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The nucleus $^{64}_{29}Cu$ accepts an orbital clectron to yield.
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The correct answer is:
$\frac{64}{28} \mathrm{Ni}$
The nucleus $^{64}_{29}Cu$ accepts an orbital electron to yield $^{64}_{28}Ni$. The atomic number of $\mathrm{Cu}$ is 29 , which is equal to the number of electrons and also equal to the number of protons. When $^{64}_{29}Cu$ accepts an orbital electron then electrons subtract from the atomic number, i.e.
$29-1=28$
$$
^{64}_{29}Cu+{ }_{-1} e^{0} \longrightarrow{ }_{28}^{64} \mathrm{Ni}
$$
$29-1=28$
$$
^{64}_{29}Cu+{ }_{-1} e^{0} \longrightarrow{ }_{28}^{64} \mathrm{Ni}
$$
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