Question

Calculate the lifetimes of ${\mathit{U}}^{\mathbf{238}}$and$\mathit{P}{\mathit{o}}^{\mathbf{212}}$, using Equations8.25and 8.28 . Hint: The density of nuclear matter is relatively constant (i.e., the same for all nuclei), so${\left(r_1\right)}^3$is proportional to (the number of neutrons plus protons). Empirically,

${\mathbf{r}}_{\mathbf{1}}\mathbf{=}\left(1.07\mathrm{fm}\right){\mathbf{A}}^{\mathbf{1}\mathbf{/}\mathbf{3}}$

Short answer

$$\begin{gathered} {\mathrm{\tau }}_{\mathrm{\upsilon }238}\approx 2.4\times {10}^{14}\mathrm{yrs}. \\ {\mathrm{\tau }}_{\mathrm{\rho o}212}\approx 3.2\times {10}^{-4}\mathrm{s}. \end{gathered}$$

Step-by-step solution

To write the relatively constant.

The density of nuclear matter relatively constant is,

$\mathrm{\gamma }={\mathrm{K}}_1\frac{\mathrm{Z}}{\sqrt{\mathrm{E}}}-{\mathrm{K}}_2\sqrt{{\mathrm{Zr}}_1}.$

Where,

role="math" localid="1658403215569" $$\begin{gathered} {\mathrm{K}}_1=1.980{\mathrm{MeV}}^{1/2} \\ {\mathrm{K}}_2=1.485{\mathrm{fm}}^{-1/2} \end{gathered}$$

Zis the atomic number.

Eis the energy.

${\mathrm{r}}_1$is the radius and ${\mathrm{r}}_1\cong \left(1.07\mathrm{fm}\right){\mathrm{A}}^{1/3}$.

Accordingly, we can write

$\mathrm{\gamma }1.980{\mathrm{MeV}}^{1/2}\frac{\mathrm{Z}}{\sqrt{\mathrm{E}}}-1.54\sqrt{{\mathrm{ZA}}^{1/3}}$

To calculate the lifetimes of U238 .

Consider first the uranium-238 nucleus. It has A=238 and Z=92 .

Accordingly, the daughter nucleus has Z=90 (which is the thorium-234).

Hence,

$$\begin{gathered} \mathrm{\gamma }1.980{\mathrm{MeV}}^{1/2}\frac{90}{\sqrt{\mathrm{E}}}-1.54\sqrt{90{\left(238\right)}^{1/3}} \\ \mathrm{\gamma }\approx 178.2{\mathrm{MeV}}^{1/2}\frac{1}{\sqrt{\mathrm{E}}}-36.37 \end{gathered}$$

To calculate E , we need to look up for the masses ${\text{m}}_{\mathrm{p}},{\mathrm{m}}_{\mathrm{d}},\mathrm{and}{\mathrm{m}}_{\infty }$in any standard nuclear physics textbook.

We see that${\mathrm{m}}_{\mathrm{U}238}=238.05078826\mathrm{u},{\mathrm{m}}_{\mathrm{\tau h}234}=234.04360123\mathrm{u},\mathrm{and}{\mathrm{m}}_{\infty }=4.00262\mathrm{u}.$

It is convenient to write the masses in u (unified atomic mass unit ) and then multiply the result by 931 to convert the answer into the MeV energy unit .

It follows that,

$$\begin{gathered} \mathrm{E}=\left(238.05078826-234.04360123-4.002602\right)\times 931 \\ \mathrm{E}\approx 4.27\mathrm{MeV} \end{gathered}$$

Substituting the result into the formula for $\gamma$yields,

$$\begin{gathered} \mathrm{\gamma }=178.2{\mathrm{MeV}}^{1/2}\frac{1}{\sqrt{4.27\mathrm{MeV}}}-36.37 \\ \mathrm{\gamma }\approx 49.87 \end{gathered}$$

It is left to estimate the speed of the alpha particle (using the relation $\mathrm{E}=\left(1/2\right){\mathrm{m}}_{\infty }{\mathrm{V}}^2)$

$$\begin{gathered} \mathrm{v}\approx \sqrt{\frac{2\mathrm{E}}{{\mathrm{m}}_{\infty }}} \\ \mathrm{v}=\sqrt{\frac{2\times 4.27\times {10}^{6}1.6\times {10}^{-19}\mathrm{J}}{6.65\times {10}^{-27}\mathrm{kg}}} \\ \mathrm{v}\approx 1.43\times {10}^7\mathrm{m}/\mathrm{s} \end{gathered}$$

Finally, we calculate the lifetime of ${\mathrm{U}}^{238}$using Equation,

$$\begin{gathered} \mathrm{\tau }=\frac{2\times \left(1.07\mathrm{fm}\right){\mathrm{A}}^{1/3}}{\mathrm{v}}{\mathrm{e}}^{2\mathrm{\gamma }} \\ \mathrm{\tau }=\frac{2\times 1.07\times {10}^{-15}\mathrm{m}\times {\left(238\right)}^{1/3}}{1.43\times {10}^7\mathrm{m}/\mathrm{s}}{\mathrm{e}}^{22\left(49.88\right)} \\ \mathrm{\tau }=\frac{2\times 6.38\times {10}^{-15}}{1.43\times {10}^7}{\mathrm{e}}^{2\left(49.88\right)}\mathrm{s} \\ \mathrm{\tau }=7.46\times {10}^{21}\mathrm{s} \\ \mathrm{\tau }==\frac{7.46\times {10}^{21}}{3.15\times {10}^7}\mathrm{yr} \\ \mathrm{\tau }=2.4\times {10}^{14}\mathrm{yrs}. \end{gathered}$$

To calculate the lifetimes of Po212.

Now consider the polonium-212 nucleus. It has A=212 and Z=84 .

Accordingly, the daughter nucleus has Z=82 (which is the lead Pb-208).

Hence,

$$\begin{gathered} \mathrm{\gamma }1.980{\mathrm{MeV}}^{1/2}\frac{82}{\sqrt{\mathrm{E}}}-1.54\sqrt{82{\left(212\right)}^{1/3}} \\ \mathrm{\gamma }\approx 162.36{\mathrm{MeV}}^{1/2}\frac{1}{\sqrt{\mathrm{E}}}-34.05 \end{gathered}$$

Again, to calculate E, we need to look up for the massess ${\text{m}}_{\mathrm{p}},{\mathrm{m}}_{\mathrm{d}},\mathrm{and}{\mathrm{m}}_{\infty }$

We see that

${\mathrm{m}}_{po212}=211.9988867969\mathrm{u},{\mathrm{m}}_{\mathrm{\tau h}234}=234.04360123\mathrm{u},\mathrm{and}{\mathrm{m}}_{\infty }=4.00262\mathrm{u}.{\mathrm{m}}_{\mathrm{U}238}=238.05078826\mathrm{u},{\mathrm{m}}_{\mathrm{pb}208}=207.976666552071\mathrm{u},\mathrm{and}{\mathrm{m}}_{\infty }=4.00262\mathrm{u}.$

It follows that,

$$\begin{gathered} \mathrm{E}=\left(211.988867969-207.9766552071-4.002602\right)\times 931 \\ \mathrm{E}\approx 8.95\mathrm{MeV} \end{gathered}$$

Substituting the result into the formula for yields,

$$\begin{gathered} \mathrm{\gamma }=162.36{\mathrm{MeV}}^{1/2}\frac{1}{\sqrt{8.95\mathrm{MeV}}}-34.05 \\ \mathrm{\gamma }\approx 20.22 \end{gathered}$$

We calculate the lifetime of $p{o}^{212}$using Equation

$$\begin{gathered} \tau =\frac{2\times \left(1.07fm\right)A^{1/3}}{v}{e}^{2\gamma } \\ \mathrm{\tau }=\frac{2\times 1.07\times {10}^{-15}\mathrm{m}\times {\left(212\right)}^{1/3}}{2.08\times {10}^7\mathrm{m}/\mathrm{s}}{\mathrm{e}}^{2\left(20.22\right)} \\ \mathrm{\tau }=\frac{2\times 6.38\times {10}^{-15}}{2.08\times {10}^7}{\mathrm{e}}^{40.44}\mathrm{s} \\ \mathrm{\tau }=3.2\times {10}^{--4}\mathrm{s}. \end{gathered}$$