00:01
This problem involves pair production of an electron position pair.
00:07
So for part a, we have energy conservation.
00:16
It would be hc divided by wavelength plus mc squared.
00:22
It's equal to 2myc squared plus m fancy y mc squared.
00:32
For part b, momentum conservation would be h divided by wavelength is equal to 2m fancy y v.
00:51
Cosine plus m, y, m v .m, v .m.
00:56
Part c, if we eliminate h divided by wavelength between the two equations, we get 2m fancy y divided by m times 1 minus v over c cosine, plus wavelength m times 1 minus vm over c equals 1.
01:18
For part d, we want vm if vm is less than c.
01:25
So fancy ym is equal to 1 plus negative 1 1⁄2 times negative vm over c squared, which is equal to 1 plus vm squared over 2c squared, and that's equal to 1.
01:44
So if we use this to rewrite the result from part c, we get 2m fancy y over m, 1 minus v over c cosine plus 1 vm over c equals 1, which if we solve for vm, that gives us 2m fancy y over m over m, c minus v cosine.
02:14
For part e, for vm to be zero, c minus v cosine has to equal zero, which is not possible...