In Rutherford's alpha-scattering experiment a narrow beam of alpha particles of the same kinetic energy from a radioactive source were targeted at a thin piece of gold foil which was only a few atomic layers thick. The alpha particles were scattered by the foil and detected on the zinc sulphide screen that was mounted in front of a microscope. Each alpha particle hitting the fluorescent screen produced a tiny speck of light. The microscope was moved in order to count the number of alpha particles scattered through the different values of angle θ per minute.
The results led to the following significant observations (which rued out the Thomson plum-pudding):
- Most of the alpha particles passed straight through the thin gold foil with little scattering (1/2000 was scattered).
- This means that most of the atom was empty space with most of the mass concentrated in a small region (the nucleus).
- Very few (1/10000) were deflected through angles greater than 90°
- This concluded that the nucleus had a positive charge (as it repelled few positive alpha particles). The charge on the nucleus is quantised to +Ze (where Z is the atomic number of the element)
The scattering of the alpha particles can be modelled in terms of Coulomb's law (any two point charges exert an electrostatic (electrical) force on each other that is directly proportional to the product of their charges and inversely proportional to the square of the distance between them).
Rutherford predicted that a fraction of alpha particles would be scattered through an angle θ. He found that more energetic alpha particles managed to get much closer to the nucleus. He concluded that the nucleus has a radius of about 10-14 m. In one experiment he used alpha particles of Ek 1.2 x 10-12 J. The distance (d) of the closest approach between and alpha particle and the gold nucleus can be calculated using the idea of conservation of energy. At this distance the alpha particle momentarily stops meaning that initial Ek = electric potential energy at d:
E = Qq/4πε0r = 1.2 x 10-12 = Qq/4πε0d
(NOTE: Q = Ze = 79e, q = 2e)
1.2 x 10-12 = (79 x 2 x e) / (4πε0d)
d = 3.0 x 10-14 ≈ 10-14 m
This gives an upper limit for the radius of the gold nucleus. More energetic alpha particles might get closer. The order of magnitude for the value of the radius of a nucleus is ~ 10-15 m. The radius of most atoms is 10-10 m so the nucleus is 105 times smaller than the atom.
The nuclear model of the atom
The nucleus of an atom contain positive protons and uncharged neutrons (a proton and a neutron approximately have the same mass). Isotopes are nuclei of the same element that have the same number of protons but different numbers of neutrons. Isotopes of an element undergo the same chemical reactions.The masses of atoms and nuclear particles are often expressed in atomic mass units (u). 1 u is one twelfth the mass of a neutral carbon-12 atom. The experimental value of 1 u is about 1.661 x 10-27 kg.
The radius of a nucleus depends on the nucleon number (A) of the nucleus. Fast moving electrons have a de Broglie wavelength of ~ 10-15 m. Diffraction of these electrons can be used to determine the radii of isotopes. The radius (R) of a nucleus is given by the equation:
R = r0 A1/3
r0 has an approximate value of 1.2fm. All nuclei have a density of about 1017 kg m-3.
Regarding the strong nuclear force, lets take a helium-r nucleus as an example. The two protons are separated by a distance of 10-15 m and exert a large electrostatic force on each other. According to Coulomb's law:
F = Qq/4πε0r2
e2/ 4πε0(10-15)2 ≈ 230N
(THIS IS NOT THE STRONG NUCLEAR FORCE - THIS IS THE REPULSIVE FORCE)
This is extremely large but the protons do not fly apart. Basically, the attractive gravitational force between the protons is very small (10-34 N) so there must be another force acting on the protons. This is the strong nuclear force. The strong nuclear force acts between all nucleons and is a very short range force (effective over only a few femtometres). The force is attractive to about 3 fm and repulsive below 0.5 fm.
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