ and $7.27\times {10}^{5}m/s$ respectively.

2. Calculate the orbital period in each of these levels.

Ans: Consider $T_1$ to be the orbital period of the electron when it is in level $n_1=1$ .

It is known that the orbital period is related to the orbital speed as

$T_1=\frac{2\pi r_1}{{\nu }_1}$

Here,

$r_1=$ Radius of the orbit in $n_1$$=\frac{n_1^2{h}^2{\in }_0}{\pi m{e}^2}$

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

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

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

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

$\Rightarrow T_1=\frac{2\pi \times {\left(1\right)}^2\times {(6.62\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{2.18\times {10}^6\times \pi \times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$

$\Rightarrow T_1=15.27\times {10}^{-17}$

$\Rightarrow T_1=1.527\times {10}^{-16}s$

For level $n_2=2$, we can write the orbital period as;

$T_2=\frac{2\pi r_2}{{\nu }_2}$

Here,

$r_2=$ Radius of the orbit in $n_2$$=\frac{n_2^2{h}^2{\in }_0}{\pi m{e}^2}$

$\Rightarrow T_1=\frac{2\pi \times {\left(2\right)}^2\times {(6.62\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{1.09\times {10}^6\times \pi \times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}=1.22\times {10}^{-15}s$

And for the level $n_3=3$, we can write the orbital period as;

$T_3=\frac{2\pi r_3}{{\nu }_3}$

Here,

$r_3=$ Radius of the orbit in $n_3$$=\frac{n_3^2{h}^2{\in }_0}{\pi m{e}^2}$

$\Rightarrow T_3=\frac{2\pi \times {\left(3\right)}^2\times {(6.62\times {10}^{-34})}^2\times 8.85\times {10}^{-12}}{7.27\times {10}^5\times \pi \times 9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}=4.12\times {10}^{-15}s$

Hence, the orbital periods in the levels $n=1,2$ and $3$ are $1.527\times {10}^{-16}s$ , $1.22\times {10}^{-15}s$ and $4.12\times {10}^{-15}s$ respectively.

8. The innermost electron orbit of a hydrogen atom has a radius of $5.3\times {10}^{-11}m$. What are the radii of the $n=2$ and $n=3$ orbits?

Ans: Provided that the innermost radius, $r_1=5.3\times {10}^{-11}m$ .

Let $r_2$ be the radius of the orbit at $n=2$ . It is related to the radius of the innermost orbit as;

$r_2={\left(n\right)}^2{r}_1$

$\Rightarrow r_2={\left(2\right)}^2\times 5.3\times {10}^{-11}=2.1\times {10}^{-10}m$

Similarly, for $n=3$;

$r_3={\left(n\right)}^2{r}_1$

$\Rightarrow r_3={\left(3\right)}^2\times 5.3\times {10}^{-11}=4.77\times {10}^{-10}m$

Clearly, the radii of the $n=2$ and $n=3$ orbits are $2.1\times {10}^{-10}m$ and $4.77\times {10}^{-10}m$ respectively.

9. A $12.5eV$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Ans: It is provided that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is $12.5eV$ .

It is also known that the energy of the gaseous hydrogen in its ground state at room temperature is $-13.6eV$ .

When gaseous hydrogen is bombarded with an electron beam at room temperature, the energy of the gaseous hydrogen becomes $$-13.6+12.5eV=-1.1eV$$.

Now, the orbital energy is related to orbit level $$\left(n\right)$$ as;

$E=\frac{-13.6}{{\left(n\right)}^2}eV$

For $n=3;E=\frac{-13.6}{9}=-1.5eV$

This energy is approximately equal to the energy of gaseous hydrogen.

So, it can be concluded that the electron has excited from $$n=1ton=3$$ level.

During its de-excitation, the electrons can jump from $$n=3ton=1$$ directly, which forms a line of the Lyman series of the hydrogen spectrum.

The formula for wave number for Lyman series is given as;

$\frac{1}{\lambda }=R_y(\frac{1}{1^2}-\frac{1}{n^2})$

Here, 

$R_y=$ Rydberg constant $=1.097\times {10}^7{m}^{-1}$

$$\lambda =$$ Wavelength of radiation emitted by the transition of the electron

Using this relation for $n=3$ we get,

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

$$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7(1-\frac{1}{9})$$

$$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7\left(\frac{8}{9}\right)$$

$\Rightarrow \lambda =\frac{9}{8\times 1.097\times {10}^7}=102.55nm$

If the transition takes place from $n=3ton=2$ , and then from $$n=2ton=1$$, then the wavelength of the radiation emitted in transition from $$n=3ton=2$$ is given as;

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

$$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{4}-\frac{1}{9})$$

$$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7\left(\frac{5}{36}\right)$$

$\Rightarrow \lambda =\frac{36}{5\times 1.097\times {10}^7}=656.33nm$

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Now, the wavelength of the radiation when the transition takes place from $$n=2ton=1$$ is given as;

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

$$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7(1-\frac{1}{4})$$

$$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7\left(\frac{3}{4}\right)$$

$\Rightarrow \lambda =\frac{4}{3\times 1.097\times {10}^7}=121.54nm$

Clearly, in the Lyman series, two wavelengths are emitted i.e., 102.5nm and 121.5nm whereas in the Balmer series, only one wavelength is emitted i.e., 656.33nm.

10. In accordance with the Bohr’s model, what is the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius $1.5\times {10}^{11}m$ with an orbital speed of $3\times {10}^{4}m/s$ . The mass of the earth is given as $6\times {10}^{24}kg$ .

Ans: Here, it is provided that,

Radius of the earth’s orbit around the sun, $r=1.5\times {10}^{11}m$

Orbital speed of the earth, $\nu =3\times {10}^{4}m/s$

Mass of the earth, $m=6\times {10}^{24}kg$

With respect to the Bohr’s model, angular momentum is quantized and is given as;

$m\nu r=\frac{nh}{2\pi }$

Here,

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

$n=$ Quantum number

$n=\frac{m\nu r^2\pi }{h}$

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

$\Rightarrow n=25.61\times {10}^{73}=2.6\times {10}^{74}$

Clearly, the quantum number that characterizes the earth’s revolution around the sun is $2.6\times {10}^{74}$ .

11. Which of the following questions help you understand the difference between Thomson's model and Rutherford's model better.

1. 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?

Ans: About the same.

The average angle of deflection of $\alpha -$particles caused by a thin gold foil considered by Thomson's model is about the same size as that considered by Rutherford's model.

This is because in both the models, the average angle was used.

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

Ans: Much less.

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

3. 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?

Ans: Scattering is mainly caused due to single collisions.

The chances of a single collision have a linearly increasing nature with the number of target atoms.

As the number of target atoms increases with an increase in thickness, the collision probability varies linearly with the thickness of the target.

4. 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?

Ans: Thomson's model.

It is incorrect to not consider 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 produces very little deflection in this model.

Thus, the observed average scattering angle can be demonstrated only by considering multiple scattering.

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: The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about ${10}^{-40}$. 

It is given that,

Radius of the first Bohr orbit is given by the relation,

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

Here,

${\in }_0=$ Permittivity of free space

$h=$ Planck’s constant $=6.63\times {10}^{-34}Js$

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

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

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

$r=$ Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given by;

$F_C=\frac{e^2}{4\pi {\in }_0{r}^2}$ …(2)

Gravitational force of attraction between an electron and a proton is given by;

$F_G=\frac{G{m}_p{m}_e}{r^2}$ …(3)

Here,

$G=$ Gravitational constant $=6.67\times {10}^{-11}N{m}^2/k{g}^2$

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

$F_G=F_C$

$\Rightarrow \frac{G{m}_p{m}_e}{r^2}=\frac{e^2}{4\pi {\in }_0{r}^2}$ …(using 2 & 3)

$\Rightarrow \frac{e^2}{4\pi {\in }_0}=G{m}_p{m}_e$ …(4)

Now, by substituting the value of equation (4) in (1),  we get; 

$r_1=\frac{{\left(\frac{h}{2\pi }\right)}^2}{G{m}_p{m}_e}$

$\Rightarrow r_1=\frac{{\left(\frac{6.63\times {10}^{-34}}{2\times 3.14}\right)}^2}{6.67\times {10}^{-11}\times 1.67\times {10}^{-27}\times {(9.1\times {10}^{-31})}^2}$

$\Rightarrow r_1\approx 1.21\times {10}^{29}m$

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

13. Derive 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: It is given that a hydrogen atom makes transition from an upper level $\left(n\right)$ to a lower level $(n-1)$. We have the relation for energy $\left(E_1\right)$ of radiation at level $n$ as;

$E_1=h{\nu }_1=\frac{hm{e}^4}{{\left(4\pi \right)}^3{\in }_0^2{\left(\frac{h}{2\pi }\right)}^3}\times \left(\frac{1}{n^2}\right)$ …(1)

Here,

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

$h=$ Planck’s constant

$m=$ Mass of hydrogen atom

$e=$ Charge on an electron

${\in }_0=$ Permittivity of free space

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

$E_2=h{\nu }_2=\frac{hm{e}^4}{{\left(4\pi \right)}^3{\in }_0^2{\left(\frac{h}{2\pi }\right)}^3}\times \frac{1}{{(n-1)}^2}$ …(2)

Here,

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

Energy $\left(E\right)$ released as a result of de-excitation;

$E=E_2-E_1$

$\Rightarrow h\nu =E_2-E_1$ …(3)

Here,

$\nu =$ Frequency of radiation emitted

Using the values of equations (1) and (2) in equation (3), we get;

$\nu =\frac{m{e}^4}{{\left(4\pi \right)}^3{\in }_0^2{\left(\frac{h}{2\pi }\right)}^3}[\frac{1}{{(n-1)}^2}-\frac{1}{n^2}]$

$\Rightarrow \nu =\frac{m{e}^4(2n-1)}{{\left(4\pi \right)}^3{\in }_0^2{\left(\frac{h}{2\pi }\right)}^3{n}^2{(n-1)}^2}$

For large $n$, we can write $(2n-1)-2n$ and $(n-1)\simeq n$

$\Rightarrow \nu =\frac{m{e}^4}{32{\pi }^3{\in }_0^2{\left(\frac{h}{2\pi }\right)}^3{n}^3}$ …(4)

Classical relation of frequency of revolution of an electron is given by;

${\nu }_c=\frac{\nu }{2\pi r}$ …(5)

Here,

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

$\nu =\frac{e^2}{4\pi {\in }_0\left(\frac{h}{2\pi }\right)n}$ …(6)

And, radius of the $n^{th}$ orbit is given by;

$r=\frac{4\pi {\in }_0{\left(\frac{h}{2\pi }\right)}^2}{m{e}^2}{n}^2$ …(7)

Substituting the values of equation (6) and (7) in equation (5), we get;

${\nu }_c=\frac{m{e}^4}{32{\pi }^3{\in }_0^2{\left(\frac{h}{2\pi }\right)}^3{n}^3}$ …(8)

Clearly, when equations (4) and (8) are compared, it can be seen that the frequency of radiation emitted by the hydrogen atom is the same as the classical orbital frequency.

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 $(\sim {10}^{-10}m)$.

1. Construct a quantity with the dimensions of length from the fundamental constants $e,m_{e}andc$. Also, determine its numerical value.

Ans: To construct a quantity with the dimensions of length from the fundamental constants $e,m_{e}andc$, take: 

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

The quantity having dimensions of length and involving the given quantities is $\left(\frac{e^2}{4\pi {\in }_0{m}_e{c}^2}\right)$

Here,

${\in }_0=$ Permittivity of free space

And, $\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N{m}^2{C}^{-2}$

$\therefore$The numerical value of the taken quantity will be

$$\frac{1}{4\pi {\in }_0}\times \frac{e^2}{m_e{c}^2}=9\times {10}^9\times \frac{{(1.6\times {10}^{-19})}^2}{9.1\times {10}^{-31}\times {(3\times {10}^8)}^2}=2.81\times {10}^{-15}m$$

Clearly, the numerical value of the taken quantity is much less than the typical size of an atom.

2. You will observe 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 $$h,⊸m_e,⊸and⊸e$$ will yield the right atomic size. Construct a quantity with the dimension of length from $$h,m_{e}ande$$ and confirm that its numerical value has indeed the correct order of magnitude.

Ans: To construct a quantity with the dimension of length from $$h,m_{e}ande$$, take,

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, $h=6.63\times {10}^{-34}Js$

Now, let us take a quantity involving the given quantities as, $\frac{4\pi {\in }_0\left(\frac{h^2}{2\pi }\right)}{m_e{e}^2}$

Here, ${\in }_0=$ Permittivity of free space

And, $\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N{m}^2{C}^{-2}$

$\therefore$The numerical value of the taken quantity would be

$$4\pi {\in }_0\times \frac{{\left(\frac{h}{2\pi }\right)}^2}{m_e{e}^2}=\frac{1}{9\times {10}^9}\times \frac{{\left(\frac{6.63\times {10}^{-34}}{2\times 3.14}\right)}^2}{9.1\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}=0.53\times {10}^{-10}m$$

Clearly, the value of the quantity taken is of the order of the atomic size.

15. The total energy of an electron in the first excited state of the hydrogen atom is about $-3.4eV$.

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

Ans: Kinetic energy of the electron is the same as the negative of the total energy.

$\Rightarrow K=-E$

$$\Rightarrow K=-(-3.4)=+3.4eV$$

Clearly, the kinetic energy of the electron in the given state is $+3.4eV$.

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

Ans: Potential energy $\left(U\right)$ of the electron is the same as the negative of twice of its kinetic energy

$\Rightarrow U=-2K$

$$\Rightarrow U=-(-3.4)=-6.8eV$$

Clearly, the potential energy of the electron in the given state is $-6.8eV$.

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

Ans: The potential energy of a system is dependent on the reference point taken. Here, the potential energy of the reference point is considered to be zero.

When the reference point is changed, then the magnitude of the potential energy of the system also changes.

As total energy is the sum of kinetic and potential energies, total energy of the system would also differ.

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: It is not much spoken about the quantization of orbits of planets around the Sun since the angular momentum associated with planetary motion is largely relative to the value of constant $\left(h\right)$. 

The angular momentum of the Earth in its orbit is of the order of ${10}^{70}h$. This causes a very high value of quantum levels $n$ of the order of ${10}^{70}$. 

When large values of $n$ are considered, successive energies and angular momenta are found to be relatively very small. Clearly, the quantum levels for planetary motion are considered continuous.

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 $\left({\mu }^{-}\right)$mass about $207m_e$. orbits around a proton).

Ans: To obtain the first Bohr's radius and the ground state energy of a muonic hydrogen atom, consider the mass of a negatively charged muon to be $m_{\mu }=207m$.

Now, according to Bohr’s model,

Bohr radius, $r_e\propto \left(\frac{1}{m_e}\right)$

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

Also, energy of a ground state muonic hydrogen atom, $E_{\mu }\propto m_e$.

It is known that the value of the first Bohr orbit, $r_e=0.53A=0.53\times {10}^{-10}m$

Consider $r_{\mu }$ to be the radius of muonic hydrogen atom.

At equilibrium, we have the relation:

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

$\Rightarrow 207m_e\times r_{\mu }=m_e{r}_e$

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

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

Now, we have, 

$E_e=-13.6eV$
Taking the ratio of these energies as $\frac{E_e}{E_{\mu }}=\frac{m_e}{m_{\mu }}$

$\Rightarrow \frac{E_e}{E_{\mu }}=\frac{m_e}{207m_e}$

$\Rightarrow E_{\mu }=207E_e$

$\Rightarrow E_{\mu }=207\times (-13.6)=-2.81keV$

Clearly, the ground state energy of a muonic hydrogen atom is $-2.81keV$.

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