Question

Verify that, as reported in Table 43-1, fissioning of the${}^{\mathbf{235}}\mathbf{U}$ in 1.0 kg of${\mathbf{UO}}_{\mathbf{2}}$ (enriched so that${}^{\mathbf{235}}\mathbf{U}$ is 3.0% of the total uranium) could keep a 100 W lamp burning for 690 y.

Short answer

It is verified that the lamp is burning for 690 y.

Step-by-step solution

Describe the expression to calculate time

The expression to calculate the time is given by,

$\mathbf{t}\mathbf{=}\frac{\mathbf{E}}{\mathbf{P}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{}$

Here,E is energy released, and P is power.

The expression to calculate the energy released is given by,

$\mathit{E}\mathbf{=}\mathit{N}\mathit{Q}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

Here,N is number of U-235 nuclei, and Q is total energy.

Verify that the lamp is burning for 690 y

The uranium oxide contains, 3% ${}^{235}U$, and 97% ${}^{235}\mathrm{U}$. Therefore, 1000 gm of$U{O}_2$ sample contain M grams ${}^{235}\mathrm{U}$.

Find the mass of${}^{235}U$ as follows.

$$\begin{gathered} M_{U-235}=\left(3\%\right)\left(1000\mathrm{gm}\right)\left[\frac{\left(97\%\right)\left(238\right)+\left(3\%\right)\left(235\right)}{\left(97\%\right)\left(238\right)+\left(3\%\right)\left(235\right)+2\left(16\right)}\right] \\ =26.4\mathrm{gm} \end{gathered}$$

Find the number of${}^{235}\mathrm{U}$ nuclei as follows.

$$\begin{gathered} N=\frac{\left(26.4\mathrm{gm}\right)\left(6.02\times {10}^{23}\mathrm{atoms}/\mathrm{mol}\right)}{\left({\displaystyle \frac{235\mathrm{g}}{1\mathrm{mol}}}\right)} \\ =6.77\times {10}^{22}atmos \end{gathered}$$

Substitute all the known values in equation (2) to find energy released.

$$\begin{gathered} E=\left(6.77\times {10}^{22}\mathrm{atmos}\right)\left(200\mathrm{MeV}\right) \\ =1.35\times {10}^{25}\mathrm{MeV} \\ =\left(1.35\times {10}^{25}\mathrm{MeV}\right)\left(\frac{1.6\times {10}^{-13}\mathrm{J}}{1\mathrm{MeV}}\right) \\ =2.17\times {10}^{12}\mathrm{J} \end{gathered}$$

Substitute all the known values in equation (1) to find time.

$$\begin{gathered} t=\frac{2.17\times {10}^{12}}{J} \\ =2.17\times {10}^{10}\mathrm{s} \\ =\left(2.17\times {10}^{10}\mathrm{s}\right)\left(\frac{1\mathrm{yr}}{12\times 30\times 24-3600\mathrm{s}}\right) \\ \approx 690\mathrm{y} \end{gathered}$$

Therefore, it is verified that the lamp is burning for 690 y.