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Rutherford's α - particle scattering experiment
Rutherford and his associates studied the scattering of the α - particles by a thin gold foil in order to investigate the structure of the atom. An α-particle is a positively charged particle having a mass equal to that of helium atom and positive charge in magnitude equal to twice the charge of an electron. They are emitted by many radioactive elements. The scattering of α-particles provide useful information about the structure of the atom.
A fine pencil of α−particles was obtained from a radioactive material like radium or radon by placing it in a lead box with narrow opening as shown in Fig.
The α-particles are emitted from the source in all possible directions, but only a narrow beam emerges from the lead box. The remaining α-particles are absorbed in the lead box itself. After passing through the diaphragms D1 and D2, a narrow beam of α-particles incident on a thin gold foil, are scattered through different angles. The scattered α-particles strike a fluorescent screen coated with zinc sulphide. When the α-particles strike the screen, tiny flashes of light are produced. The observations can be made with the help of a low power microscope.
Observations and conclusions
(i) Most of the α particles either passed straight through the gold foil or were scattered by only small angles of the order of a few degrees.
This observation led to the conclusion that an atom has a lot of empty space as shown in Fig.
(ii) A few α particles were scattered in the backward direction, which led Rutherford to conclude that the whole of the positive charge was concentrated in a tiny space of about 10-14m. This region of the atom was named as nucleus. Only a small number of particles approaches the nucleus of the atom and they were deflected at large angles.
Distance of closest approach
An α particle directed towards the centre of the nucleus will move close upto a distance ro as shown in Fig, where its kinetic energy will appear as electrostatic potential energy. After this, the α particle begins to retrace its path. This distance ro is known as the distance of the closest approach.
Let m and v be the mass and velocity of the α particle directed towards the centre of the nucleus. Then, the kinetic energy of the particle,
Ek = ½ mv2 …………… (1)
Since, charge of an α-particle is 2e and that of the nucleus of the atom is Ze, the electrostatic potential energy of the α particle, when at a distance ro from the centre of the nucleus is given by,
Ep = ( 1/ 4πε0 ) . (2e )( Ze) / r0 ............(2)
where Z is the atomic number of the atom and εo, the permittivity of free space.
On reaching the distance of the closest approach ro, the kinetic energy of the α particle appears as its potential energy.
Ep = Ek
( 1/ 4πε0 ) . (2e )( Ze) / r0 = ½ mv2
r0 = ( 1/ 4πε0 ) 4Ze2 / mv2
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