Calculate the neutrino flux from the sun

[warfreak131]

Homework Statement

Every reaction in the sun has the energy equivalent to 0.03 mp, and generates 2 neutrinos per reaction. Calculate the number of neutrinos per second, and calculate the neutrino flux at Earth.

Astronomy generally uses the CGS (centimeter gram second) system, so just be aware of that when I do my calculations.

Homework Equations

The Attempt at a Solution

n_reactions * E_reaction = Luminosity_sun

n_reactions * mc² = Luminosity_sun

n_reactions * (.03 * m_p)(3*10¹⁰)² = Luminosity_sun

n_reactions * (.03 * 1.67*10^-24)(3*10¹⁰)² = Luminosity_sun

n_reactions * (.03 * 1.67*10^-24)(3*10¹⁰)² = 3.9*10³³

n_reactions * 4.51*10^-5 erg = 3.9*10³³ erg s^-1

n_reactions = 3.9*10³³ erg s^-1 / 4.51*10^-5 erg

n_reactions = 8.67*10³⁷ reactions/s

n_reactions / s * 2 neutrinos / reaction = 1.73*10³⁸ neutrinos

N_neutrinos / Area = F_neutrino

Area = 4*pi*R² [R = 1 AU = 1.5*10¹³cm]

Area = 2.83*10²⁷cm²

So divide N_neutrinos and Area, and I get 6.1*10¹⁰ neutrinos/cm²

I looked online, and found answers which were also approx 6, but were a different order of magnitude. Did I do anything wrong?

 

[ideasrule]

Your answer's right. Where did you find the values online?

 

[warfreak131]

thanks'

http://www.cosmicrays.org/muon-solar-neutrinos.php

 

[ideasrule]

That link gives 6.296541*10^14 /(m^2*s). Your answer is in /(cm^2*s).

 

[warfreak131]

That link gives 6.296541*10^14 /(m^2*s). Your answer is in /(cm^2*s).

right right, right you are, thanks