Identify the missing particle X in the following nuclear reaction:...

Let the missing particle $X$ have mass number $A$ and atomic number $Z$.

The nuclear reaction is:
$^{27}_{13}\text{Al} + ^{4}_{2}\text{He} \rightarrow ^{30}_{15}\text{P} + ^{A}_{Z}X$

Applying the law of conservation of mass number:
$27 + 4 = 30 + A$
$31 = 30 + A$
$A = 1$

Applying the law of conservation of atomic number:
$13 + 2 = 15 + Z$
$15 = 15 + Z$
$Z = 0$

The particle with mass number $A = 1$ and atomic number $Z = 0$ is a neutron ($^{1}_{0}\text{n}$).

Answer: Neutron