(a) Identifying the Particles

Based on charge and baryon number conservation in the standard p-p chain:

• R-1: $2p \to X_1 + e^+ + X_2$. Two protons fuse to form Deuterium, a positron, and a neutrino.
  $X_1 \equiv {}^2H$ (Deuterium), $X_2 \equiv \nu_e$ (Neutrino).
• R-2: $p + {}^2H \to X_3 + \gamma$. Proton + Deuterium forms Helium-3.
  $X_3 \equiv {}^3He$.
• R-4: $2(^3He) \to {}^4He + 2X_4$. Two He-3 fuse to He-4 and release two protons.
  $X_4 \equiv {}^1H$ (Proton).
• R-5 (pp-II): $^3He + {}^4He \to X_5 + \gamma$. Fusion creates Beryllium-7.
  $X_5 \equiv {}^7Be$.
• R-6: $^7Be + e^- \to X_6 + X$. Electron capture converts Be-7 to Lithium-7 and a neutrino.
  $X_6 \equiv {}^7Li$, $X_7 \equiv \nu_e$ (Neutrino).

(b) Total Energy Released

The net reaction converts 4 protons and 2 electrons (which annihilate with the positrons) into one Helium-4 nucleus.

$$ 4p + 2e^- \to {}^4He + 2\nu + 6\gamma + \text{Energy} $$

Calculating the mass defect in terms of energy ($\Delta m c^2$):

$$ Q_{total} = [4m_p + 2m_e – m_{He4}]c^2 $$

Using the given values ($m_p=938.27$ MeV, $m_e=0.51$ MeV, $m_{He}=3727.38$ MeV):

$$ Q = 4(938.27) + 2(0.51) – 3727.38 $$

$$ Q = 3753.08 + 1.02 – 3727.38 = 26.72 \text{ MeV} $$

(c) Energies x and y

Finding x (Reaction R-3): Electron-positron annihilation.

$$ x = 2m_e c^2 = 2 \times 0.51 = 1.02 \text{ MeV} $$

Finding y (Reaction R-4):

The total energy of the pp-I branch (26.72 MeV) is the sum of energies of all steps. To make one $^4He$, we need:
2 sets of (R-1 + R-2) to create two $^3He$ nuclei, followed by one R-4.

$$ Q_{total} = 2(Q_{R1} + Q_{R2} + Q_{annihilation}) + y $$

Given energies in text: $Q_{R1} = 0.42$ MeV, $Q_{R2} = 5.49$ MeV, Annihilation $x=1.02$ MeV.

$$ 26.72 = 2(0.42 + 5.49 + 1.02) + y $$

$$ 26.72 = 2(6.93) + y \implies 26.72 = 13.86 + y $$

$$ y = 12.86 \text{ MeV} $$

(d) Why Reaction R-6 does not produce a positron

The reaction $X_5 \to X_6 + e^+ + X_7$ would be $\beta^+$ decay: ${}^7Be \to {}^7Li + e^+ + \nu$.

For this to occur spontaneously, the mass difference must exceed $2m_e$ (due to atomic mass definitions involving electron clouds). The mass difference between ${}^7Be$ and ${}^7Li$ is insufficient to create a positron. Therefore, the nucleus captures an orbital electron (Electron Capture) which is energetically favorable.

(e) Unique vs. Variable Energy

Variable Energy (Continuous Spectrum): Occurs in 3-body decay processes where energy is shared between three particles.
$\to$ Reaction R-1 ($p+p \to D + e^+ + \nu$)

Unique Energy: Occurs in 2-body decay where momentum conservation dictates specific energies for the products.
$\to$ Reaction R-2, R-3, R-5, R-6, R-7.

(Note: In R-6, the neutrino gets a specific energy because it is a 2-body final state: $^7Li + \nu$).