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The nucleus ${ }_{10}^{23}$ Ne decays by $\boldsymbol{\beta}^{-}$emission. Write down the $\beta$-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: $\mathrm{m}\left({ }_{10}^{23} \mathrm{Ne}\right)=\mathbf{2 2 . 9 9 4 4 6 6 ~ a m u}$ $\mathrm{m}\left({ }_{11}^{23} \mathrm{Na}\right)=22.089770$ amu
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Verified Answer
$\beta$-decay equation is
$$
{ }_{10}^{23} \mathrm{Ne} \longrightarrow{ }_{11}^{23} \mathrm{Na}+{ }_{-1}^0 \mathrm{e}+\bar{v}+\mathrm{Q}
$$
Neglect rest mass of antineutrino and electron,
$$
\begin{aligned}
\mathrm{Q} &=(\Delta \mathrm{m}) \mathrm{c}^2=\left(\mathrm{m}_{\mathrm{Ne}}-\mathrm{m}_{\mathrm{Na}}\right) \mathrm{c}^2 \\
&=(22.994466-22.989770) \times 931.5 \\
&=4.372 \mathrm{MeV}
\end{aligned}
$$
This energy is shared by $\mathrm{e}^{-}$and $\overline{\mathrm{v}}$ pair. The max energy of $\mathrm{e}^{-}$will be $4.372 \mathrm{MeV}$ when energy of $\bar{\nu}$ will be zero.
$$
{ }_{10}^{23} \mathrm{Ne} \longrightarrow{ }_{11}^{23} \mathrm{Na}+{ }_{-1}^0 \mathrm{e}+\bar{v}+\mathrm{Q}
$$
Neglect rest mass of antineutrino and electron,
$$
\begin{aligned}
\mathrm{Q} &=(\Delta \mathrm{m}) \mathrm{c}^2=\left(\mathrm{m}_{\mathrm{Ne}}-\mathrm{m}_{\mathrm{Na}}\right) \mathrm{c}^2 \\
&=(22.994466-22.989770) \times 931.5 \\
&=4.372 \mathrm{MeV}
\end{aligned}
$$
This energy is shared by $\mathrm{e}^{-}$and $\overline{\mathrm{v}}$ pair. The max energy of $\mathrm{e}^{-}$will be $4.372 \mathrm{MeV}$ when energy of $\bar{\nu}$ will be zero.
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