Question

In the following nuclear reaction, what is $X$?
${}_{13}^{27}Al{+}_2^{4}He{\to }_{15}^{30}P+X$

• A. $\beta -$ particle
• B. Positron
• C. $\alpha -$ particle
• D. Neutron

Answer 1

Answer: (D)

Neutron

In the following nuclear reaction, ${}_{13}^{27}Al{+}_2^{4}He{\to }_{15}^{30}P{+}_0^{1}X$, $X$ is neutron. Neutron has zero charge and mass of 1 amu approximately.
On Left hand side, sum of atomic numbers $=13+3=15$.
On right hand side, sum of atomic numbers $=15+0=15$ .
Hence, the sum of the atomic numbers is same on both sides.

On Left hand side, sum of mass numbers $=27+4=31$.
On right hand side, sum of mass numbers $=30+1=31$.
Hence, the sum of the mass numbers is same on both sides.