00:01
Okay, so for this question, they want us to first determine the wavelength of this gamma emission, and then for part b, they want us to determine the difference and mass.
00:09
So they tell us that the gamma photon release corresponds to an energy of 0 .143 -4 -3 mega -electron volts.
00:17
We need to convert this all the way to joules.
00:20
So one mega -electron volt corresponds to 1 times 10 to 6 electron volts, and 1 electron volt is equal to right here.
00:30
1 .602 times 10 to negative 19 joules.
00:35
We plug that into the calculator and we should get 2 .29 times 10 to negative 14 joules.
00:45
Now we need to use this equation right here.
00:50
So delta e, which is change in energy, is equal to h, which is the plane constant...