List three quantities that are conserved in nuclear reactions - Leaving Cert Physics - Question 10 - 2011

The initial confusion arose because, in beta decay, it appeared as though energy and momentum were not conserved. The resolution to this contradiction was the proposal of the neutrino, a nearly massless and electrically neutral particle, which carries away energy and momentum during beta decay. This discovery allowed physicists to account for the conservation laws that seem violated in the reaction.

The fundamental forces of nature ordered by increasing strength are:

1. Gravitational - Weakest force, governing the attraction between masses.
2. Weak nuclear - Responsible for processes like beta decay.
3. Electromagnetic - Governs interaction between charged particles.
4. Strong nuclear - Strongest force, holding protons and neutrons together in the nucleus.

The fundamental force involved in beta decay is the weak nuclear force. This force is responsible for the transformation of neutrons into protons (and vice versa) through the emission of beta particles.

New particles are produced in the collision due to the conversion of energy into mass, as described by Einstein's equation $E=mc^2$. When two protons collide with sufficient energy, they can produce additional particles, such as the charged pi mesons, through the interactions governed by the strong nuclear force.

The equation representing the collision can be written as: $p + p \rightarrow p + p + \pi^+ + \pi^-$
This equation shows that two protons collide to produce two protons and two charged pi mesons.

Using the equation for energy conservation: $E_{total} = mc^2$ The rest mass energy of the two protons is: $2 \times 938.27 \text{ MeV} = 1876.54 \text{ MeV}$ For two pi mesons, the rest mass energy is: $2 \times 139.57 \text{ MeV} = 279.14 \text{ MeV}$ Thus, to conserve energy: $1876.54 ext{ MeV} \text{ (initial)} = 279.14 ext{ MeV (produced)} + KE_{protons}$ This simplifies to: $KE_{protons} = 1876.54 ext{ MeV} - 279.14 ext{ MeV} = 1597.4 ext{ MeV}$ Since we have two protons: $KE_{per proton} \geq \frac{1597.4 ext{ MeV}}{2} = 798.7 ext{ MeV}$ However, for particle production, considering round-off or threshold levels, each proton's kinetic energy must be at least 140 MeV to ensure the collision occurs.