CBSE Class 12 Chemistry Atoms Exams Notes

CBSE Class 12 Chemistry Atoms Exams Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Class 12 Physics: Atoms Exam Notes

1. ALPHA PARTICLE SCATTERING EXPERIMENT

Thomson’s Atom Model

1. J. J. Thomson presented an atomic model in which positive charge is distributed uniformly in a small spherical space of the atom and electrons are embedded inside it like the seeds of a watermelon embedded in its pulp. The model was called the watermelon model or plum pudding model.
2. The magnitude of positive charge was taken equal to the total negative charge of electrons to explain the electrical neutrality of the atom.
3. The model failed to explain the emission of discrete wavelengths from atoms. Also such a model of atom cannot form a stable structure.

Geiger–Marsden Experiment

Geiger and Marsden used a bismuth source, placed inside a thick block of lead, in order to get α-particles. The α-particles having energy of 5.5 MeV coming out of a slit in the block were scattered when incident on a thin foil of gold of thickness \(2.1\times10^{-7}\) m. Scintillations were observed on a screen of zinc sulphide mounted on a (detector) microscope. The whole arrangement was enclosed in a cylinder having thick walls. The cylinder was mounted on a thick disc which could be rotated. The source and foil were kept steady on the base but could be rotated around the microscope. Each α-particle striking the screen created a bright spot in it. By counting the bright spots formed, the number of α-particles striking in a given time interval can be decided.

The graph shows the number of α-particles scattered at different angles in a given time interval. The number of α-particles scattered is about \(10^5\) at \(15^\circ\) and about 80 at \(150^\circ\). (Dots indicate experimental values and the continuous line was obtained theoretically by Rutherford).

Rutherford’s Calculations

1. An α-particle can undergo only one scattering while passing through the thin foil.
2. An atom of gold remains almost stationary during the collision as it is about 50 times heavier than the α-particle.
3. Most of the α-particles pass through the foil without deviation. Of the scattered particles, most of them do so at small angles. It indicates that the atom contains a lot of empty space.
4. A few α-particles are scattered at large angles. Further, a very small number of particles are reflected back. This suggests that the total positive charge and total mass of the atom are concentrated in a very small central region of the atom, which he called the nucleus.

The repulsive force on an α-particle when incident on gold foil due to the nucleus of gold is

\[F = \frac{1}{4\pi\varepsilon_0}\frac{(Ze)(2e)}{r^2}\]

where \(2e\) = charge of the α-particle and \(Ze\) = charge of the gold nucleus (\(Z = 79\) for gold).

Obviously, the trajectory depends on the initial perpendicular distance of the velocity vector from the nucleus. This minimum perpendicular distance is called the distance of closest approach or impact parameter (\(b\)).

2. RUTHERFORD’S ATOM MODEL

1. The atom has a central part called the nucleus, which contains the entire positive charge and almost all the mass of the atom.
2. Electrons revolve round the nucleus and the required centripetal force is supplied by the electrostatic force of attraction between the electron and the nucleus.
3. The atom includes a wide empty area.
4. The number of electrons is such that the effect of the positive charge of the nucleus is nullified.

Failure of Rutherford’s Model

1. An electron moving around the nucleus is in accelerated motion and so it should emit an electromagnetic wave.
2. The electron must lose energy, go out of balance, spiral inward, and finally collapse on the nucleus (very quickly, within \(10^{-8}\) s). In such a case, the atom would be highly unstable. But this does not happen.
3. According to this model, the radiations emitted by the electron should be continuous. But analysis of radiations emitted by hydrogen shows spectral lines instead of a continuous band.

3. BOHR MODEL OF HYDROGEN ATOM

Atomic Spectra

Emission spectra: When an atomic gas or vapour is excited at low pressure, it emits radiations of certain specific wavelengths only. A spectrum of this kind is termed an emission line spectrum, as spectral analysis shows a few bright lines on a dark background.

The spectrum emitted by atomic hydrogen is shown in the figure.

Each element has a characteristic spectrum of radiation, which it emits. Therefore, the study of emission line spectra of a material can be used for the identification of the material.

Absorption spectra: When white light is passed through a gas and the transmitted light is analysed using a spectrometer, a few dark lines are observed in the spectrum. These dark lines correspond to the wavelengths absorbed by the gas and the set of these dark lines is called the absorption spectrum of the gas.

Spectral Series

The emission spectrum of a material consists of the wavelengths emitted by it. There is no regular pattern for the entire spectrum. But the spacing between lines within certain sets of the spectrum may change in a regular way. Each of these sets can be called a spectral series.

The atomic hydrogen emits a line spectrum consisting of various series. The first series observed was the Balmer series in the visible region. The wavelengths in the Balmer series follow a simple formula, which is given below.

\[\frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{n^2}\right),\quad n = 3,4,5,6,\ldots\qquad\text{Balmer series is observed in the visible region}\]

\(\lambda\) is the wavelength and \(R\) is a constant called the Rydberg constant with value \(1.097\times10^{7}\ \text{m}^{-1}\). Other series in the hydrogen spectrum and their respective formulae are given below:

Lyman series (UV):

\[\frac{1}{\lambda} = R\left(\frac{1}{1^2}-\frac{1}{n^2}\right),\quad n = 2,3,4,\ldots\]

Paschen series (IR):

\[\frac{1}{\lambda} = R\left(\frac{1}{3^2}-\frac{1}{n^2}\right),\quad n = 4,5,6,\ldots\]

Brackett series (IR):

\[\frac{1}{\lambda} = R\left(\frac{1}{4^2}-\frac{1}{n^2}\right),\quad n = 5,6,7,\ldots\]

Pfund series (IR):

\[\frac{1}{\lambda} = R\left(\frac{1}{5^2}-\frac{1}{n^2}\right),\quad n = 6,7,8,\ldots\]

Rutherford’s atom model could not explain why radiations are emitted in certain wavelengths only.

Bohr Model

Niels Bohr made modifications to Rutherford’s atom model. The postulates of the Bohr model are:

1. An electron in an atom could revolve in certain stable orbits without the emission of radiant energy.
2. Bohr quantisation condition: The electron revolves around the nucleus only in those orbits for which the angular momentum is an integral multiple of \(\dfrac{h}{2\pi}\), where \(h\) is Planck’s constant. The angular momentum (\(L\)) of the orbiting electron is quantised: \(L = \dfrac{nh}{2\pi}\).
3. Bohr frequency condition: An electron could make a transition from one of its specified non-radiating orbits to another. When the transition takes place to another orbit of lower energy, a photon is emitted having energy equal to the energy difference between the initial and final states. If \(\nu\) is the frequency of the emitted photon, then \(h\nu = E_i - E_f\). Energy is to be supplied to transfer an electron to a higher energy orbit.

Theory

Let an electron having mass \(m\) and charge \(e\) revolve around a nucleus having charge \(Ze\) in a circular orbit of radius \(r\) with a linear speed \(v\) as shown in the figure. The centripetal force is provided by the Coulomb force of attraction between the electron and the nucleus.

\[\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2}\qquad\text{…(i)}\]

The angular momentum of the electron in an orbit is \(L = mvr = \dfrac{nh}{2\pi}\), where \(n = 1, 2, 3, \ldots\) is the principal quantum number. Then

\[m^2v^2r^2 = \frac{n^2h^2}{4\pi^2}\qquad\text{…(ii)}\]

From (i) and (ii), the radius of the orbit is

\[r = \frac{n^2h^2\varepsilon_0}{\pi Ze^2m}\qquad\text{…(iii)}\]

The speed of the electron in the orbit:

\[v = \frac{Ze^2}{2\varepsilon_0nh}\qquad\text{…(iv)}\]

The energy of the electron in the orbit is given by:

\[\text{Kinetic energy, } K = \frac{1}{8\pi\varepsilon_0}\frac{Ze^2}{r}\qquad \text{Potential energy, } U = -\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r}\]

Total energy (in the \(n\)th orbit),

\[E_n = K+U = -\frac{1}{8\pi\varepsilon_0}\frac{Ze^2}{r}\qquad\text{…(v)}\]

The above equations show that K.E. = −(T.E.) and P.E. = 2 T.E. The negative sign indicates that the electron is bound to the nucleus.

From (iii) and (v) (using the value of \(r\)),

\[E_n = -\frac{Z^2e^4m}{8\varepsilon_0^2n^2h^2}\qquad\text{…(vi)}\]

Hydrogen Atom

For the hydrogen atom, \(Z = 1\). The radius of the \(n\)th orbit from (iii) is

\[r_n = \frac{n^2h^2\varepsilon_0}{\pi e^2m}\qquad\text{…(vii)}\]

The radius of the first (innermost) orbit is called the Bohr radius. It is denoted by \(a_0\) and can be obtained by putting \(n = 1\) and substituting for the symbols in (vii). The Bohr radius is calculated to be 0.53 Å.

Energy in the \(n\)th orbit from (vi):

\[E_n = -\frac{me^4}{8\varepsilon_0^2h^2}\cdot\frac{1}{n^2} = -\frac{13.6}{n^2}\ \text{eV}\qquad\text{…(viii)}\]

Using equation (viii), the energy of an electron in different orbits can be calculated for \(n = 1, 2, 3, \ldots\)

4. LINE SPECTRA OF HYDROGEN ATOM

Energy Levels: The total energy of the electron in the \(n\)th orbit of the hydrogen atom is given by:

\[E_n = -\frac{13.6}{n^2}\ \text{eV}\qquad\text{…(i)}\]

The negative sign indicates that the electron is bound to the nucleus by a force of attraction. Additional energy is to be supplied from outside to release an electron. The maximum energy of an electron is taken as zero (for \(n = \infty\)). The various energy levels for the electron are given below:

The energy level diagram is shown below:

Line Spectra of hydrogen atom

The line spectra of the hydrogen atom arise from the transition of the electron from higher energy levels to lower energy levels. The excess energy is emitted as a photon. Transition from a higher energy level \(n_i\) to a lower energy level \(n_f\) causes the emission of a photon of frequency \(\nu_{fi}\) such that:

\[h\nu_{fi} = E_{n_i}-E_{n_f} = \frac{me^4}{8\varepsilon_0^2h^2}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\]

\[\nu_{fi} = \frac{me^4}{8\varepsilon_0^2h^3}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\qquad\text{…(ii)}\]

Equation (ii) is referred to as the Rydberg formula for the spectrum of the hydrogen atom. But \(\nu_{fi} = \dfrac{c}{\lambda_{fi}}\). Then

\[\frac{1}{\lambda_{fi}} = \frac{me^4}{8\varepsilon_0^2h^3c}\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) = R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\]

\(R = \dfrac{me^4}{8\varepsilon_0^2h^3c}\) is called the Rydberg constant.

Various spectral series emitted by the hydrogen atom are explained as below:

Lyman series: arise due to the transition from any higher energy level to the ground state (\(n = 1\)). These are UV radiations.

\[\frac{1}{\lambda} = R\left(\frac{1}{1^2}-\frac{1}{n_i^2}\right),\qquad n_i = 2,3,4,\ldots\]

Balmer series: arise due to the transition from any higher energy level to the first excited state (\(n = 2\)). This is visible light.

\[\frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{n_i^2}\right),\qquad n_i = 3,4,5,\ldots\]

Paschen series: arise due to the transition from any higher energy level to the second excited state (\(n = 3\)). These are IR radiations.

\[\frac{1}{\lambda} = R\left(\frac{1}{3^2}-\frac{1}{n_i^2}\right),\qquad n_i = 4,5,6,\ldots\]

Brackett series: arise due to the transition from any higher energy level to the third excited state (\(n = 4\)). These are also IR radiations.

\[\frac{1}{\lambda} = R\left(\frac{1}{4^2}-\frac{1}{n_i^2}\right),\qquad n_i = 5,6,7,\ldots\]

Pfund series: arise due to the transition from any higher energy level to the fourth excited state (\(n = 5\)). These are also IR radiations.

\[\frac{1}{\lambda} = R\left(\frac{1}{5^2}-\frac{1}{n_i^2}\right),\qquad n_i = 6,7,8,\ldots\]

Hydrogen gas can give rise to absorption spectra also. If photons with a continuous range of frequencies pass through rarefied hydrogen gas and the transmitted photons are analysed with a spectrometer, then a series of dark spectral absorption lines appear in the spectrum. These dark lines indicate the frequencies that have been absorbed. The absorbed energy is used to transfer electrons from lower energy levels to higher energy levels.

Note: A material absorbs the same wavelengths which it can emit.

Limitations of Bohr Model

1. Orbits of electrons need not be circular as assumed in the Bohr model.
2. The calculation of energies in the orbits involves a complex combination of classical mechanics and quantum principles.
3. Additional lines are seen with many of the spectral lines of hydrogen spectra, which cannot be explained on the basis of the Bohr model.
4. The Bohr model fails to explain the relative intensities of the spectral lines observed in the actual spectra.

5. DE BROGLIE’S EXPLANATION OF BOHR’S QUANTISATION

Going with the wave nature of electrons, the condition of non-radiation of energy can be linked to the formation of stationary waves. For the vibrations of a wire loop, the stationary waves would be formed if each wave joins smoothly with the next, i.e. the number of wavelengths on the loop must be an integer. Similarly, an electron can circle around an atomic nucleus indefinitely without radiating energy if the circumference of its orbit is an integral multiple of the wavelength of the electron.

\[2\pi r = n\lambda = \frac{nh}{p}\qquad\left(\text{as } \lambda = \frac{h}{p},\ \text{the de Broglie wavelength}\right)\]

\[\text{Or, } pr = n\frac{h}{2\pi}\ \Rightarrow\ mvr = \frac{nh}{2\pi}\qquad(\text{as } p = mv)\]