91影视

The density of nuclear matter relatively constant is,

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

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 ${\mathrm{r}}_1鈮�\left(1.07\mathrm{fm}\right){\mathrm{A}}^{1/3}$.

Accordingly, we can write

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

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,

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

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

We see that${\mathrm{m}}_{\mathrm{U}238}=238.05078826\mathrm{u},{\mathrm{m}}_{\mathrm{蟿丑}234}=234.04360123\mathrm{u},\mathrm{and}{\mathrm{m}}_{鈭�}=4.00262\mathrm{u}.$

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,

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

Substituting the result into the formula for $纬$yields,

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

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

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

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

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

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,

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

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

We see that

It follows that,

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

Substituting the result into the formula for yields,

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

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

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