Calculating Neutrino Energy Released in Supernova 1987A using Flux and Distance

[Qyzren]

The neutrino flux from supernova 1987A was estimated to be 1.3e14 m^-2 at the Earth. SN 1987A occurred in the Large Magellanic cloud, which is located at a distance of 50 kpc. If the average energy per neutrino was 4.2 MeV, estimate the amount of energy released in the form of neutrinos during the supernova explosion

Homework Equations

electron volt = 1.602e-19 J
pc = 3.086e16 m

The Attempt at a Solution

Answer says: 2.6e45 J

I can't seem to get the answer... :(
any help will be appreciated :)
i was thinking since flux falls off ~ 1/d².
so i tried E = flux *d²*avg energy of neutrino
1.3e14*(50000*3.086e16)²*4.2e6*1.602e-19
=3.258e44 J.
So I'm off by more than a factor of 10... :(
What am i doing wrong??

 

[malawi_glenn]

the area of a spehere is $$4 \pi r^2$$

So you should multiply with 4pi = 12,.. =)

also check the numericals, 1.3e14*(50000*3.086e16)²*4.2e6*1.602e-19 = 2.08*10^44 J

 

[Qyzren]

oh! Thank you
(yeh i think i typed something wrong in the calculator before).