ALPHA PARTICLE SCATTERING AND RUTHERFORD'S NUCLEAR MODEL OF ATOM -

THOMSON’S MODEL OF ATOM

According to this an atom is a positively charged sphere $\fn_cm \large \left ( r\approx 10^{-10}m \right )$ in which the entire mass and +ve charge of the atom is uniformly distributed with -ve electrons embedded in it (plum pudding model).

Science, the atom is electrically neutral, so the number of electrons is such that their -ve charge is equal to the +ve charge of the atom.

DRAWBACKS OF THIS MODEL

It could not explain the large-angle scattering of α particles given by Rutherford.
It could not explain the discrete spectral lines emitted by H- atom (and other)

RUTHERFORD’S α- PARTICLE SCATTERING EXPERIMENT

At the suggestion of Ernst Rutherford in 1911, H.Geiger and E. Marsden performed some experiment.

In this experiment, they directed a beam of 5.5 Mev α particles emitted from a $\fn_cm \large _{83}^{214}\textrm{Bi}$ (bismuth)radioactive source at a thin metal foil made of gold.

The beam was allowed to fall on a thin foil of gold of thickness $\fn_cm 2.1 \times 10^{-7}m$ . The scattered α particles were observed through a rotatable detector consisting of a Zinc Sulphide (Zns) screen and a microscope. The scattered α particles on striking the screen produced light flashes.

the following graph shows the number of scattered particles detected as a function of the angle of scattering.

OBSERVATION:-

Most of the α particles pass straight through the gold foil. It means that they do not suffer any collision.
only about 0.14 % of the incident α particles scatter by more than $\fn_cm \large 1^{0}$ .
a small number of α particles suffered fairly large deflections.
about 1 in 8000 deflect by more than $\fn_cm \large 90^{0}$ or suffered deflection of nearly $\fn_cm \large 180^{0}$

CONCLUSION:-

Most of the portion of an atom is empty because it is easy to see why most α particles go right through the metal foil.
Every atom consists of a tiny central core called the nucleus which contains all the atom’s +ve charge and most of its mass (99.9 %). Rutherford argued that to deflect the α particle backward, it must experience a large repulsive force. This force could be provided if the greater part of the mass of the atom and its +ve charge were concentrated tightly at its center known as the nucleus.
From the experiment, the size of the nucleus to be about $\fn_cm \large 10^{-15}m$ to $\fn_cm \large 10^{-14}m$ and the size of the atom to be about 10,000 to 100,000 times larger than the size of the nucleus.
Since an atom is electrically neutral, so the number of electrons must be the same as the number of protons(in the nucleus) and the electrons occupy the space outside the nucleus. The atomic electrons being so light, do not appreciably affect the α particles.
Electrons are not stationary but revolve around the nucleus in various circular orbits.

IMPACT PARAMETER:-

Impact parameter defined as the perpendicular distance of the velocity vector of the α particles from the central line of the nucleus.

Rutherford calculated the relationship between the impact parameter b and scattering angle θ.

$\fn_cm \large \left [ b=\frac{1}{4\pi\epsilon_{0}}\frac{Ze^{2}cot(\frac{\theta}{2})}{K_{\alpha}} \right ]$

where $\fn_cm \large K_{\alpha}$ is K.E of α particle

a. if $\fn_cm \large b$ is large, $\fn_cm \large cot\left ( \frac{\theta }{2} \right )$ will be large i.e scattering angle will be small.

b. if $\fn_cm \large b$ is small, $\fn_cm \large cot\left ( \frac{\theta }{2} \right )$ will be small i.e scattering angle will be large.

c. if $\fn_cm \large b=0,\;\;cot\left ( \frac{\theta }{2} \right ),\;\;\frac{\theta }{2}=90^0\;\;and\;\;\theta =180^0$ , i.e α particle is directed towards the center of the nucleus, it retraces its path.

ELECTRON ORBITS:-

Electrons are not stationary but revolve around the nucleus in various circular orbits as do the planets around the sun. The electrostatic force of attraction between the revolving electrons and the nucleus provides the necessary centripetal force to keep them in orbits.

i.e for H- atom

$\fn_cm \large \frac{k\; e.e}{r^2}=\frac{mv^2}{r}$

$\fn_cm \large \left [\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}=mv^2 \right ]$

The K.E of the electrons in H-atom is

$\fn_cm \large K.E=\frac{1}{2}mv^2=\frac{1}{8\pi\epsilon_0}\frac{e^2}{r}$

The P.E of the electron in H-atom is

$\fn_cm \large U=-\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}\;\;\;(\because U=\frac{kq_1 q_2}{r})$

$\fn_cm \large \therefore$ Total energy E of the electron in H-atom is

$\fn_cm \large E=K+U$

$\fn_cm \large E=\frac{1}{8\pi\epsilon_0}\frac{e^2}{r}-\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}$

$\fn_cm \large E=-\frac{1}{8\pi\epsilon_0}\frac{e^2}{r}$

the total energy of the electron is -ve. This implies the fact that the electron is bound to the nucleus.

IMPORTANT LINKS OF WAVE OPTICS
Introduction	Alpha Particle Scattering and Rutherford’s Nuclear Model of Atom
Atomic Spectra	Bohr Model of the Hydrogen Atom
The Line Spectra of the Hydrogen Atom	De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation