Question

In the nuclear reaction ${}_{13}A{l}^{27}{+}_{2}H{e}^4--{>}_{15}{P}^{30}$+$X$, $X$ will be :

• A. Proton
• B. Electron
• C. Neutron
• D. $\alpha$ -particle

Answer 1

The correct option is C Neutron

The two basic laws used are 1] conservation of mass
2] conservation of charge
${}_{13}A{l}^{27}{+}_{2}H{e}^4{\to }_{15}{P}^{30}+X$
as we can see that no. of protons on both sides are balanced already.
$13+2=15$
but masses are not balanced,
reactant side has more mass by one unit than $P^{30}$
therefore $X$ carries $1$ unit mass.
$X$ is a particle with $0$ charge and $1$ unit mass.
Hence, it is a neutron.