Question

The power output of the Sun is \(4 \times {10^{26}}{\rm{ }}W\).

(a) If \(90\% \) of this is supplied by the proton-proton cycle, how many protons are consumed per second?

(b) How many neutrinos per second should there be per square meter at the Earth from this process? This huge number is indicative of how rarely a neutrino interacts, since large detectors observe very few per day.

Short answer

(a) If\(90\% \)of power output of the Sun is supplied by the proton-proton cycle, the number of protons consumed per second is\(N = 4.47 \times {10^{38}}{\rm{ }}protons/s\).

(b) There should be \(n = 5.96 \times {10^{14}}{\rm{ }}neutrinos{\rm{ }}/{m^2} \cdot s\) at the Earth from the process.

Step-by-step solution

Concept Introduction

The proton-proton chain, also referred to as the\({\rm{p - p}}\)chain, is one of two known sets of nuclear fusion events that stars use to convert hydrogen to helium. The second known reaction, the\({\rm{CNO}}\)cycle, is dominant in stars with masses greater than or equal to\({\rm{1}}{\rm{.3}}\)times that of the Sun, according to theoretical models, but it is dominating in stars with masses less than or equal to that of the Sun.

Information Provided

• Power Output for the Sun is:\(4 \times {10^{26}}W\).
• Amount of power supplied by proton-proton chain is: \(90\% \).

Number of Protons consumed

To calculate the number of photons first find how much energy is coming from the proton-proton cycle. The relation \(32.16\) is giving final nuclear reaction and there is a release of \(26.7{\rm{ }}MeV\) from the following –

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

So, to get the number of photons use the total output power because \(W\) can be expressed also as \(J/s\) so the power output can be scaled per second, relative to the energy used in the proton-proton cycle.

The value of \(1eV = 1.60 \times {10^{ - 19}}J\), so converting (\(26.7{\rm{ }}MeV\)) to \(J\) gives \(4.27 \times {10^{ - 12}}J\). Also, in each proton-proton cycle, \(4\) protons are used, so the total number of protons used for this power output (of \(90\% \)) can be calculated by the following relation –

\(N = \frac{{P \cdot 90\% \cdot 4{\rm{ }}protons{\rm{ }}}}{E}\)

Plugging in the values –

\(N = \frac{{4 \times {{10}^{26}}\;J/s \cdot 0.9 \cdot 4{\rm{ }}protons{\rm{ }}}}{{4.27 \times {{10}^{ - 12}}\;J}}\)

Hence, the result is obtained as –

\(N = 4.47 \times {10^{38}}{\rm{ }}protons/s\)

Therefore, the value for number of protons is obtained as \(N = 4.47 \times {10^{38}}{\rm{ }}protons/s\).

Number of Neutrinos

Look at Relation\(32.13 - 32.16\), it can be seen that as\(4\)protons are destroyed two neutrinos have been created.

To calculate the number of neutrinos per square meter, think that neutrons are traveling through the sphere that has a radius equal to the distance Sun-Earth, consulting that the distance from Earth to Sun is around\(1.5 \times {10^{11}}{\rm{ }}m\). So, the number of neutrinos per unit area will be –

\(n = \frac{{\left( {2{v_a}/4{\rm{ }}protons{\rm{ }}} \right)N}}{{4\pi {r^2}}}\)

Plugging in the values –

\(n = \frac{{\left( {2{v_a}/4{\rm{ }}protons{\rm{ }}} \right)4.47 \times {{10}^{38}}{\rm{ }}protons{\rm{ }}/s}}{{4\pi 1.5 \times {{10}^{11}}\;m}}\)

Hence, the result is obtained as –

\(n = 5.96 \times {10^{14}}{\rm{ }}neutrinos{\rm{ }}/{m^2} \cdot s\)

Therefore, the value for number of neutrons is obtained as \(n = 5.96 \times {10^{14}}{\rm{ }}neutrinos{\rm{ }}/{m^2} \cdot s\).