Chapter 30, Problem 8P

To determine

The stable nucleus has half the radius of uranium.

Answer to Problem 8P

Solution:

${}_{15}^{31}P$

Explanation of Solution

Let $R_1$ is the radius of uranium nucleus and $R_2$ is the nucleus of atom having approximately half radius than uranium.

Atomic mass number of uranium is 238.

Radius of uranium atom is given by,

$R_1=r_0{A}_1{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Where A is atomic mass number and r₀=1.25fermi

$\begin{array}{l}{R}_2=\frac{R_1}{2}\mathrm{...}(\text{given})\\ r_0{A}_2{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{r_0{A}_1{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2}\\ A_2{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=0.5A_1{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ A_2{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=0.5{(}^{238}\\ A_2{}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=3.09855\\ A_2=29.7492\\ A_2\approx 30\end{array}$

Thus, atom with atomic mass number 30.94 is phosphorus ${}_{15}^{31}P$