Neutron decay

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A neutron at rest in the laboratory spontaneously decays into a proton, an electron, and a small essentially massless particle called a neutrino. Calculate the kinetic energy of the proton and the electron in each of the following cases:
a) the neutrino has no kinetic energy
b) the neutrino has 300keV of kinetic energy and is traveling opposite the proton and the same direction as the electron
c) the neutrino has 300keV of kinetic energy and is traveling perpendicular to the proton and electron, which are traveling opposite of each other.

I have solved part a. But part b I am having a problem with. I understand that the neutrino’s E = K + E0 = K + mc2 = K = 300keV (where mc2 = 0 because it has no mass) but I don’t know how to use this with the energies of the other two. Do I simply subtract the energies of the proton, electron and the given K of the neutrino to get an amount that is a new total energy? On part c how do I handle the 300keV of the neutrino that is traveling perpendicularly to the proton and electron?

Any hints would be appreciated…

moondog
 

Answers and Replies

  • #2
dextercioby
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Though it's obviouly a relativistic problem,maybe some of the ole tricks from Newtonian mechanics would help.Do a vector (momenta) diagram and pay attention when u write the conservation of total momentum,ie.write it in vector form & pay attention with the projections.
This valid for point "b" and especially for point "c".

For the record:it's a (massless) electronic ANTIneutrino.

Daniel.
 
  • #3
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For relativistic mechanics you have to use
[tex]
E^2 = p^2 +m^2
[/tex]
In natrual units, instead of the old E1 + E2 ..En = Etot, you get:
[tex]
E_{after}=E_{before}][/tex]
[tex]
\sqrt{p_{after 1}^2 + m_{after 1}^2} +\sqrt{p_{after 2}^2 + m_{after 2}^2}... = \sqrt{p_1^2 + m_1^2} +\sqrt{p_2^2 + m_2^2}...
[/tex]
Also you still have momentum conservation,
[tex]
P_{after \ 1} + P_{after \ 2} ... = P_1 + p_2 ...
[/tex]

Hope this helps. There is the four vector notation for this math that simplifies some of this maybe someone else can post it.
 
  • #4
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My results so far...

For part a I calculated a kinetic energy of 0.782MeV for the electron and 0.752keV for the proton.
For part b I calculated a kinetic energy of 0.483MeV for the proton and 0.387keV for the proton.
Lastly, for part c I calculated 1.08MeV for the electron and 1.21keV for the proton.
Does that make sense?

moondog
 
  • #5
dextercioby
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Are you sure with the numbers,i mean,taking into account the mass ratio,the electron would be 1 million times faster than the proton... :eek:

Daniel.
 
  • #6
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Perpindicular neutrino travel

The main part I am not understanding is how to deal with the perpendicular travel of the neutrino. Since E=K+E0=K+mc^2 and the neutrino is massless this means that E=K=.300MeV. How do I deal with this in the perpendicular direction?

moondog
 
  • #7
Astronuc
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There is total (kinetic + rest) energy, a scalar quantity.

An then there is momentum, a vector quantity, with components in 2 dimensions, e.g. x-direction and y-direction. The neutrino has momentum, in the direction of travel, which must equal the momentum components of the electron and proton, which are in the opposite direction.

See also - Neutrino Nuclear Physics (1.04 Mb, 50 pages, Japanese language support not necessary - download with save target as).
 

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