P-P Cycle and CNO Cycle

 The source of energy of Sun’s and other Stars is nuclear fusion. There are two possible cycles:

1. Proton – Proton Cycle

2. Carbon – Nitrogen - Oxygen Cycle

Proton – Proton Cycle:

         P-P cycle is one of the two known set of nuclear fusion reaction in which stars converts hydrogen into helium. This cycle can occur only if the KE of the protons is high enough to overcome their mutual electrostatic repulsion.

          The conversion of hydrogen to helium is slow in the Sun. So the complete conversion of the hydrogen in the core of the Sun is calculated to take more than ten billion years. 

           In 1939, Hans Bethe calculated the rates of reactions in various stars. In first step, two protons fuse together and produces deuterium, positron and electron neutrino.

₁H¹ + ₁H¹ → ₁H² + ₁e⁰ + ν_e + 1.442 MeV

In second step, the deuterium can fuse with another proton to produce the light isotope of helium (₂He³).

₁H² + ₁H¹ → ₂He³ + γ + 5.49 MeV

In the next step, two ₂He³ nuclei interact and form ₂He⁴ and two protons.

 ₂He³ + ₂He³ → ₂He⁴ + ₁H¹ + ₁H¹ + 12.8 MeV

Net result is

₁H¹ + ₁H¹ + ₁H¹ + ₁H¹ → ₂He⁴ + 2 ₁e⁰ + 2 ν_e + 2γ + 26.732 MeV

The complete Proton-Proton chain reaction releases a net energy of 26.732 MeV. Two percent of this total energy is lost to the neutrinos that are produced. The P-P reaction is dominant at temperatures of 10 to 14 MK. Below 10 MK, the P-P chain does not produce much ₂He⁴.

Carbon – Nitrogen - Oxygen Cycle:

          CNO cycle is sometimes called Bethe-Weizsacker cycle. Unlike P-P cycle, the CNO cycle is a catalytic cycle (multistep reaction mechanism). Here four hydrogen nuclei fuse into one Helium nucleus using carbon, nitrogen and oxygen as catalysts. It generates less than 10% of the total solar energy. 

₆C¹² + ₁H¹ → ₇N¹³ + γ

₇N¹³ → ₆C¹³ + ₁e⁰ + ν_e

₆C¹³ + ₁H¹ → ₇N¹⁴ + γ

₇N¹⁴ + ₁H¹ → ₈O¹⁵ + γ

₈O¹⁵ → ₇N¹⁵ + ₁e⁰ + ν_e

₇N¹⁵ + ₁H¹ → ₆C¹² + ₂He⁴

In this cycle acts ₆C¹² like a catalyst.

Net result is

₁H¹ + ₁H¹ + ₁H¹ + ₁H¹ → ₂He⁴ + 2 ₁e⁰ + 2 ν_e + Energy

Calculation of energy release:

4 ₁H¹ = 4.031300 u; ₂He⁴ = 4.002603 u and 2 ₁e⁰ = 0.001098 u

Loss in mass Δm = {4.031300 – (4.002603+0.001098)}

Δm = 0.02756 u

so the energy released is E = 0.02756 × 931 MeV = 25.65 MeV

[1 MeV = 1.6×10^-13 J]

E = 25.65 × 1.6×10^-13 J = 41.04 ×10^-13 J