Question

In the equation
${}_{13}^{27}Al+{}_2^{4}He⟶{}_{15}^{30}P+X$
The correct symbol for $X$ is

• A. ${}_{-1}^0$
• B. ${}_1^{1}H$
• C. ${}_2^{4}He$
• D. ${}_0^{1}n$

Answer 1

Answer: (D)

Conserving mass and charge, net mass reduces by $1$ by charge does not change.
Hence, a neutron must have been released.