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A positron is emitted from ${ }_{11}^{23} \mathrm{Na}$. The ratio of the atomic mass and atomic number of the resulting nuclide is
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23/10
$$
{ }_{11}^{23} \mathrm{Na} \longrightarrow{ }_{10}^{23} \mathrm{Na}+{ }_{-1}^0 e
$$
{ }_{11}^{23} \mathrm{Na} \longrightarrow{ }_{10}^{23} \mathrm{Na}+{ }_{-1}^0 e
$$
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