Answer: The correct option is B.
Explanation: This is an example of radioactive decay and all the radioactive decay processes follow First order of kinetics.
Expression for the half life of first order kinetics is:
\(t_{1/2}=\frac{0.693}{k}\)
We are given:
\(t_{1/2}=12.32years\)
Putting in above equation, we get:
\(12.32=\frac{0.693}{k}\\k=0.05625year^{-1}\)
Expression to calculate the amount of sample which is unchanged is:
\(N=N_oe^{-kt}\)
where,
N = Amount left after time t
\(N_o\) = Initial amount
k = Rate constant
t = time period
Putting value of k = 0.05625 and t = 24.6 in above equation, we get:
\(N=N_oe^{-0.05625\times 24.6}\)
\(\frac{N}{N_o}=0.25\)
The above fraction is the amount of sample unchanged and that is equal to \(\frac{1}{4}\)
Hence, the correct option is B.
