NCERT Class 12 Physics Chapter 12: Complete Resource for Atoms
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Since they are never able to understand the basics and concepts completely, they try to escape from studying the subject. NCERT Solutions try to make learning easy and simplified for the students to understand the basics and solve questions easily. Chapter 12 of NCERT Solutions discusses Atoms. The structure of atoms, the origin of spectra, Bohr's Theory of Hydrogen Atoms, and XRays are the main topics of focus in the chapter.
In detail, Chapter 12 highlights the different models of Atom  Thomson's, Rutherford’s, and Bohr's atomic models; their pros, cons, and limitations; production, properties, and application of XRays and the use of Mosley's law in it. So, when we study about Atoms, it is going back to studying chemistry, but this time its application in machines. NCERT Solutions provide you with great content that is simplified as per the students' convenience and help them gain good marks.
Topics Covered in CBSE Chapter 12  Atoms Class 12
In chapter 12  Atoms following topics are covered:
Introduction to Atoms
Alphaparticle Scattering and Rutherford's Nuclear Model of Atom
Alphaparticle trajectory
Electron orbits
Atomic Spectra
Spectral series
Bohr Model of the Hydrogen Atom
Energy levels
The Line Spectra of the Hydrogen Atoms
De Broglie’s Explanation of Bohr's Second Postulate of Quantisation
Would you like to view a summarized version of this chapter? Check out the 'Chapter at a glance' section below the PDF of NCERT Solutions.
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Atoms Chapter at a Glance  Class 12 NCERT Solutions
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 plumpudding 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 αparticles through large angles as in case of Rutherford’s αscattering experiment.
1.2. Rutherford’s Alpha Scattering Experiment Observations
Most of αparticles were found to pass through the gold foil without any appreciable deflection.
The different αparticles in passing through the gold foil undergo different amounts of deflections. A large number of αparticles suffer fairly large deflections.
A very small number of αparticles (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 αparticles passed undeviated, the atom has a lot of empty space in it.
Since fast and the heavy αparticles 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 αparticles 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 αscattering 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 αparticle 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 αparticle is completely converted into potential energy. Mathematically.
$1/2mu^2=\frac{1}{4\pi \varepsilon _0}.\frac{(2e)(Ze)}{r_0}\;\;\;\therefore r_0=\frac{1}{4\pi\varepsilon_0}.\frac{2Ze^2}{1/2mu^2}$
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. $\frac{mv^2}{r}=\frac{kze^2}{r^2}$
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 = 1010 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:
Total energy of the electron in an atom is negative, indicating that it is bound. 
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) $\left [ \frac{1}{n^2_1}\frac{1}{n^2_2} \right ].$
ΔE = hν ; ν = frequency of spectral line emitted .
$\frac{1}{\lambda}=\bar{v}=$ wave no. ( no. of waves in unit length (1m))
$\frac{1}{\lambda} = R\left [ \frac{1}{n^2_1}\frac{1}{n^2_2} \right ].$
Where R = Rydberg's constant for hydrogen = 1.097 × 107 m1 .
For hydrogen like atom/species of atomic number Z:
$r_{nz}=\dfrac{Bohr\;radius}{Z}n^{2}=(0.529A^{^{\circ}})\dfrac{n^{2}}{Z}$
$E_{nz}=(13.6)\frac{Z^2}{n^2}ev$
Rz = RZ2 – Rydberg's constant for element of atomic no. Z
2.2 Spectral Series
Lyman Series: (Landing orbit n = 1).
Ultraviolet region $\bar{v}=R\left [ \frac{1}{1^2}\frac{1}{n^2_2} \right ];$ n2 > 1
Balmer Series: (Landing orbit n = 2)
Visible region $\bar{v}=R\left [ \frac{1}{2^2}\frac{1}{n^2_2} \right ];$ n2 > 2
Paschan Series: (Landing orbit n = 3)
In the near infrared region $\bar{v}=R\left [ \frac{1}{3^2}\frac{1}{n^2_2} \right ];$ n2 > 3
Bracket Series: (Landing orbit n = 4)
In the mid infrared region $\bar{v}=R\left [ \frac{1}{4^2}\frac{1}{n^2_2} \right ];$ n2 > 4
Pfund Series: (Landing orbit n = 5)
In far infrared region $\bar{v}=R\left [ \frac{1}{5^2}\frac{1}{n^2_2} \right ];$ 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:
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.
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.
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.
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 nonuniform mass distribution in Rutherford's model.
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;
$\frac{hc}{\lambda }=21.76\times {{10}^{19}}\left[ \frac{1}{n_{1}^{2}}\frac{1}{n_{2}^{2}} \right]$
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}}=\infty $.
$\Rightarrow$ $\frac{hc}{\lambda }=21.76\times {{10}^{19}}\left[ \frac{1}{{3}^{2}}\frac{1}{{\infty}^{2}} \right]$
$\Rightarrow$ $\lambda =\frac{6.6\times {{10}^{34}}\times 3\times {{10}^{8}}\times 9}{21.76\times {{10}^{19}}}$
$\Rightarrow$ $\lambda =8.189\times {{10}^{7}}m$
$\Rightarrow$ $\lambda =818.9nm$
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 $=E=\left( 13.6 \right)=13.6eV$
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.
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 $=6.6\times {{10}^{34}}Js$
$\Rightarrow {{\nu }_{1}}=\frac{{{\left( 1.6\times {{10}^{19}} \right)}^{2}}}{2\times 8.85\times {{10}^{12}}\times 6.62\times {{10}^{34}}}$
$\Rightarrow {{\nu }_{1}}=0.0218\times {{10}^{8}}$
$\Rightarrow {{\nu }_{1}}=2.18\times {{10}^{6}}m/s$
For level ${{n}_{2}}=2$ , we can write the relation for the corresponding orbital speed as;
${{\nu }_{2}}=\frac{{{e}^{2}}}{{{n}_{2}}2{{\in }_{0}}h}$
$\Rightarrow {{\nu }_{2}}=\frac{{{\left( 1.16\times {{10}^{19}} \right)}^{2}}}{2\times 2\times 8.85\times {{10}^{12}}\times 6.62\times {{10}^{34}}}$
$\Rightarrow {{\nu }_{2}}=1.09\times {{10}^{6}}m/s$
And, for ${{n}_{3}}=3$ , we can write the relation for the corresponding orbital speed as;
${{\nu }_{3}}=\frac{{{e}^{2}}}{{{n}_{3}}2{{\in }_{0}}h}$
$\Rightarrow {{\nu }_{3}}=\frac{{{\left( 1.16\times {{10}^{19}} \right)}^{2}}}{3\times 2\times 8.85\times {{10}^{12}}\times 6.62\times {{10}^{34}}}$
$\Rightarrow {{\nu }_{3}}=7.27\times {{10}^{5}}m/s$
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.
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}}$$=\frac{n_{1}^{2}{{h}^{2}}{{\in }_{0}}}{\pi m{{e}^{2}}}$
$h=$ Planck’s constant $=6.6\times {{10}^{34}}Js$
$e=$ Charge of an electron $=1.6\times {{10}^{19}}C$
${{\in }_{0}}=$ Permittivity of free space $=8.85\times {{10}^{12}}{{N}^{1}}{{C}^{2}}{{m}^{2}}$
$m=$ Mass of an electron $=9.1\times {{10}^{31}}kg$
$\Rightarrow {{T}_{1}}=\frac{2\pi \times {{\left( 1 \right)}^{2}}\times {{\left( 6.62\times {{10}^{34}} \right)}^{2}}\times 8.85\times {{10}^{12}}}{2.18\times {{10}^{6}}\times \pi \times 9.1\times {{10}^{31}}\times {{\left( 1.6\times {{10}^{19}} \right)}^{2}}}$
$\Rightarrow {{T}_{1}}=15.27\times {{10}^{17}}$
$\Rightarrow {{T}_{1}}=1.527\times {{10}^{16}}s$
For level ${{n}_{2}}=2$, we can write the orbital period as;
${{T}_{2}}=\frac{2\pi {{r}_{2}}}{{{\nu }_{2}}}$
Here,
${{r}_{2}}=$ Radius of the orbit in ${{n}_{2}}$$=\frac{n_{2}^{2}{{h}^{2}}{{\in }_{0}}}{\pi m{{e}^{2}}}$
$\Rightarrow {{T}_{1}}=\frac{2\pi \times {{\left( 2 \right)}^{2}}\times {{\left( 6.62\times {{10}^{34}} \right)}^{2}}\times 8.85\times {{10}^{12}}}{1.09\times {{10}^{6}}\times \pi \times 9.1\times {{10}^{31}}\times {{\left( 1.6\times {{10}^{19}} \right)}^{2}}}=1.22\times {{10}^{15}}s$
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}}}$
$\Rightarrow {{T}_{3}}=\frac{2\pi \times {{\left( 3 \right)}^{2}}\times {{\left( 6.62\times {{10}^{34}} \right)}^{2}}\times 8.85\times {{10}^{12}}}{7.27\times {{10}^{5}}\times \pi \times 9.1\times {{10}^{31}}\times {{\left( 1.6\times {{10}^{19}} \right)}^{2}}}=4.12\times {{10}^{15}}s$
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 \[n=1\text{ }to\text{ }n=3\] level.
During its deexcitation, the electrons can jump from \[n=3\text{ }to\text{ }n=1\] directly, which forms a line of the Lyman series of the hydrogen spectrum.
The formula for wave number for Lyman series is given as;
$\frac{1}{\lambda }={{R}_{y}}\left( \frac{1}{{{1}^{2}}}\frac{1}{{{n}^{2}}} \right)$
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,
$\frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{1}{{{1}^{2}}}\frac{1}{{{3}^{2}}} \right)$
\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( 1\frac{1}{9} \right)\]
\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{8}{9} \right)\]
$\Rightarrow \lambda =\frac{9}{8\times 1.097\times {{10}^{7}}}=102.55nm$
If the transition takes place from $n=3\text{ }to\text{ }n=2$ , and then from \[n=2\text{ }to\text{ }n=1\], then the wavelength of the radiation emitted in transition from \[n=3\text{ }to\text{ }n=2\] is given as;
$\frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{1}{{{2}^{2}}}\frac{1}{{{3}^{2}}} \right)$
\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{1}{4}\frac{1}{9} \right)\]
\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{5}{36} \right)\]
$\Rightarrow \lambda =\frac{36}{5\times 1.097\times {{10}^{7}}}=656.33nm$
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Now, the wavelength of the radiation when the transition takes place from \[n=2\text{ }to\text{ }n=1\] is given as;
$\frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{1}{{{1}^{2}}}\frac{1}{{{2}^{2}}} \right)$
\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( 1\frac{1}{4} \right)\]
\[\Rightarrow \frac{1}{\lambda }=1.097\times {{10}^{7}}\left( \frac{3}{4} \right)\]
$\Rightarrow \lambda =\frac{4}{3\times 1.097\times {{10}^{7}}}=121.54nm$
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.
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.
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.
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.
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;
${{r}_{1}}=\frac{{{\left( \frac{h}{2\pi } \right)}^{2}}}{G{{m}_{p}}{{m}_{e}}}$
$\Rightarrow {{r}_{1}}=\frac{{{\left( \frac{6.63\times {{10}^{34}}}{2\times 3.14} \right)}^{2}}}{6.67\times {{10}^{11}}\times 1.67\times {{10}^{27}}\times {{\left( 9.1\times {{10}^{31}} \right)}^{2}}}$
$\Rightarrow {{r}_{1}}\approx 1.21\times {{10}^{29}}m$
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 deexcites from level $n$ to level $\left( n1 \right)$. 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 $\left( n1 \right)$. 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 $\left( n1 \right)$ is given as;
${{E}_{2}}=h{{\nu }_{2}}=\frac{hm{{e}^{4}}}{{{\left( 4\pi \right)}^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}}\times \frac{1}{{{\left( n1 \right)}^{2}}}$ …(2)
Here,
${{\nu }_{2}}=$ Frequency of radiation at level $\left( n1 \right)$
Energy $\left( E \right)$ released as a result of deexcitation;
$E={{E}_{2}}{{E}_{1}}$
$\Rightarrow h\nu ={{E}_{2}}{{E}_{1}}$ …(3)
Here,
$\nu =$ Frequency of radiation emitted
Using the values of equations (1) and (2) in equation (3), we get;
$\nu =\frac{m{{e}^{4}}}{{{\left( 4\pi \right)}^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}}\left[ \frac{1}{{{\left( n1 \right)}^{2}}}\frac{1}{{{n}^{2}}} \right]$
$\Rightarrow \nu =\frac{m{{e}^{4}}\left( 2n1 \right)}{{{\left( 4\pi \right)}^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}{{n}^{2}}{{\left( n1 \right)}^{2}}}$
For large $n$, we can write $\left( 2n1 \right)2n$ and $\left( n1 \right)\simeq n$
$\Rightarrow \nu =\frac{m{{e}^{4}}}{32{{\pi }^{3}}\in _{0}^{2}{{\left( \frac{h}{2\pi } \right)}^{3}}{{n}^{3}}}$ …(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 $\left( \sim {{10}^{10}}m \right)$.
Construct a quantity with the dimensions of length from the fundamental constants $e,{{m}_{e}}\text{ }and\text{ }c$. Also, determine its numerical value.
Ans: To construct a quantity with the dimensions of length from the fundamental constants $e,{{m}_{e}}\text{ }and\text{ }c$, 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
\[\frac{1}{4\pi {{\in }_{0}}}\times \frac{{{e}^{2}}}{{{m}_{e}}{{c}^{2}}}=9\times {{10}^{9}}\times \frac{{{\left( 1.6\times {{10}^{19}} \right)}^{2}}}{9.1\times {{10}^{31}}\times {{\left( 3\times {{10}^{8}} \right)}^{2}}}=2.81\times {{10}^{15}}m\]
Clearly, the numerical value of the taken quantity is much less than the typical size of an atom.
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 nonrelativistic 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 \[h,\multimap {{m}_{e}},\multimap and\multimap e\] will yield the right atomic size. Construct a quantity with the dimension of length from \[h,{{m}_{e}}\text{ }and\text{ }e\] and confirm that its numerical value has indeed the correct order of magnitude.
Ans: To construct a quantity with the dimension of length from \[h,{{m}_{e}}\text{ }and\text{ }e\], 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
\[4\pi {{\in }_{0}}\times \frac{{{\left( \frac{h}{2\pi } \right)}^{2}}}{{{m}_{e}}{{e}^{2}}}=\frac{1}{9\times {{10}^{9}}}\times \frac{{{\left( \frac{6.63\times {{10}^{34}}}{2\times 3.14} \right)}^{2}}}{9.1\times {{10}^{31}}\times {{\left( 1.6\times {{10}^{19}} \right)}^{2}}}=0.53\times {{10}^{10}}m\]
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 $3.4eV$.
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.
$\Rightarrow K=E$
\[\Rightarrow K=\left( 3.4 \right)=+3.4eV\]
Clearly, the kinetic energy of the electron in the given state is $+3.4eV$.
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
$\Rightarrow U=2K$
\[\Rightarrow U=\left( 3.4 \right)=6.8eV\]
Clearly, the potential energy of the electron in the given state is $6.8eV$.
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 $207{{m}_{e}}$. 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:
${{m}_{\mu }}{{r}_{\mu }}={{m}_{e}}{{r}_{e}}$
$\Rightarrow 207{{m}_{e}}\times {{r}_{\mu }}={{m}_{e}}{{r}_{e}}$
$\Rightarrow {{r}_{\mu }}=\frac{0.53\times {{10}^{10}}}{207}=2.56\times {{10}^{13}}m$
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 $\frac{{{E}_{e}}}{{{E}_{\mu }}}=\frac{{{m}_{e}}}{{{m}_{\mu }}}$
$\Rightarrow \frac{{{E}_{e}}}{{{E}_{\mu }}}=\frac{{{m}_{e}}}{207{{m}_{e}}}$
$\Rightarrow {{E}_{\mu }}=207{{E}_{e}}$
$\Rightarrow {{E}_{\mu }}=207\times \left( 13.6 \right)=2.81keV$
Clearly, the ground state energy of a muonic hydrogen atom is $2.81keV$.
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NCERT Solutions help you to improve your scores by clearing out the basics and other concepts of all the chapters involved, topic by topic. The topics are arranged so that you can easily understand all the basics as you proceed further. You can download the PDF at any time and study Physics offline for free. The study material is formulated and designed by professional teachers to gain maximum benefits and seamlessly put yourself into studying and understanding all the chapters topic by topic so that Physics does not seem like a burden to you anymore. The PDF of Chapter 12 includes numerical questions, multiplechoice questions, short and long answer questions so that you can practice and improve your knowledge.
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NCERT Solutions for Class 12 Physics Chapter 12 – Atoms
Based on plenty of equations by Bohr, Chapter 12 Atom of Class 12 NCERT Solutions discusses how spectra are formed by emission and absorption of energy by various atoms. It discusses and defines the energy levels of different atoms and how and why they emit and absorb wavelengths of light, hence forming a spectrum.
It also focuses on the production of Xrays by Coolidge Tubes and its properties and applications and how to control its penetration power and intensity. It's all about how an atom, the smallest unit of matter, has its applications in machines that we use in day to day lives. Since this chapter includes many equations and constants to be learned and understood to be able to apply them in numerical, NCERT Solutions prove to be useful.
Benefits of NCERT Solutions For Class 12 Physics
Studying from a book can be tiring and difficult for students due to the lack of explanations of basic concepts.
Many students lose interest in Physics because they struggle to understand the concepts and theories presented in books.
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Answering practice questions successfully boosts students' morale and motivates them to study further.
If students fail to comprehend the explanations, they may lose hope and give up on the subject.
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The key concepts covered in this chapter include the alphaparticle scattering experiment, the Bohr model, hydrogen spectrum, Rutherford's model of an atom, and energy levels.
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Conclusion
NCERT Solutions for Class 12 Physics Chapter 12 Atoms provide invaluable support and resources for students studying this important topic. The chapter covers essential concepts such as the alphaparticle scattering experiment, the Bohr model, hydrogen spectrum, Rutherford's model of an atom, and energy levels. By offering comprehensive solutions and explanations, NCERT Solutions enable students to understand and solve the questions effectively. Additionally, the availability of revision notes, important questions, sample papers, and related study materials further enhances the students' understanding of the chapter. With NCERT Solutions, students can overcome the challenges of studying atoms and gain confidence in their knowledge and problemsolving abilities. These solutions serve as a valuable tool in helping students succeed in their Class 12 Physics examinations.
FAQs on NCERT Solutions for Class 12 Physics Chapter 12  Atoms
1. What sort of Questions Does NCERT Solutions Class 12 Physics Offer for Chapter 12?
NCERT Solutions for Class 12 Physics Chapter Atoms provides you with a variety of questions carefully selected by experts of the subject that cover all the topics that have been discussed in the chapter so that you understand and practice all sorts of questions. The solutions provided are perfectly correct and accurate, so you do not need to bother yourself with that.
2. Do NCERT Solutions Prove to be Helpful in Cracking Competitive Exams?
Competitive exams try to sum up their question from the basics, and students tend to focus on complex studies and end up scoring low in competitive exams. Since NCERT Solutions provides you with a wide variety of questions covering all the chapters and subjects' topics, it is a very good option to choose if you are preparing for competitive exams. Suppose you study from the NCERT Solutions of class 10 and 12 while you are doing your schooling. In that case, you will have already gained good knowledge about the basic concepts of your subjects, and then you can spend your time practising questions framed by qualified and experienced teachers.
3. Are NCERT Solutions Sufficient Enough to help Score High Marks in 10th and 12th?
Yes, NCERT Solutions are well modified and curated according to the CBSE guidelines and study pattern. So you can trust your exams with NCERT Solutions, provided that you make good use of them. It provides you with a generous amount of questions to solve, which improves your knowledge and timing. The more the number of questions you solve, the quicker you become at solving them. NCERT Solutions provide accurate and reliable solutions to all the given questions. Formulated by experts, it includes all sorts of questions ranging from easy to tough to test your understanding and where you currently stand.
4. Can you please provide a Stepwise Study Plan to ace Class 12 Physics?
Class 12 is considered to be the major turning point in any student's life. If you are a science student, you’ll be studying Physics. A few things you should keep in mind while studying this subject are :
Adhere to the syllabus given by CBSE.
Select the books wisely.
Once the concept is clear, revise and practice regularly.
Solve mock papers and previous years question papers.
5. What are the concepts covered in Class 12 Physics Chapter 12?
The concepts covered in Class 12 Physics Chapter 12 are:
Introduction
AlphaParticle Scattering and Rutherford’s Nuclear Model Of Atom
Alphaparticle Trajectory
Electron Orbits
Atomic Spectra
Spectral Series
Bohr Model of The Hydrogen Atom
Energy Levels
The Line Spectra of The Hydrogen Atom
De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation
6. What is an atom Class 12 Physics?
An atom is a matter particle that specifically defines a chemical element. An atom is composed of a central nucleus surrounded by one or more electrons. Each electron carries a negative charge. The nucleus is positively charged and contains one or more protons and neutrons, which are relatively heavy particles. If you want to know more about atoms indepth, refer to the study material available on Class 12 Physics Chapter 12, and increase your knowledge. The solutions provided are free of cost. They are also available on the Vedantu Mobile app.
7. What are Energy levels Class 12 Physics?
In Class 12 Physics Chapter 12, stationary states or energy levels refer to the various orbits in which electrons circulate. These stationary states/energy levels for an electron are denoted by the numbers n = 1, 2, 3........... These integers are often referred to as the primary quantum numbers. The energy of the stationary state in which an electron is placed is given by:
En=−Rh1/n2
Where RH is called the Rydberg constant whose value is 2.18 × 10–18 J.
8. What are some muststudy chapters in Physics Class 12 to score good marks in the CBSE board exams?
When you prepare for CBSE Board exams, you must keep in mind the chapters that hold the maximum weightage of marks. If you focus on these chapters and practice their sample questions, you can surely expect to score good marks. The units which are easy and hold maximum marks in Physics Class 12 are Electrostatics, Magnetism and Optics. Focus on clearing the concepts of the chapters in these units.