Complete the following nuclear reaction : `._13Al^27 + ._2He^4 to ._15P^30 ` + ?

Explanation

Let A and Z be mass and atomic number of the unknown particle respectively. From the law of conservation of mass number
27+4=30+A or, A=1
From the law of conservation of atomic number,
13+2=15+Z or, Z=0
The particle is therefore a neutron `._0n^1`
`therefore` The complete equation of the reaction : `._13Al^27 + ._2He^4 =._15P^30 + ._0n^1`