Physicists are studying a process in which a particle (6.11 x 10^-30 kg) traveli
ID: 2099761 • Letter: P
Question
Physicists are studying a process in which a particle (6.11 x 10^-30 kg) traveling at a speed of 0.760c decays, creating two less-massive particles ( 1.44 x 10^-30 kg, 4.21 x 10^-30 kg ) with speeds of 0.850c and 0.640c, respectively. Since the total energy of the two decay products is less than that of the initial particle, experimenters suspect that a third particle, who identity is unknown, was created during the decay. Assuming that a third particle was created, could it be more massive than an electron?
I know there isn't enough information to apply relativistic momentum conservation but the relativistic principle of the conservation of energy limts the amount of mass the third particle can have. With this in mind I used the mass energy equivalence equation:
E = mc^2 / square-root 1 - v^2 / c^2
The total energy I calculated for the initial particle (7.24 x 10^-13) ended up being less than the combined total energy of the two less massive particles (1.52 x 10^-13 + 5.92 x 10^-13 = 7.44 x 10^-13).
If someone could tell me where I am in error, giving me the proper calculation and how to proceed from there I would appreciate it.
Explanation / Answer
hey you just missed the energy due to loss of mass. the loss of mass is 0.46*10^-30 kg
so E=mc^2
=0.414*10^-13 J
now the net energy given out=(7.24+0.414-7.44)*10^-13
=2.14*10^-14 J.
hopw this helps