Question

Calculate the mass, radius, and density of the nucleus of $\text{(a)}{}^7\mathrm{Li}\text{and}(\mathrm{b}{)}^{207}\mathrm{Pb}$.

Give all answers in SI units.

Short answer

Hence, the mass, radius and density is:

$$\begin{gathered} \text{a.)}m=9.988\times {10}^{-27}\mathrm{kg},r=2.18\times {10}^{-15}\mathrm{m},\rho =2.3\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3 \\ \text{b.)}\mathrm{m}=3.437\times {10}^{-25}\mathrm{kg},\mathrm{r}=7.1\times {10}^{-15}\mathrm{m},\mathrm{\rho }=2.3\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Step-by-step solution

Given information

$\text{(a)}{}^7\mathrm{Li}\text{and}(\mathrm{b}{)}^{207}\mathrm{Pb}$

$$\begin{gathered} m\left({}^6\mathrm{Li}\right)=6.0135\mathrm{u} \\ \mathrm{m}\left({}^{207}\mathrm{Pb}\right)=206.9309\mathrm{u} \\ {\mathrm{r}}_{\mathrm{o}}=1.2\mathrm{fm}\to 1.2\times {10}^{-15}\mathrm{m} \end{gathered}$$

Explanation

a.) We have to find value of $m,r\text{and}\rho \text{for}{}^6\mathrm{Li}:$

The mass is:

$$\begin{gathered} m=6.0135\mathrm{u}\left(\frac{1.661\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right) \\ \mathrm{m}=9.988\times {10}^{-27}\mathrm{kg} \end{gathered}$$

The radius is:

$$\begin{gathered} r=r_o{A}^{1/3} \\ r=\left(1.2\times {10}^{-15}\mathrm{m}\right)(6{)}^{1/3} \\ r=2.18\times {10}^{-15}\mathrm{m} \end{gathered}$$

The density is

$$\begin{gathered} \rho =\frac{m}{\frac{4}{3}\pi r^3} \\ \rho =\frac{9.988\times {10}^{-27}\mathrm{kg}}{\frac{4}{3}\pi {\left(2.18\times {10}^{-15}\mathrm{m}\right)}^3} \\ \rho =2.3\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

Explanation

b.) We have to find value of $m,r\text{and}\rho \text{for}{}^{207}\mathrm{Pb}:$

The mass is:

$$\begin{gathered} m=206.9309\mathrm{u}\left(\frac{1.661\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right) \\ \mathrm{m}=3.437\times {10}^{-25}\mathrm{kg} \end{gathered}$$

The radius is:

$$\begin{gathered} r=r_o{A}^{1/3} \\ r=\left(1.2\times {10}^{-15}\mathrm{m}\right)(207{)}^{1/3} \\ r=7.1\times {10}^{-15}\mathrm{m} \end{gathered}$$

The density is:

$$\begin{gathered} \rho =\frac{m}{\frac{4}{3}\pi r^3} \\ \rho =\frac{3.437\times {10}^{-25}\mathrm{kg}}{\frac{4}{3}\pi {\left(7.1\times {10}^{-15}\mathrm{m}\right)}^3} \\ \rho =2.3\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$