History of discovery of neutron
History of discovery of neutron: Up to 1920, only three elementary particles electron, proton, and alpha-particle were known. They all were charged particles. In order to explain the tritium (₁H³) and deuterium (₁H²) nuclides which are isotopes of hydrogen and have one proton in them, Rutherford In 1920 suggested that there are must be some particle in these nuclides whose mass is nearly equal to that of proton and must be charged less so that approximate ratio 1:2:3 of ₁H¹, ₁H², ₁H³ masses obtained experimentally could find out theoretically. But Rutherford himself could not find this particle experimentally.
In 1930 Bothe and Becker during artificial transmutation experiment observed that then when a beryllium (Be) target was bombarded by alpha particle obtained from radioactive source polonium, a highly penetrating and least ionization power radiations are emitted by a target. They also found that these could not be deflected by electric or magnetic fields. They assumed that this radiation could be γ-rays produced in the following reaction.
₄Be⁹ + ₂He⁴ → ₆C¹³ + γ-rays
The energy of this radiation was calculated by measuring its absorption coefficient in lead. The energy of the assumed γ-rays was found to be about 7 MeV.
In 1932 Madam Curie and her husband Joliot Curie repeated this experiment and observed that when these radiations were incident on a block of paraffin wax, protons of energy up to about 5.7 MeV were emitted from the paraffin. From calculation they estimation by assuming radiations coming from beryllium target as γ-rays, the energy of incident radiations must be about 55 MeV for the exclusion of protons of energy 5.7 MeV from paraffin. This theoretical result contradicts the assumption.
For solving this difficulty Chadwick in 1932 proposed a suggestion that the radiations coming from the beryllium target may not be γ-rays, but maybe the electrically neutral particle of mass nearly equal to that of the proton. He called the particle neutron and emission of this particle was described by the following reaction:
₂He⁴(α-particle) + ₄Be⁹→ ₆C¹² + ₀n¹ (neutron)
In a series of experiments, Chadwick allowed the radiation emitted by the beryllium target to pass through the nitrogen contained in a Wilson cloud chamber. The cloud chamber was placed in a magnetic field. In some cloud chamber photographs, two tracks were observed starting from a common point. By measurement on the tracks, the particles causing the tracks were identified as a proton and a residual nitrogen nuclide. Chadwick measured the maximum values of the velocities of proton and residual nitrogen. The mass of the neutron was then calculated from these values assuming elastic collision of the neutron with hydrogen nucleus and nitrogen nuclide.
Let a neutron of mass m moving with a velocity V collides with a stationary hydrogen nuclide of mass M. Let the velocity of the neutron after the collision is V’ and the velocity of the hydrogen nucleus be Vº.
∴ From the law of conservation of momentum,
mV = mV’ + MVº
or m(V-V’) = MVº
or (V-V’) = MVº/m (1)
From the law of conservation of kinetic energy,
(1/2)mV² = (1/2)mV’² + (1/2)M(Vº)²
or m(V²-V’²)= M(Vº)²
or (V²-V’²)= MV²/m (2)
Dividing equation (2) by equation (1),
V+V’ = Vº
Adding equation (1) and (3),
2V = Vº(1+(M/m)
or , Vº = 2mV/(m+M)
Similarly when the neutron collides with the nitrogen of mass Mn,
Vn = 2mV/(m+Mn)
Dividing equation (4) by equation (5),
Vº/ Vn = (m+Mn)/(m+M)
Chadwick measured maximum velocities of hydrogen and nitrogen nucleus. These values were Vº = 3.3×10⁷ m/s and Vn = 4.7 ×10⁶ m/s. And hence by Putting M = 1 amu and Mn= 14 amu we get m= 1.17 much higher than the correct value. A more accurate value of the mass of neutron was obtained later by other methods.
The mass of neutron accepted at present is,
m= 1.008665 amu = 1.6750× 10¯²⁷ kg = 939.37 MeV/c²
For the discovery of the second particle neutron the nuclide, Chadwick was awarded Noble prize in 1935.
Properties of the neutron :
- Neutron is an elementary particle which exists in the nucleus along with proton. And its mass is slightly greater than the mass of the proton.
- It is a neutral particle. And hence it is not deflected by the electric and magnetic fields.
- Neutron being electrically neutral does not ionize the gases and as a result, it cannot be detected in the ionization chamber.
- Its penetrating power is very high. It passes through the thick plate of lead and it can penetrate into the nucleus very easily. The reason is that being neutral it is not affected by the charge of the atom.
- The neutron cannot stay freely outside the nucleus and decays into proton, electron and antineutrino particles.
₀n¹→ ₁p¹ + ₋₁β⁰ + ν‾