$_{Z}X^{M} + _{2}He^{4} \to _{15}P^{30} + _{0}n^{1}$. Then

Identify $X$ in $_{16}S^{32} + X \to _{15}P^{30} + _{2}He^{4}$.

$_{5}B^{10} + _{2}He^{4} \rightarrow X + _{0}n^{1}$
In the above nuclear reaction '$X$' will be

$_{13}Al^{28}$ when radiated by a suitable projectile gives $_{15}P^{31}$ and a neutron. The projectile used is:

$A$ closed vessel with rigid walls contains $1 \ mol$ of ${ }_{92}^{238} U$ and $1 \ mol$ of air at $298 \ K$. Considering the complete decay of ${ }_{92}^{238} U$ to ${ }_{82}^{206} Pb$, the ratio of the final pressure to the initial pressure of the system at $298 \ K$ is:

For the fission reaction $_{92}U^{235} + _0n^1 \to _{56}Ba^{140} + _yE^x + 2_0n^1$,the values of $x$ and $y$ are: