Revision Notes for Class 12 Physics Chapter 12 Atoms

Revision Notes for Class 12 Physics Chapter 12 Atoms

Thomson’s model of atom:

It was proposed by J. J. Thomson in 1898. According to this model, the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon.

Rutherford’s a-scattering experiment

Rutherford and his two associates, Geiger and Marsden, studies the scattering of the a-particles from a thin gold foil in order to investigate the structure of the atom.

Rutherford’s observations and results:

– Most of the a -particles pass through the gold foil without any deflection. This shows that most of the space in an atom is empty.
– Few a-particles got scattered, deflecting at various angles from 0 to θ. This shows that atom has a small positively charged core called ‘nucleus’ at centre of atom, which deflects the positively charged a-particles at different angles depending on their distance from centre of nucleus.
– Very few a-particles (1 in 8000) suffers deflection of 180°. This shows that size of nucleus is very small, nearly 1/8000 times the size of atom.

Graph of deflection of number of particles with angle of deflection θ:

Rutherford’s α-scattering formulae

Number of a particles scattered per unit area, N(θ) at scattering angle q varies inversely as sin⁴(θ/2), i.e.,

$$
N(\theta) \propto \frac{1}{\sin ^4(\theta / 2)}
$$

Impact parameter:

It is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the centre of the nucleus, when the particle is far away from the nucleus of the atom.

– The scattering angle θ of the α particle and impact parameter b are related as

$$
b=\frac{Z e^2 \cot (\theta / 2)}{4 \pi \varepsilon_0 K}
$$

where K is the kinetic energy of the α-particle and Z is the atomic number of the nucleus.

– Smaller the impact parameter, larger the angle of scattering θ.

Distance of closest approach: At the distance of closest approach whole kinetic energy of the alpha particles is converted into potential energy.

– Distance of closest approach

$$
r_0=\frac{2 Z e^2}{4 \pi \varepsilon_0 K}
$$

Rutherford’s nuclear model of the atom:

According to this the entire positive charge and most of the mass of the atom is concentrated in a small volume known as the nucleus with electrons revolving around it just as planets revolve around the sun.

Bohr’s model:

Bohr combined classical and early quantum concepts and gave his theory of hydrogen and hydrogen-like atoms which have only one orbital electron. His postulates are

(i) An electron can revolve around the nucleus only in certain allowed circular orbits of definite energy and in these orbits it does not radiate.
(ii) These orbits are known as stationary orbits.
(iii) Angular momentum of the electron in a stationary orbit is an integral multiple of h/2π.

$$
\text { i.e., } L=\frac{n h}{2 \pi} \quad \text { or, } \quad m v r=\frac{n h}{2 \pi}
$$

This is known as Bohr’s quantisation condition.

The emission of radiation takes place when an electron makes a transition from a higher to a lower orbit. The frequency of the radiation is given by

$$
v=\frac{E_2-E_1}{h}
$$

where E₂ and E₁ are the energies of the electron in the higher and lower orbits respectively.

Bohr’s formulae

Radius of nth orbit

$$
r_n=\frac{4 \pi \varepsilon_0 n^2 h^2}{4 \pi^2 m Z e^2} ; \quad r_n=\frac{0.53 n^2}{Z} \
$$

Velocity of the electron in the nth orbit

$$
v_n=\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}=\frac{2.2 \times 10^6 Z}{n} \mathrm{~m} / \mathrm{s}
$$

The kinetic energy of the electron in the nth orbit

$$
\begin{aligned}
K_n & =\frac{1}{4 \pi \varepsilon_0} \frac{Z e^2}{2 r_n}=\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2} \\
& =\frac{13.6 Z^2}{n^2} \mathrm{eV}
\end{aligned}
$$

The potential energy of the electron in the nth orbit

$$
\begin{aligned}
U_n & =-\frac{1}{4 \pi \varepsilon_0} \frac{Z e^2}{r_n}=-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 m e^4 Z^2}{n^2 h^2} \\
& =\frac{-27.2 Z^2}{n^2} \mathrm{eV}
\end{aligned}
$$

Total energy of electron in the nth orbit

$$
\begin{aligned}
E_n & =U_n+K_n=-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2} \\
& =-\frac{13.6 Z^2}{n^2} \mathrm{eV} \\
K_n & =-E_n, U_n=2 E_n=-2 K_n
\end{aligned}
$$

Frequency of the electron in the nth orbit

$$
v_n=\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}=\frac{6.62 \times 10^{15} Z^2}{n^3}
$$

Wavelength of radiation in the transition from

$$n_2 \rightarrow n_1$$ is given by $$\frac{1}{\lambda}=R Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$$ where ‘R’ is called Rydberg’s constant.
$$
R=\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{2 \pi^2 m e^4}{c h^3}=1.097 \times 10^7 \mathrm{~m}^{-1}
$$

Spectral series of hydrogen atom:

When the electron in a H-atom jumps from higher energy level to lower energy level, the difference of energies of the two energy levels is emitted as radiation of particular wavelength, known as spectral line. Spectral lines of different wavelengths are obtained for transition of electron between two different energy levels,
which are found to fall in a number of spectral series given by

Lyman series

– Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 2, 3, …,∞) to first energy level (n1 = 1) constitute Lyman series.

$$
\frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{n_2^2}\right]
$$
where $$n_2=2,3,4, \ldots \ldots, \infty$$

– Series limit line (shortest wavelength) of Lyman series is given by

$$
\frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]=R \quad \text { or } \quad \lambda=\frac{1}{R}
$$

– The first line (longest wavelength) of the Lyman series is given by

$$
\frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3 R}{4} \quad \text { or } \quad \lambda=\frac{4}{3 R}
$$

– Lyman series lie in the ultraviolet region of electromagnetic spectrum.
– Lyman series is obtained in emission as well as in absorption spectrum.

Balmer series

– Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 3, 4, ….∞) to second energy level (n1 = 2) constitute Balmer series.

$$
\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{n_2^2}\right]
$$
where $$n_2=3,4,5 \ldots \ldots, \infty$$

– Series limit line (shortest wavelength) of Balmer series is given by

$$
\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{\infty^2}\right]=R \quad \text { or } \quad \lambda=\frac{4}{R}
$$

– The first line (longest wavelength) of the Balmer series is given by

$$
\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5 R}{36} \quad \text { or } \quad \lambda=\frac{36}{5 R}
$$

– Balmer series lie in the visible region of electromagnetic spectrum.
– This series is obtained only in emission spectrum.

Paschen series

– Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 4, 5, …..,∞) to third energy level (n1 = 3) constitute Paschen series.

$$
\frac{1}{\lambda}=R\left[\frac{1}{3^2}-\frac{1}{n_2^2}\right]
$$
where $$n_2=4,5,6 \ldots \ldots, \infty$$

• Series limit line (shortest wavelength) of the Paschen series is given by:

$$
\frac{1}{\lambda}=R\left[\frac{1}{3^2}-\frac{1}{\infty^2}\right]=R \quad \text { or } \quad \lambda=\frac{9}{R}
$$

• The first line (longest wavelength) of the Paschen series is given by:

$$
\frac{1}{\lambda}=R\left[\frac{1}{3^2}-\frac{1}{4^2}\right]=\frac{7 R}{144} \quad \text { or } \quad \lambda=\frac{144}{7 R}
$$

– Paschen series lie in the infrared region of the electromagnetic spectrum.
– This series is obtained only in the emission spectrum.

Brackett Series

– Emission spectral lines corresponding to the transition of electron from higher energy levels (n2 = 5, 6, 7, …..,∞) to fourth energy level (n1 = 4) constitute Brackett series.

$$
\frac{1}{\lambda}=R\left[\frac{1}{4^2}-\frac{1}{n_2^2}\right]
$$
where $$n_2=5,6,7 \ldots \ldots, \infty$$

Series limit line (shortest wavelength) of Brackett series is given by

$$
\frac{1}{\lambda}=R\left[\frac{1}{4^2}-\frac{1}{\infty^2}\right]=R \quad \text { or } \quad \lambda=\frac{16}{R}
$$

– The first line (longest wavelength) of Brackett series is given by

$$
\frac{1}{\lambda}=R\left[\frac{1}{4^2}-\frac{1}{5^2}\right]=\frac{9 R}{400} \quad \text { or } \quad \lambda=\frac{400}{9 R}
$$

– Brackett series lie in the infrared region of the electromagnetic spectrum.
– This series is obtained only in the emission spectrum.

Pfund Series

– Emission spectral lines corresponding to the transition of electron from higher energy levels (_n2 = 6, 7, 8,…….,∞) to 5th energy level (n₁ = 5) constitute Pfund series.

$$
\frac{1}{\lambda}=R\left[\frac{1}{5^2}-\frac{1}{n_2^2}\right]
$$
where $$n_2= 6,7,8 \ldots \ldots, \infty$$

– Series limit line (shortest wavelength) of Pfund series is given by

$$
\frac{1}{\lambda}=R\left[\frac{1}{5^2}-\frac{1}{\infty^2}\right]=R \quad \text { or } \quad \lambda=\frac{25}{R}
$$

– The first line (longest wavelength) of the Pfund series is given by

$$
\frac{1}{\lambda}=R\left[\frac{1}{5^2}-\frac{1}{6^2}\right]=\frac{11 R}{900} \quad \text { or } \quad \lambda=\frac{900}{11 R}
$$

– Pfund series also lie in the infrared region of electromagnetic spectrum.
– This series is obtained only in the emission spectrum.

Download Revision Notes for CBSE Class 12 Physics Chapter 12 Atoms

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Revision notes are an essential tool for students preparing for the CBSE Class 12 Physics exam. Here are some of the reasons why revision notes are important:

1. Concise and Organized: Revision notes are a condensed version of the course material that have been organized in a clear and concise manner. This makes it easier for students to review the key concepts and important topics quickly, which can be helpful when preparing for exams.
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