The proton–proton chain is one of two known sets of nuclear fusion reactions that converts hydrogen to helium in stars. It dominates in stars with $M \lesssim 1.3 \, {\rm M_{\odot}}$ ![[ppchain.png|align:center]] $ \textbf{The Net Reaction:} \hspace{1cm} 4 \, p + 2 \,e^{-} \longrightarrow \ce{^{4}He} + 2 \, \nu_{e} \hspace{1cm} \left( 26.4 \, {\rm MeV} \right) $ > [!key-idea] Important Notes > - The *rate limiting reaction* is the proton fusion (p-p reaction) to form deuteron ($D$), due to being mediated by the weak nuclear force. > - In all branches, the net energy released is the same: $E_{\rm gen} = 26.731 \; {\rm MeV}$. However, some of that energy escapes through neutrinos: $E_{\nu} \approx 0.262 \; {\rm MeV}$. > - This is because the overall net reaction is the same (i.e. same initial reactants and final products) for all three branches. $E_{\rm nuc} = \left(m_{\rm reactants} - m_{\rm products} \right) c^{2}$ > - For the production of one $\ce{^{4}He}$ nuclei... > - The [[#PP I]] branch requires **two*** *p-p reactions* > - In the [[#PP II]]/[[#PP III]] branches requires **one** *p-p reactions* > > - In the [[#PP II]]/[[#PP III]] branches, the initial $\ce{^{4}He}$ nuclei acts as a catalyst, allowing the $\ce{^{3}He} + p \longrightarrow \ce{^{4}He} + \nu_{e}$ net reaction to occur > - If a significant $\ce{^{4}He}$ is present in the system, then all three branches will operate simultaneously. > - Neutrinos from the [[#PP III]] branch (from the $\ce{^{8}B}$ decay) are emitted on the energy continuum up to $15 \; {\rm MeV}$ and are easily detectable. *(This is important for the [[Sun#Solar Neutrino Problem]])* ^important-notes > [!sun] In the Sun... > - In the [[Sun|Sun]], the [[#PP I]] branch dominates ([[#PP I]] : $\sim 69 \%$ , [[#PP II]] : $\sim 30.9 \%$ , [[#PP III]] : $\sim 0.1 \%$) > - Reaction Timescales: > - The average proton in the core of the Sun waits $\sim 10^{9} \; {\rm year}$ before it fuses with another proton. > - Each newly created deuterium ($D$) nucleus exists for only $\sim 1 \, {\rm s}$ before it is converted into helium-3 ($\ce{^{3}He}$). > - The $\ce{^{3}He}$ isotope then only exists for $\sim 400 \, {\rm yr}$ before it is converted into helium-4 ($\ce{^{4}He}$). > - $99\%$ of the energy output of the sun comes from the p–p chains, with the other $1\%$ coming from the [[Carbon-Nitrogen-Oxygen Cycle|CNO cycle]]. ^in-the-sun ## Steps of Reaction 1) Two protons ($p$) fuse together to produce deuteron ($D$, stable). *(This is known as the p-p reaction)* - As the protons fuse, one of them undergoes [[Beta Decay#Beta-Plus Decay]], converting into a neutron by emitting a positron and an electron neutrino. - The positron will annihilate with an electron from the environment into two gamma rays. - The energy produced in this reaction is emitted through the photons ($\sim 1.18 \; {\rm MeV}$) and the neutrino ($\sim 0.262 \; {\rm MeV}$). - This *p-p reaction* is the rate-limiting reaction for the p-p chain due to being mediated by the weak nuclear force (therefore, it is extremely slow) $ p + p \longrightarrow D + e^{+} + \nu_{e} \hspace{1.5cm} \textcolor{gray}{\left[ e^{+} + e^{-} \longrightarrow 2 \gamma \right]} \hspace{1.5cm} \left( 1.442 \, {\rm MeV} \right) $ > [!atom] PEP Reaction > This net reaction produces the same amount of energy as the rare *proton-electron-proton (p-e-p) reaction* (electron capture). > $p + e^{-} + p \longrightarrow D + \nu_{e} \hspace{1cm} \left( 1.442 \, {\rm MeV} \right)$ > The frequency ratio of the *p-e-p reaction* vs. the *p–p reaction* is 1:400. However, the neutrinos released by the *p-e-p reaction* are far more energetic. > - Neutrinos produced in the *p–p reaction* range in energy up to $0.42 \, {\rm MeV}$ > - Neutrinos produced in the *p-e-p reaction* creates sharp-energy-line neutrinos at $1.442 \, {\rm MeV}$ 2) A deuteron ($D$) fuses with another proton to produce the helium-3 isotope ($\ce{^{3}He}$, stable) *(also see [[Deuteron Fusion]])* $p + D \longrightarrow \ce{^{3}He} + \gamma$ This process is extremely faster than the proton fusion in step 1, due to being mediated by the strong nuclear force. 3) For the production of helium-4 ($\ce{^{4}He}$), the fusion reaction chain splits into **three branches** ([[#PP I]], [[#PP II]], [[#PP III]]) - In [[#PP I]], the production of $\ce{^{4}He}$ requires two *p-p reactions* - In [[#PP II]] and [[#PP III]], the production of $\ce{^{4}He}$ requires only one *p-p reaction*. - The $\ce{^{4}He}$ nucleus is a catalyst, allowing the net reaction to occur $ \ce{^{3}He} + p \longrightarrow \ce{^{4}He} + e^{+} + \nu_{e} \hspace{1.5cm} \textcolor{gray}{\left[ e^{+} + e^{-} \longrightarrow 2 \gamma \right]} $ - If there is a significant $\ce{^{4}He}$ already present, then all branches will operate simultaneously. ### PP I 4) Using two $\ce{^{3}He}$ produced through the previous reactions, they can immediately form a $\ce{^{4}He}$ nuclei $\ce{^{3}He} + \ce{^{3}He} \longrightarrow \ce{^{4}He} + 2 p$ > [!note] > This branch is dominant at temperatures $T \in \left[ 10 \, , \; 18 \right] \; \times 10^{6} \; {\rm K}$ . ### PP II 4) Using an already present $\ce{^{4}He}$ nuclei, a $\ce{^{3}He}$ nuclei and $\ce{^{4}He}$ nuclei will fuse with produce a $\ce{^{7}Be}$ nuclei. *(same reaction in both the [[#PP II]] and [[#PP III]] branches)* $ \ce{^{3}He} + \ce{^{4}He} \longrightarrow \ce{^{7}Be + \gamma} $ 5) The $\ce{^{7}Be}$ nuclei then uses [[Electron Capture]] to convert a proton ($p$) into a neutron ($n$) to generate a $\ce{^{7}Li}$ nuclei. $ \ce{^{7}Be} + e^{-} \longrightarrow \ce{^{7}Li + \nu_{e}} $ 6) The $\ce{^{7}Li}$ nuclei then uses a proton to generate initiate a quick [[Alpha Decay|alpha decay]] into two $\ce{^{4}He}$ nuclei. *($\ce{^{8}Be}$ has a half-life $\sim 10^{-17} \; {\rm s}$)* $ \ce{^{7}Li} + p \longrightarrow 2 \, \ce{^{4}He} $ > [!note] > This branch is dominant at temperatures $T \in \left[ 18 \, , \; 25 \right] \; \times 10^{6} \; {\rm K}$ . ### PP III 4) Using an already present $\ce{^{4}He}$ nuclei, a $\ce{^{3}He}$ nuclei and $\ce{^{4}He}$ nuclei will fuse with produce a $\ce{^{7}Be}$ nuclei. *(same reaction in both the [[#PP II]] and [[#PP III]] branches)* $ \ce{^{3}He} + \ce{^{4}He} \longrightarrow \ce{^{7}Be + \gamma} $ 5) The $\ce{^{7}Be}$ nuclei then uses [[Proton Capture]] to generate a $\ce{^{8}B}$ nuclei. $ \ce{^{7}Be} + p \longrightarrow \ce{^{8}B + \gamma} $ 6) The $\ce{^{8}B}$ isotope (being highly unstable) quickly [[Beta Decay#Beta-Plus Decay]] into a $\ce{^{8}Be}$ isotope. $ \ce{^{8}B} \longrightarrow \ce{^{8}Be} + e^{+} + \nu_{e} $ 7) With a half-life $\sim 10^{-17} \; {\rm s}$, the $\ce{^{8}Be}$ nuclei very quickly undergoes [[Alpha Decay]] to generate two $\ce{^{4}He}$ nuclei. $ \ce{^{8}Be} \longrightarrow 2 \, \ce{^{4}He} $ > [!note] > This branch is dominant at temperatures $T > 25 \times 10^{6} \; {\rm K}$ . ## Energy Generation Rate Using the [[Nuclear Energy Generation Rate|energy generation rate (per volume)]]... $\epsilon_{\rm v} = \rho \, \epsilon_{m} = Q_{\rm nuc} \, n_{A} \, n_{B} \, \sigma v \left( \frac{1}{1 + \delta_{AB}} \right) \quad \propto \quad Q_{\rm nuc} \, n_{A} \, n_{B} \, S(E_{0}) \; T_{6}^{-2/3} \, e^{-B T_{6}^{-1/3}}$ ...where $T_{6} = 10^{6} \; {\rm K}$. We can linearize the relationship around the dominate temperature for [[#PP I]] branch ($T_{6} \approx 15$) such that $\epsilon \propto T^{4}$.