Question

In carbon cycle of fusion, $4$ protons combine to yield one alpha particle and
A. One positron
B. Two positrons
C. Ten positrons
D. Three positrons

Answer 1

The carbon cycle of fusion is a process of stellar nucleosynthesis in which the main sequence of stars fuse hydrogen into helium through a six stage sequence of protons. We can say that the carbon cycle is the sequence of thermonuclear reactions that provides most of the energy that is radiated by the stars. Here, we will discuss each reaction one by one.

Complete answer:
The carbon cycle of fusion refers to the carbon-nitrogen-oxygen cycle which is a process of stellar nucleosynthesis in which the main sequence of stars fuse hydrogen into helium through a six stage sequence of protons. Now, let us discuss the sequence of the carbon cycle of fusion, which is shown below When the hydrogen is fused into the carbon $- 12$ , it will emit a gamma ray and will produce a nitrogen $- 13$ as shown below,
${}_6^{12}C\, + \,{}_1^1H\, \Rightarrow \,{}_7^{13}N\, + \gamma$
Now, as we know that nitrogen $- 13$ is unstable, therefore, it will emit a beta particle and decays to carbon$- 13$ as shown below
${}_7^{13}N\, \Rightarrow \,{}_6^{13}C\, + \,{}_1^0e\, + \beta$
Now, hydrogen is fused into carbon $- 13$ , it will emit a gamma ray and will produce a nitrogen$- 14$ as shown below
${}_6^{13}C\, + \,{}_1^1H\, \Rightarrow \,{}_7^{14}N\, + \,\gamma$
Now, when the hydrogen is fused into nitrogen $- 14$ , it will become oxygen $- 15$ and emit gamma ray as shown below
${}_7^{14}N\, + \,{}_1^1H\, \Rightarrow \,{}_8^{15}O\, + \,\gamma$
Now, oxygen$- 15$ will decay into nitrogen$- 15$ by emitting beta as shown below
${}_8^{15}O\, \Rightarrow \,{}_7^{15}N\, + \,{}_1^0e\, + \beta$
Now, when the hydrogen is fused into nitrogen $- 15$ , it will emit helium nucleus and will produce carbon $- 12$ as shown below
${}_7^{15}N\, + \,{}_1^1H\, \Rightarrow \,{}_6^{12}C\, + \,{}_2^4He\, + \,\beta$
This where the cycle started.
Now, when we will add all the equations, we get
\eqalign{ & 4{}_1^1H\,\,\,\,\,\,\,\,\,\, \to \,{}_2^4He\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,2{}_1^0e\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,E \cr & \left( {proton} \right)\,\,\,\,\,\,\left( {\alpha - particle} \right)\,\,\,\,\,\left( {positron} \right)\,\,\,\,\left( {energy} \right) \cr}
Here, $E = \gamma + \beta + \gamma + \gamma + \beta + \beta$
Therefore, in the carbon cycle, $4$ protons will combine to yield one alpha particle and two positrons.

Hence, option B is the correct option.

Note: As we can see that the carbon $- 12$ nucleus used in the first reaction is regenerated in the last reaction, therefore, it acts as a catalyst in the whole process. The carbon cycle of fusion is the primary source of energy in stars of mass which is greater than $1.5$ . In the above cycle, four hydrogens are converted into one helium nucleus.