Question

Calculate the nuclear diameters of$\text{(a)}{}^4\mathrm{He},\text{(b)}{}^{56}\mathrm{Fe}\text{, and(c)}{}^{238}\mathrm{U}$.

Short answer

Therefore, the nuclear radius and diameters are:

$$\begin{gathered} \text{a.)}r=1.904\mathrm{fm},d=3.8097\mathrm{fm} \\ \text{b.)}r=4.59103\mathrm{fm},d=9.18206\mathrm{fm} \\ \text{c.)}r=7.43658\mathrm{fm},d=14.87317\mathrm{fm} \end{gathered}$$

Step-by-step solution

Given information

$r_o=1.2\mathrm{fm}$

For radius we have,

$r=r_o{A}^{1/3}$

For diameter we have,

$d=2·r$

Explanation

(a)The radius and diameter of the nucleus of ⁴He must be determined:

The radius is:

$$\begin{gathered} r=r_o{A}^{1/3} \\ r=(1.2\mathrm{fm})(4{)}^{1/3} \\ r=(1.2\mathrm{fm}\left)\right(1.587) \\ r=1.904\mathrm{fm} \end{gathered}$$

The diameter is:

$$\begin{gathered} d=2·r \\ d=2·1.904\mathrm{fm} \\ d=3.8097\mathrm{fm} \end{gathered}$$

(b)The radius and diameter of the nucleus of ${}^{56}\mathrm{Fe}$must be determined:

The radius is:

$$\begin{gathered} r=r_o{A}^{1/3} \\ r=(1.2\mathrm{fm})(56{)}^{1/3} \\ r=(1.2\mathrm{fm}\left)\right(3.82586) \\ r=4.59103\mathrm{fm} \end{gathered}$$

The diameter is:

$$\begin{gathered} d=2r \\ d=2\times 4.59103\mathrm{fm} \\ d=9.18206\mathrm{fm} \end{gathered}$$

Explanation

(c)The radius and diameter of the nucleus of ${}^{238}\mathrm{U}$must be determined:

The radius is:

$$\begin{gathered} r=r_o{A}^{1/3} \\ r=(1.2\mathrm{fm})(238{)}^{1/3} \\ r=(1.2\mathrm{fm}\left)\right(6.19715) \\ r=7.43658\mathrm{fm} \end{gathered}$$

The diameter is:

$$\begin{gathered} d=2r \\ d=2\times 7.43658\mathrm{fm} \\ d=14.87317\mathrm{fm} \end{gathered}$$