In the equation ${ }_{13}^{27} Al + { }_2^4 He \longrightarrow { }_{15}^{30} P + X$,the correct symbol for $X$ is:

Nuclear reactions are given as $(i)$ $(n, p) {}_{16}S^{32} \to {}_{15}P^{32}$,$(ii)$ $(p, \alpha) {}_{8}O^{16} \to {}_{9}F^{19}$,$(iii)$ ${}_{7}N^{14} + ? \to {}_{6}C^{14} + {}_{1}H^{1}$. The missing particles or nuclides in these reactions are respectively:

$A$ nucleus has mass number $\alpha$ and radius $R_{\alpha}$. Another nucleus has mass number $\beta$ and radius $R_{\beta}$. If $\beta = 8\alpha$, then $R_{\alpha}/R_{\beta}$ is:

The radius of an atomic nucleus of mass number $64$ is $4.8 \text{ fermi}$. Then the mass number of another atomic nucleus of radius $6 \text{ fermi}$ is

Two nuclei have their mass numbers in the ratio of $1: 3$. The ratio of their nuclear densities would be

The size of the nucleus is of the order of: