Answer
(a) $32$
(b) $P^{32}_{15}$
Work Step by Step
(a) We know that
$r_1=(1.2\times 10^{-15})A_1^{\frac{1}{3}}$.......eq(1)
Where $r_1$ is the radius of the nucleus of an alpha particle and $A_1$ is the mass number.
If we double the radius, then the above equation becomes
$2r_1=(1.2\times 10^{-15})A_2^{\frac{1}{3}}$......eq(2)
Dividing eq(1) by eq(2), we have
$\frac{r_1}{2r_1}=\frac{(1.2\times 10^{-15})A_1^{\frac{1}{3}}}{(1.2\times 10^{-15})A_2^{\frac{1}{3}}}$
This simplifies to:
$\frac{1}{2}=\frac{4^{\frac{1}{3}}}{A_2^{\frac{1}{3}}}$
$\implies A_2=(2\times 4^{\frac{1}{3}})^3$
$A_2=32$
As the number of nucleons is equal to the mass number of the element, thus the required number of nucleons in the nucleus is $32$.
(b) We know that phosphorus has 23 isotopes and they have the same atomic number but different mass number. The mass number of the phosphorus which has a radius twice that of alpha particle is $32$, hence the required symbol can be written as $P^{32}_{15}$.
