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  • The nucleus _10 ^ 23 Ne decays by beta ^ - emission. Write down the beta-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m _10 ^23 Ne = 22.994466 u

Q.13.14  The nucleus _{10}^{23}\textrm{Ne} decays by \beta ^{-} emission. Write down the \beta-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

                  (i) m (_{10}^{23}\textrm{Ne} ) = 22.994466 \; u

                 (ii) m (_{11}^{23}\textrm{Na} ) = 22.089770 \; u
                

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The \beta decay equation is

_{10}^{23}\textrm{Ne}\rightarrow _{11}^{23}\textrm{Na}+e^{-}+\bar{\nu }+Q

\\\Delta m=m(_{10}^{23}\textrm{Ne})-_{11}^{23}\textrm{Na}-m_{e}\\ \Delta m=22.994466-22.989770\\ \Delta m=0.004696u

 (we did not subtract the mass of the electron as it is cancelled because of the presence of one more electron in the sodium atom)

Q=0.004696\times931.5

Q=4.3743 eV

The emitted nucleus is way heavier than the \beta 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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