Unreasonable Results

1. Repeat exercise but include the \({\rm{0}}{\rm{.0055 \% }}\) natural abundance of \(^{{\rm{234}}}{\rm{U}}\) with its \(2.45 \times {10^5}\) y half-life.
2. What is unreasonable about this result?
3. What assumption is responsible?
4. Where does the \(^{{\rm{234}}}{\rm{U}}\) come from if it is not primordial?

The following is the relationship between activity, half-life, and the number of atoms:

\({\rm{R = }}\frac{{{\rm{0}}{\rm{.693N}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\)

Where,

\({{\rm{t}}_{{\rm{1/2}}}}{\rm{ = }}\)Half life

\({\rm{R = }}\)Activity

\({\rm{N = }}\)Number of atoms

\(N = {N_0}{e^{ - \lambda t}} \Rightarrow \frac{{{N_0}}}{N} = {e^{ - \lambda t}}\)

Where\({N_0} = \) initial activity,

\(N = \)final activity

\(\lambda = \frac{{0.693}}{{{t_{1/2}}}}\)and

\(t = \)time

\(\lambda = \frac{{0.693}}{{{t_{1/2}}}},{t_{1/2}} = 2.45 \times {10^5}y\)

Also,\({\rm{t = 4}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{y }}\)

Substitute the values in the above equation,

\(\begin{aligned}\frac{{{N_0}}}{N} = {e^{\frac{{0.693\left( {4.5 \times {{10}^9}y} \right)}}{{2.45 \times {{10}^5}y}}}}\\ = {e^{12728}}\\ = {10^{5527}}\end{aligned}\)

Therefore, the value of \(\frac{{{N_0}}}{N} = {10^{5527}}\).