If SN 1987A produces 10⁵⁷ neutrinos per second (uniformly in all directions), and it is at a distance of 51,500 pc (=1.59 × 10²¹ m), what would be the flux of neutrinos at an Earth-based detector with an area of 1 square meter?

Question

Quizplus · Accepted Answer

The flux of neutrinos, $F$, at a distance $d$ from a source emitting $N$ neutrinos per second isotropically is given by $F = \frac{N}{4\pi d^2}$. Substituting $N = 10^{57}$ neutrinos per second and $d = 1.59 \times 10^{21}$ meters, the flux at Earth is calculated as $F = \frac{10^{57}}{4\pi (1.59 \times 10^{21})^2}$ meters$^{-2}$s$^{-1}$, which simplifies to approximately $3.15 \times 10^{13}$ neutrinos m$^{-2}$s$^{-1}$, matching choice A.

If SN 1987A Produces 1057 Neutrinos Per Second (Uniformly in All

If SN 1987A Produces 10⁵⁷ Neutrinos Per Second (Uniformly in All