Question

Show that the total energy released in the proton-proton cycle is \({\rm{26}}{\rm{.7 MeV}}\), considering the overall effect in \(^1H{ + ^1}H{ \to ^2}H + {e^ + } + {\nu _e}{,^1}H{ + ^2}H{ \to ^3}He + \gamma \) and \(^3He{ + ^3}He{ \to ^4}He{ + ^1}H{ + ^1}H\) being certain to include the annihilation energy.

Short answer

If the annihilation energy is included the total energy released in the proton-proton cycle is \({\rm{26}}{\rm{.7 MeV}}\). The first two reactions must occur twice for the third to happen.

Step-by-step solution

Concept Introduction

Nuclear energy is the energy stored in an atom's core. An atom is a little particle that makes up all matter in the universe. The mass of an atom is normally concentrated at the nucleus's centre. The nucleus is made up of two subatomic particles: neutrons and protons. Bonds that connect atoms together contain a significant amount of energy.

Information Provided

• Total energy to be released:\({\rm{26}}{\rm{.7 MeV}}\).
• Proton-Proton reactions are:

\(\begin{aligned} ^1H{ + ^1}H{ \to ^2}H + {e^ + } + {\nu _e}\\^1H{ + ^2}H{ \to ^3}He + \gamma \\^3He{ + ^3}He{ \to ^4}He{ + ^1}H{ + ^1}H\end{aligned}\).

Energy Released

There are three reactions –

\(\begin{aligned} ^1H{ + ^1}H{ \to ^2}H + {e^ + } + {\nu _e}{\rm{ }}(0.42MeV)\\^1H{ + ^2}H{ \to ^3}He + \gamma {\rm{ }}(5.49MeV)\\^3He{ + ^3}He{ \to ^4}He{ + ^1}H{ + ^1}H{\rm{ }}(12.86MeV)\end{aligned}\)

Energy is released in each reaction, but for the third reaction to being possibly the first two must occur twice. Then if the total energy is calculated, it will be –

\(\begin{aligned} {E_{tot}} &= (20.42 + 25.49 + 12.86)\,{\rm{MeV}}\\ &= 26.7{\rm{ MeV}}\end{aligned}\)

The overall effect, beside the helium is that positrons will bump into electrons and annihilate producing\(\gamma \)rays, so the following resulting effect is obtained –

\(2{e^ - } + {4^1}H{ \to ^4}He + 2{\nu _e} + 6\gamma \)

With the annihilation of electron and positron it is obtained that –

\({e^ + } + {e^ - } \to 2\gamma \to 2{m_e}{c^2}\)

Therefore, the value for total energy is obtained as \({\rm{26}}{\rm{.7 MeV}}\)..