00:01
So let's say we have some 500 kev photons incident on lead.
00:09
And we want to know in the rate of energy transmission.
00:13
So let's put this e.
00:15
Dot that is transmitted.
00:18
We're told is four times 10 to the fourth mega electron volts per second.
00:24
And part a asks us how many or what fraction of the incident photon energy? is absorbed in the sheet.
00:32
So the energy that is absorbed over the incident energy is going to be 1 minus e to the minus mu times x, where x is the thickness of our slab.
00:44
I forgot to mention that.
00:44
So this is 8 millimeters.
00:47
So if you consult some tables to determine what the linear attenuation coefficient is, it's going to be the density of lead times the the mass let's see what letter should we use for this we'll just call it the mass attenuation coefficient so kappa is a mass attenuation coefficient and so if we calculate this the density of lead is about 11 .29 grams per cubic centimeter and the mass attenuation coefficient for lead at 500 kv or half and m eve is about 0 .1 and the units are centimeter squared per gram so that means our mass attenuation coefficient is just 1 .129 and the units would be inverse centimeters so then we can calculate what this absorbed fraction is it's going to be 1 minus e to the minus 1 .129 inverse centimeters times 0 .8 centimeters.
02:12
And this should come out to about 0 .595 approximately.
02:19
Part b then asks, how many photons per second are incident on the sheet? so the energy that is incident on the sheet, we're right it this way, is just going to be the energy that's transmitted, divided by, one minus this number we just calculated.
02:46
Actually, we can just write it as e to the minus me with x.
02:49
We just calculated one minus that.
02:52
So this should come out to four times 10 to the negative fourth m .ev per second, divided by 0 .405...