Similarly Proton has anti-Proton. The two have identical masses but opposite charge.
All three generations of quarks, leptons and neutrinos have their anti-particle counterpart. When two material anti-particles combine they get annihilated to give equivalent amount of energy by mass equivalence law given by special theory of relativity:
e + anti e ↔ 2γ (soft gamma photons)
p + anti p ↔ 2γ (hard gamma photons)
| Up quark | Strange antiquark | K + Meson | |
| Charge | (2/3) e 0 | (1/3) e 0 | e 0 |
| Equivalent mass | 310MeV | -505MeV | -195MeV |
| Spin | (1/2)ћ | (-1/2) ћ | 0ћ |
……………….1.129
| Up quark | Down anti-quark | Π + | |
| Charge | (2/3) e 0 | (1/3) e 0 | +e 0 |
| Mass | 310MeV | -310MeV | 0 eV |
| Spin | (1/2) ћ | -(1/2) ћ | 0 ћ |
……………………..1.130
From Eq.(1.129) and Eq.(1.130) we see that both K + and П + have 0ћ spin angular momentum hence they are Bosons. But K + has mass equivalence of -195MeV and П + has mass equivalent of 0eV. Therefore in a nucleus where equal mass particles are present there П + mediate as carrier particles but where unequal mass particles are present there K + mediate as carrier particles.
When particle and antiparticle meet they annihilate and mass equivalent amount of energy is released as photon of appropriate wavelength.
Consider electron and positron: m e c 2 /q = 0.5MeV;
Therefore e + e(anti)= 1MeV↔ 1.24/(1×10 ^6 eV)=1.24×10 ^-6 μm=1.24pm(pico meter)
Pico to Femto meter corresponds to gamma rays(γ rays).
Pico is soft gamma rays whereas Femto is hard gamma rays.
Consider proton and anti-proton: m p c 2 = 0.9396GeV;
Therefore p + p(anti)= (2×0.9396)GeV ↔ 1.24/(2×0.9396×10 9 eV)=0.67×10 -15 m= 0.66fm;
Thus we conclude that electron and its antiparticle positron collide to give soft γ photons and proton and antiproton annihilate to give hard γ photons.
Amongst the nuclear particles proton and neutron are the lightest(mass equivalent of 1GeV). Therefore at t = 10 -6 sec the temperature is 10 13 K which is equivalent to 1GeV. Therefore with expansion of the space-time fabric, temperature further falls. The hot soup of electrons-quarks-photons undergoes a phase transition to electron-nucleon soup. This is known as quark-nucleon phase transition. Below 1 GeV strong nuclear force freezes out, relic quarks get decoupled and we are left with a hot soup of hydrogen-helium nucleons, electrons and photons equivalent to 10 13 K. This is known as the fourth state of matter namely plasma. In this hot plasma the photons are energetic enough to prevent Hydrogen and Helium from staying in deionized state or in neutral state . From t = 10 -6 second (temperature 10 13 K ) to t = 1 second (temperature 10 10 K)= 1000 times Sun’s core temperature , hydrogen-helium nucleons are continuously inter- transmuting between protons and neutrons i.e.
p ↔ n …………………………………… 1.131
From t = 10 -4 second (temperature 10 12 K ) to t = 1second ( temperature = 10 10 K) the inter-transmutation is leading to a steady state value of p (proton particles) : n (neutron particles) = 7:1.
After t = 1second i.e. below 10 10 K, quarks condense to a steady value of p : n = 7:1.
The steady state value is given by:
Where
After t = 100 seconds ( 1.7 minutes) temperature comes down to 10 9 K and around 3 minutes after the Big Bang the elemental form of matter has the following composition: 25% Helium , 75% Hydrogen and traces of Deuterium and Lithium.