- #1
crimsonn
- 30
- 0
Momentum Conservation Problem? Help please!
A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of 9.30 X 10^ -23 and 5.40 X 10^ -23 kg * m/s, respectively. What is the magnitude and direction of momentum of the second (recoiling) nucleus?
Here is my attempt at a solution:
Mn -- mass of first nucleus Mn' -- mass of second nucleus
Vn -- velocity of first nucleus. Vn' -- velocity of second nucleus
Me & Ve -- electron Mne & Vne --neutrino
MnVn = Mn'Vn' + MeVe + MneVne
MnVn = 0 (because it is at rest)
so
Mn'Vn' = -MeVe - MneVne
which, when you break it down into it's components
Mn'Vn' = -MeVe (sin 90) - MneVne (sin 90)
I know that might be a mouthful. Is it even correct? I have no idea how to do this problem with the mass of the nucleus...
answer is 1.08 X 10 ^-22, 30.1 degrees from the direction opposite to the electron's
A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of 9.30 X 10^ -23 and 5.40 X 10^ -23 kg * m/s, respectively. What is the magnitude and direction of momentum of the second (recoiling) nucleus?
Here is my attempt at a solution:
Mn -- mass of first nucleus Mn' -- mass of second nucleus
Vn -- velocity of first nucleus. Vn' -- velocity of second nucleus
Me & Ve -- electron Mne & Vne --neutrino
MnVn = Mn'Vn' + MeVe + MneVne
MnVn = 0 (because it is at rest)
so
Mn'Vn' = -MeVe - MneVne
which, when you break it down into it's components
Mn'Vn' = -MeVe (sin 90) - MneVne (sin 90)
I know that might be a mouthful. Is it even correct? I have no idea how to do this problem with the mass of the nucleus...
answer is 1.08 X 10 ^-22, 30.1 degrees from the direction opposite to the electron's