1.) The nucleus 10Be decays by electron emission. Determine the maximum energy of the electron emitted in this process.
Solution:
The electron emission or -
decay equation,
104Be 105Be
Mass defect:
m
= m[104Be ] - m[105Be
]
= 10.01353470 - 10.0129376
= 5.971*10-4 u
Q =
mc2
= 5.971*10-4 *931.5
= 0.5562 MeV
= 556.2 keV.
The emitted nucleus is way heavier than the -
particle and the energy of the antineutrino is also negligible and
therefore the maximum energy of the emitted electron is equal to
the Q value.
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