Size of The Nucleus – Rutherford Gold Foil Experiment
Physics requires an understanding of matter’s underlying structure. Without the Rutherford gold foil experiment, it would be impossible to determine the size of the nucleus, which is the subject of this article. The Rutherford atom model was the first proper interpretation of the atom, and it served as the foundation for Bohr’s interpretation.
Rutherford demonstrated from his experiments that the radius of a nucleus is smaller than the radius of an atom by a factor of about 10 4 and the atomic nucleus is the central core of every atom. The nucleus contains enter a positive charge and more than 99.9 % of the mass of the atom, in other words, an atom is almost empty.
An atom consists of a central positively charged region called the nucleus which consists of positively charged particles called protons and neutral particles called neutrons.
Rutherford Gold Foil Experiment
J.J Thompson in 1898, proposed a model of the atom which looked more or less like plum pudding are raisin pudding. He assumed an atom to be a spherical body in which electrons are unevenly distributed in a sphere having a positive charge which balances the electron’s charge. It is called the Plum Pudding model.
The mass of the atom is assumed to be uniformly distributed over the whole atom. It failed after rutherford’s α scattering experiment which proved the atom to be quite different.
In this experiment Rutherford allows a narrow beam of α particles to fall on a very thin gold foil. This gold foil had a circular fluorescent Zinc sulphide screen around it. The α particles emitted by radioactive substances are dipositive Helium ions (He+2 ) having a mass of 4 units and 2 units of positive charge. A tiny flash of light was produced at the point where α particles truck.
Observations of Rutherford Gold Foil Experiment:
- Most of the Alpha particles passed through the foil without any deflection.
- Few α particles were deflected by small angles.
- Very few Alpha particles (1 out of 20,000 particles) completely rebound. i.e. deflected at ∼180.
Conclusion of Rutherford Gold Foil Experiment:
- The presence of largely empty space in the atom.
- The positive charge is concentrated at a very small region and is not uniformly distributed in the whole atom (If not then a large number of α particles would have been deflected by experiencing the enormous repulsive force from the positive charge of the atom).
- The positively charged core is known as the nucleus.
- Based on the Alpha scattering experiment Rutherford gave the nuclear model of an atom
Features of Rutherford Gold Foil Experiment:
- In an atom, the mass and positive charge are centrally located in an extremely small region called the nucleus.
- The volume of the nucleus is negligible as compared to the total volume of the atom.
- Both protons and neutrons present in the nucleus are collectively called nucleons.
- Extranuclear part: the nucleus is surrounded by revolving electrons. electrons revolve with very high speed in circular paths called orbits, so as in counterbalance of the electrostatic force of attraction between protons and electrons.
- Rutherford’s model of atom resembles the solar system in which the nucleus plays the role of the sun and the electrons of revolving planets.
- The centrifugal force acting outwards balances the inward coulombic attraction by the nucleus. This stabilizes the electrons in their orbits.
A new model of the atom was required to explain these findings. The positive stuff was thought to be concentrated in a tiny but vast location dubbed the nucleus in the new concept. To complete this picture, electrons were thought to be rotating around the nucleus, preventing one atom from intruding on its neighbour’s space.
Size of Nucleus
The first experimental determination of a size of a nucleus was made from the results of Rutherford scattering of α particles. Distance of closest approach was found to be read into 3 ×10-14 m for 7.7 MeV energetic α particles. This fact indicated that the size of the nucleus should be less than 3 ×10-14 m. For α particles having a kinetic energy of more than 7.7 MeV, the distance of the closest approach will be smaller.
At K.E more than 5.5 MeV distance of closest approach will be smaller. At K.E more than 5.5 MeV, attractive nuclear forces start affecting the Coulomb’s repulsive force between α particles and gold nucleus. The size of the nucleus can be measured by using fast electrons instead of α particles for the scattering experiment. The nuclear size was found to vary linearly with the mass number (A). Since the nucleus is supposed to be spherical, having radius R.
R = R0A1/3
where R0 = 1.2×10-15 m
A nucleus’ density (ρ) is equal to its mass divided by its overall volume. Nucleons are the number of protons and neutrons in a nucleus, and their mass is A times the mass of the nucleon (A is the number of nucleons in the atom).
Nuclear density, ρ = mass of nucleus /volume of nucleus = (mass of proton or neutron)(mass number) / Volume of the nucleus.
Consider A nucleus with the mass number A and radius R.
The volume of nucleus = 4/3 πR3
R = R0A1/3
where R0 is a constant.
The volume of nucleus=4/3 π R03A
ρ = (1.66×12-27)A/4/3 π R03A
= (1.66×10-27)/4/3π (1.2×10-15)3
R0 = 1.2×10-15m
ρ = 3 ×1.66×10-27/4×3.14(1.2×10-15)3
= 2.38×1017kg m-3
Thus nuclear density ρ is constant. That is independent of mass number for all nuclei.
- Heavier nuclei are bigger in size than lighter nuclei.
- Nuclear density is independent of the mass number of an atom.
- Nuclear density (=1017 kgm-3)is about 1013 times the average density of earth (=104 kg m-3)
Question 1: Calculate the radius of a nucleus of mass number 8.
The radius of a nucleus is given by,
R = R0A1/3
R0 =1.2×10-15 m; A=8
=1.2 ×10-15 ×2 =2.4 ×10-15 m.
Question 2: What is the nuclear radius of 125Fe, if that of 27Al is 3.6 Fermi?
R = R0A1/3
RFe/RAl = (AFe/AAl)1/3
RFe = 5/3 RAl
= 5/3 × 3.6
= 6.0 Fermi
Question 3: Find the mass density of the oxygen nucleus 8O16.
Since, Volume, V = 4/3π R3 = 4/3 π (1.2×10-15)3 A
=1.16 ×10-43 m3
Mass of oxygen atoms (A =16) is approximately 16 u.
Therefore density is,
ρ = m/V
= 16 × 1.66 × 10-27 / 1.16 × 10-43
= 2.3 × 1017 kg/m3.
Question 4: Find the ratio of nuclear radii of tungsten isotope 74W186 and iron isotope 26Fe56.
Mass number of tungsten, A1 = 186
Nuclear radius of tungsten R1 = R0A11/3
Mass number of iron, A2 = 56
Nuclear radius of iron, R2 = R0A21/3
R1/R2 = (A1/A2)1/3
R1/R2 = 1.491
Taking log of equation (i) ,we get
log (R1/R2) = 1/3 log (3.32)
= 1/3(0.5211) = 0.1737
Taking antilog both sides,
R1/R2. = 1.491
Question 5: Given the mass of the iron nucleus as 55.85 u with A = 56. Find the nuclear density.
M = 55 × 85 u
= 55.85 × 1.67 × 10-27 kg
Volume, V = 4/3 π R3 = 4/3 (R0A1/3)3
= 4/3π R03 × A
Nuclear density = M/V = 3M/4π R03A
= 3 × 55.85 (1.67×10-27) / 4 × 3.14(1.2 × 10-15)3 × 56
= 2.29×1017 kg/m3