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Exercises · 13.14

Q.The nucleus 1023Ne^{23}_{10}\text{Ne} decays by β−\beta^{-} emission. Write down the β\beta-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
m(1023Ne)=22.994466m(^{23}_{10}\text{Ne}) = 22.994466 u
m(1123Na)=22.989770m(^{23}_{11}\text{Na}) = 22.989770 u.

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Write the decay equation, then compute Q=[M(Ne-23)−M(Na-23)]c2Q=[M(\text{Ne-23})-M(\text{Na-23})]c^2 directly from atomic masses (no electron-mass correction needed for β−\beta^- decay). The maximum electron kinetic energy equals this Q-value, about 4.37 MeV.

Decay equation

In β−\beta^{-} decay a neutron converts to a proton, electron, and antineutrino, so ZZ increases by 1 while AA stays fixed:

1023Ne→1123Na+e−+νˉ^{23}_{10}\text{Ne} \rightarrow {}^{23}_{11}\text{Na} + e^{-} + \bar{\nu}

Q-value

For β−\beta^{-} decay, atomic masses can be used directly without any electron-mass correction: the daughter atom (Na, Z=11Z=11) already carries one more bound electron than the parent (Ne, Z=10Z=10) in its neutral atomic mass, which exactly accounts for the emitted electron's mass.

Q=[M(23Ne)−M(23Na)]c2Q = \left[M(^{23}\text{Ne}) - M(^{23}\text{Na})\right]c^2

=[22.994466−22.989770]×931.5 MeV= [22.994466 - 22.989770] \times 931.5\ \text{MeV}

=0.004696×931.5=4.374 MeV= 0.004696 \times 931.5 = 4.374\ \text{MeV} …

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