What Is the Maximum Speed of the Daughter Nucleus in Negative Beta Decay?

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SUMMARY

The maximum speed of the daughter nucleus in negative beta decay of a He(6,2) nucleus, initially at rest, is determined using conservation of momentum and energy principles. The decay process is represented as He(6,2) → Li(6,3) + electron + antineutrino. The correct maximum speed calculation, considering momentum conservation, shows that the daughter nucleus can achieve a speed of 1x10^5 m/sec, contrary to the incorrect calculation of 1x10^7 m/sec. The error in the initial approach was neglecting momentum conservation, which is crucial in decay scenarios.

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Homework Statement


1)Determine the maximum possible speed of the daughter nucleus in negative beta decay of a He(6,2) nucleus which is initially at rest.


Homework Equations





The Attempt at a Solution



I solved in the following way:
He(6,2) --> Li(6,3) + electron + antineutrino
The daughter nucleus will have maximum speed when the electron and antineutrino has zero kinetic energy. Applying mass-energy equivalence,
M(He)c^2 = M(Li)c^2 + K(Li)
Where M(He)=atomic mass of helium isotope, M(Li)= atomic mass of Lithium isotope,
K(Li)=Kinetic energy of daughter nucleus
On simplification I get,
K(Li)=3.52 MeV
(1/2)M(Li)v^2=3.52 MeV
On simplification I get,
v=1x10^7 m/sec
But the answer given in my book is 1x10^5 m/sec. Could someone please tell me where I have gone wrong?
 
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Your decay scenario doesn't conserve momentum! Use 4 vectors.
 
The daughter nucleus will have maximum speed when the electron and antineutrino has zero kinetic energy.

You must always consider momentum conservation!
 
Max daughter speed is when the e and antineutrino both go in the same direction. Then use E and p conservation.
 

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