The actuall intensity of the radiation from the nova can be calculated... Intesity=power/area, if i remember right. As the wave, assuming it is just one wave, spreads out from the supernova, it should (in theory) expand as a sphere. Therefor the power a point on the sphere will be total nova power divided by surface area of sphere from shockwave... my maths aint great but that should be roughly right.
r(radius) = 5ly = 3*10^8 X 60X60X24X365.25 X 5 =4.733*10^16m
Surface area of sphere = 4*pie*radius^2
Surface area of shockwave = 4*pie*(4.733*10^16)^2
=2.8157*10^34 square meters.
Therefor, the power of radiation upon one square meter is 1/2.81*10^34 times the total radiation.
Now, ultra violet radiation is made of photons, which travel at the speed of light... here's where my physics knowledge stops... so i'll use the equation E=MC^2 to calculate the energy produced by the star.
...the mass of the sun is 2*10^30 KG. A star three times the mass of our sun is required to start a supernova, or atleast a black hole.... or maybee that was density... so cappella
should have weighed roughly 6*10^30KG.
Therefor the energy of the radiation produced, assuming all of the star was converted to energy, which wouldnt happen cause there would still be mass for a small neutron star left behind (i think), E=6.0*10^30 X (3*10^8)^2
=5.4*10^49 J
And so the energy of the radiation falling upon one square meter at 5lys distance is... 5.4*10^49 J X 1/2.81*10^34
=1.9217*10^15 J
Hmmm. Thats a ****-load of energy. That cant ve right. The main source of error would be in the E=MC^2 equation, particularly the mass used. Lets say only 1/100000000000 of the mass is converted to radiation energy, and the rest buggers off to form a black hole or something.
Then the mass used is 6*10^19KG, which is roughly 1/10000th the mass of the earth.
New energy = 5.4*10^36J
...over one square meter at 5ly away :
192J ...which seems a better figure
So the intesity of radiation on a planet 5ly away would be 192W/M^2 (watts per square meter)This is roughly as much radiation as standing near a powerful light bulb, though most of that wouldn't be deadly gamma rays...
yes, i know, a monkey picking numbers out of a hat could probably get a more accurate answer, but if someone who knows there stuff would like to improve on that feel free
...try using the inverse square law for light intensity, i don't know.
My brain hurts