Beta Decay Processes of 5525Mn & 5526Fe

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Homework Statement


The atomic mass of 55 25Mn = 54.938047u and that of 55 26Fe=54.938296u. Explain which of
the three possible β-decay processes involving these two nuclides are forbidden, and which
can occur. For decays which can occur, describe the energy spectrum of the products.


Homework Equations


In all the cases where β+
decay is allowed energetically (and the proton is a part of a nucleus with electron shells), it is accompanied by the electron capture process) .
However, in proton-rich nuclei where the energy difference between initial and final states is less than 2mec2, then β+
decay is not energetically possible, and electron capture is the sole decay mode.



The Attempt at a Solution



β+ decay can only happen inside nuclei when the value of the binding energy of the mother nucleus is less than that of the daughter nucleus.
B.E of Fe = 481.050477 Mev B.E of Mn = 482.033 Mev
So β+ decay seems possible.
Also there is no reason to exclude b- and electron capture processes .
Im also confused about the energy spectrum of created nuclei .
How would be possible to know their K.E ??
 
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Remember that positron decay competes with electron capture, so if the positron decay is possible, you have to look to see if if the positron can be emitted (remember its rest mass energy is 511keV). If nuclide A decays by positron (or electron capture capture) to nuclide B, then B cannot decay by beta minus to nuclide A.

As for the energy spectrum, in any of the processes, if you have multiple particles emitted, Conservation of Energy along with Conservation of Momentum dictates that the energy is shared between the particles, each particle with a minimum decay energy and a maximum decay energy, so which modes emit multiple particles?
 
multiple particles are emitted in b minus and beta plus decays .
Since the positron can be emitted from energy point of view as well

it should be a beta plus decay.
 
i don't know if you had this in your mind when you say devay modes emit multiple partivles but thanks for the answer
 

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