Question

Suppose an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron. Complete the nuclear chemical equation below so that it describes this nuclear reaction.
____ → 30P15 + 1n0

Answer 1

Explanation:

An alpha particles is basically a helium nucleus and it contains 2 protons and 2 neutrons.  

Symbol of an alpha particle is $^{4}_{2}\alpha$. Whereas a neutron is represented by a symbol $^{1}_{0}n$, that is, it has zero protons and only 1 neutron.

Therefore, reaction equation when an aluminum- nuclide transforms into a phosphorus- nuclide by absorbing an alpha particle and emitting a neutron is as follows.

        $^{27}_{13}Al + ^{1}_{0}n \rightarrow ^{30}_{15}P + ^{1}_{0}n$