(a) Calculate the rate at which the Sun generates neutrinos. Assume that energy production is entirely by the proton-proton fusion cycle. (b) At what rate do solar neutrinos reach Earth?

The expression for the rate at which the Sun generates neutrinos is given by,

${\mathbf{R}}_{\mathbf{v}}\mathbf{=}\frac{\mathbf{2}\mathbf{P}}{\mathbf{Q}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The energy per fusion is$Q=26.7MeV$ , and the power of the sun is$P=3.9\times {10}^{26}W$ .

Substitute all the known values in equation (1).

$$\begin{gathered} R_v=\frac{2\left(3.9\times {10}^{26}W\right)}{26.7MeV} \\ =\frac{2\left(3.9\times {10}^{26}W\right)}{\left(26.7MeV\right)\left(1.602\times {10}^{-13}J/MeV\right)} \\ =1.8\times {10}^{38}{s}^{-1} \end{gathered}$$

Therefore, the rate at which the Sun generates neutrinos is$1.8\times {10}^{38}{s}^{-1}$ .

Let$R_{v,e}$ be the rate at which the solar neutrinos reachthe earth,d is the distance between the earth and the sun and r is the radius of the earth.

The expression of the rate is given by,

$$\begin{gathered} R_{v,e}=R_v\left(\frac{{\mathrm{\pi r}}^2}{4{\mathrm{\pi d}}^2}\right) \\ =R_v\left(\frac{r^2}{4d^2}\right).......\left(2\right) \end{gathered}$$

The radius of the earth is$r=6.4\times {10}^{6}m$ , and the distance between the earth and the sun is$d=1.5\times {10}^{11}m$ .

Substitute all the known values in equation (2).

$$\begin{gathered} R_{v,e}=\left(1.8\times {10}^{38}{s}^{-1}\right)\left(\frac{{\left(6.4\times {10}^{6}m\right)}^2}{2{\left(1.5\times {10}^{11}m\right)}^2}\right) \\ =8.2\times {10}^{28}{s}^{-1} \end{gathered}$$

Therefore, the rate at which the solar neutrinos reach Earth is$8.2\times {10}^{28}{s}^{-1}$ .