Question

How long ago was the ratio$\frac{{}^{\mathbf{235}}\mathbf{U}}{{}^{\mathbf{238}}\mathbf{U}}$in natural uranium deposits equal to 0.15?

Short answer

The ratio $\frac{{}^{235}\mathrm{U}}{{}^{238}\mathrm{U}}$in natural uranium deposits equal to 0.15 after $3.6\times {10}^{9}y$.

Step-by-step solution

Given data

a) The ratio of the uranium deposits initially, $\frac{N_5\left(0\right)}{N_8\left(0\right)}=0.15$.

b) The ratio of the uranium deposits now $\frac{N_5\left(t\right)}{N_8\left(t\right)}=0.0072$.

c) Disintegration constant of ${}^{235}\mathrm{U}{\mathrm{\lambda }}_{\frac{1}{2}\left({}^{235}\mathrm{U}\right)}=\frac{9.85\times {10}^{-10}}{\mathrm{y}}$.

d) Disintegration constant of ${}^{235}\mathrm{U}{\mathrm{\lambda }}_{\frac{1}{2}\left({}^{238}\mathrm{U}\right)}=1.55\times {10}^{-10}$.

Determine the concept of decay equation

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, determine the required value of the time taken for the decay.

Formula:

The amount of undecayed nuclei in a sample is as follows:

$N=N_0{e}^{-\lambda t}$ ….. (i)

Calculate the time taken for decay

Now, using 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:

$$\begin{gathered} \frac{N_5\left(t\right)}{N_8\left(t\right)}=\frac{N_5\left(0\right)}{N_8\left(0\right)}{e}^{-\left({\lambda }_5-{\lambda }_8\right)t} \\ t=\frac{1}{{\lambda }_8-{\lambda }_5}\mathrm{In}\left[\left(\frac{{\mathrm{N}}_5\left(\mathrm{t}\right)}{{\mathrm{N}}_8\left(\mathrm{t}\right)}\right)\left(\frac{{\mathrm{N}}_8\left(\mathrm{t}\right)}{{\mathrm{N}}_5\left(\mathrm{t}\right)}\right)\right] \end{gathered}$$

Solve further as:

$$\begin{gathered} t=\frac{1}{\left(1.55-9.85\right)\times {10}^{-10}/y}\mathrm{In}\left[\left(0.0072\right){\left(0.15\right)}^{-1}\right] \\ \mathrm{t}=3.6\times {10}^9\mathrm{y} \end{gathered}$$

Hence, the required value of the time is $3.6\times {10}^9\mathrm{y}$.