00:01
So in this problem, we are looking at the decay of the pion into a positron and electron neutrino.
00:07
And it wants us to find the kinetic energy of the positron from the energy release from the decay process.
00:16
So the q value is equal to the connect energy of the positron plus the connect energy of the neutrino.
00:23
The neutrino rest mass is approximately zero.
00:26
So q is defined as the mass.
00:31
For the pion subtracted the positron plus the mass the neutrino which that is sorry so that's approximately 0 times c squared so these are mc square terms so that can be equivalently written as 139 that 6 minus 0 .511 minus 0 m .v.
01:15
C squared times c squared.
01:19
So that cancels.
01:23
And we end up with 139 .089 m .e .v.
01:33
So q is approximately 139 .1m .m .v.
01:41
So the two connect guarantees will end up having this sum up to that value of q.
01:46
So then we note that we're given that the pion began at rest, so there's no extra kinetic energy in the system, no extra momentum, so p total equals zero, and also is the momentum of the positron plus the momentum of the electron neutrino.
02:20
So then we can say, you either as magnitude you could write as squared, so there's some i'll write it squared, positron momentum, neutrino momentum.
02:40
And so that's p squared.
02:42
They're the same value, so we'll just replace it with p.
02:45
So in the kinetic energy equations, where you have the p -c -squared term for the different particles, we can just write, replace those with the lower p, and it'll be equivalent.
02:58
So then the kinetic energy of the positron becomes pc squared plus 0 .511 mep squared minus 0 .511 mep and k of the neutrino is pc squared plus 0 .11 .m...