Answer:
12.53 % of a sample of Uranium-238 will remain after 13.4 billion years.
Explanation:
The half life of the uranium-238 = \(t_{\frac{1}{2}}\)=4.47 billion years
All the radioactive reaction are of first order kinetics. The rate constant and t half of the reaction are related as:
\(k=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{4.47 \text{billion years}}=0.1550 (\text{billion years})^{-1}\)
\(k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}\)
where,
k = rate constant = \(0.1550 (\text{billion years})^{-1}\)
t = time taken during radio decay = 4.47 billion years
\([A_o]\) = initial amount of the reactant = \(1.45\times 10^{-6}mol/L\)
[A] = amount left left after time t.
\(\log \frac{[A]}{[A_o]}=-\frac{kt}{2.303}\)
\(\frac{[A]}{[A_o]}=0.1253=\frac{12.53}{100}=12.53\%\)
12.53 % of a sample of Uranium-238 will remain after 13.4 billion years.
