A nucleus 
undergoes β- decay and becomes 
, Calculate the maximum kinetic energy of electrons emitted assuming that the daughter nucleus and anti- neutrino carry negligible kinetic energy.
Asked by Topperlearning User | 9th Jul, 2014, 12:48: PM
Expert Answer:
β- decay of 
is given as:
Where,
Q = kinetic energy of the daughter nucleus 
Ignoring the rest mass of the anti-neutrino 
and the mass of the electron 
,
the mass defect involved in the nuclear reaction is given as:
Δm = m
-m 
= 22.994466 - 22.989770
= 0.004696 u = 0.004696 x 931.5 MeV/c2 = 4.374 MeV/c2
Q = Δ mc2 = 4.374 MeV
Hence, when the energy carried by 
is zero, the maximum kinetic energy of β particle is 4.374 MeV.
Answered by | 9th Jul, 2014, 02:48: PM
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-
A neutron is absorbed by a 3 L
nucleus with subsequent emission of 
- particle Write the corresponding nuclear reaction. Calculate the energy released in this reaction.
Given mass of 3 L = 6.015126 a.m.u.
Mass of 2
e = 4.000 26044 a.m.u.
Mass of neutron 
= 1.0086654 a.m.u.
Mass of tritium 
= 3.016049 a.m.u.
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Atomic mass of 
= 238.05079 u
Atomic mass of 
= 234.04363 u
Atomic mass of 
= 4.00260 u
1u = 931.5 Me V/c2
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