Question

In the following nuclear reaction $A{I}^{27}+H{e}^4\to {}_{15}{P}^{30}+X,X$ wii be

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

Answer 1

Answer: (C)

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.