The uranium ore mined today contains only 0.72% of fissionable ${\mathbf{}}^{\mathbf{235}}\mathbf{U}$, too little to make reactor fuel for thermal-neutron fission. For this reason, the mined ore must be enriched with ${\mathbf{}}^{\mathbf{235}}\mathbf{U}$. Both role="math" localid="1661753958684" ${\mathbf{}}^{\mathbf{235}}\mathbf{U}\mathbf{(}{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{y}\mathbf{)}$and ${\mathbf{}}^{\mathbf{238}}\mathbf{U}\mathbf{(}{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^9\mathbf{y}\mathbf{)}$are radioactive. How far back in time would natural uranium ore have been a practical reactor fuel, with a $\frac{{\mathbf{}}^{\mathbf{235}}\mathbf{U}}{{\mathbf{}}^{\mathbf{238}}\mathbf{U}}$ratio of 3.0%?

1. The ratio of the uranium deposits initially,$\frac{N_5\left(0\right)}{N_8\left(0\right)}=0.03$
2. The ratio of the uranium deposits now,$\frac{N_5\left(t\right)}{N_8\left(t\right)}=0.0072$
3. Half-life of,${}^{235}\mathrm{U},{\mathrm{T}}_{1/2\left({}^{235}\mathrm{U}\right)}=7.0\times {10}^8\mathrm{y}$
4. Half-life of,${}^{238}\mathrm{U},{\mathrm{T}}_{1/2\left({}^{238}\mathrm{U}\right)}=4.5\times {10}^9\mathrm{y}$

Being a highly unstable radioactive nuclide, the uranium isotopes undergo a decay process. This decay results in the disintegration of the deposits. Now, using the decay equation, and determine the required value of the time taken for the decay. With a shorter half-life, uranium-235 has a greater decay rate than uranium-238. Thus, if the ore contains only 0.72% of uranium-235 today, then the concentration must be higher than in the far distant past.

Formulae:

The amount of undecayed nuclei in a sample as follows:

$N=N_0{e}^{-\lambda t}$ …… (i)

The disintegration constant is as follows:

$\lambda =\frac{\ln 2}{T_{1/2}}$ ……. (ii)

Here, $T_{\frac{1}{2}}$is the half-life of the substance.

Now, using equation (ii) in equation (i) and substitute the given data as per required, the time taken for the initially present uranium deposits to decay can be given as follows:

$\frac{N_5\left(t\right)}{N_8\left(t\right)}=\frac{N_5\left(0\right)}{N_8\left(0\right)}{e}^{-\left(\frac{\ln 2}{T_{1/2,235},}-\frac{\ln 2}{T_{1/2,238},}\right)t}$

Substitute the values and solve as:

$$\begin{gathered} t=\frac{T_{1/2,235}{T}_{1/2,238}}{\left(T_{1/2,235}{T}_{1/2,238}\right)ln2}\ln \left[\left(\frac{N_5\left(t\right)}{N_8\left(t\right)}\right)\left(\frac{N_8\left(t\right)}{N_5\left(t\right)}\right)\right] \\ t=\frac{(7.0\times {10}^{8}y)(4.5\times {10}^{9}y)}{\left(\left(4.5\times {10}^{9}y\right)-\left(7.0\times {10}^{8}y\right)\right)\ln 2}\ln \left[\left(0.0072\right){\left(0.03\right)}^{-1}\right] \\ t=1.7\times {10}^{9}y \end{gathered}$$

Hence, the required value of the time is $1.7\times {10}^{9}y$.