Q67PE.
Question
Unreasonable Results
- 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.
- What is unreasonable about this result?
- What assumption is responsible?
- Where does the \(^{{\rm{234}}}{\rm{U}}\) come from if it is not primordial?
Step-by-Step Solution
Verified- The value of \(\frac{{{N_0}}}{N} = {10^{5527}}\).
- Because the ratio is so high, the result is illogical.
- The implausible finding is due to the assumption that the \(^{{\rm{234}}}{\rm{U}}\) was not generated as a daughter nucleus and instead existed primordially.
- \(^{{\rm{234}}}{\rm{U}}\) is part of the \(^{{\rm{238}}}{\rm{U}}\)chain decay.
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
- The following is the relationship between the starting number and the final number of radioactive substances after time t:
\(N = {N_0}{e^{ - \lambda t}} \Rightarrow \frac{{{N_0}}}{N} = {e^{ - \lambda t}}\)
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 \(\frac{{{N_0}}}{N} = {10^{5527}}\).
- Because the ratio is so high, the result is illogical.
- The implausible finding is due to the assumption that the \(^{{\rm{234}}}{\rm{U}}\) was not generated as a daughter nucleus and instead existed primordially.
- The \(^{{\rm{234}}}{\rm{U}}\) is part of the \(^{{\rm{238}}}{\rm{U}}\) chain decay.