Question

A radioactive isotope has a half-life of T years.
A) How long will it take the activity to reduce to 3.125%?

A radioactive isotope has a half-life of T years.
B) How long will it take the activity to reduce to 1 % of its original value?

Answer 1

A) Given,
The half-life of the radioactive isotope
=T years
$N_0$ is the actual amount of radioactive isotope.
After decay, the amount of the radioactive isotope = N
Let 𝑡 be the time taken by activity to reduce $N_0$ to $3.125\%$ of its initial value.
For radioactive substance:
$N=N_0{e}^{-\lambda t}=\frac{ln2}{T}$
$N=N_0{e}^{-\frac{ln2}{T}t}$
As $\frac{N}{N_0}\%=3.125\%$
$\frac{N}{N_0}=\frac{1}{32}$
$\frac{1}{32}=e^{-\frac{ln2}{T}t}$
$ln\left(\frac{1}{32}\right)=-\frac{0.693}{T}t$
t = 5T years
Final Answer : 5T years

B) Given, The half-life of the radioactive isotope = T years
$N_0$ is the actual amount of radioactive isotope.
After decay, the amount of the radioactive isotope =N
Let t be the time taken by activity to reduce $N_0$ to $3.125\%$ of its initial value.
For radioactive decay:
$N=N_0{e}^{-\lambda t}\Rightarrow \lambda =\frac{ln2}{T}$
$N=N_0{e}^{-\frac{ln2}{T}t}$
As $\frac{N}{N_0}\%=1\%$
$\frac{N}{N_0}=\frac{1}{100}$
$\frac{1}{100}=e^{-\frac{ln2}{T}t}$
$ln\left(\frac{1}{100}\right)=-\frac{0.693}{T}t$
t = 6.65 T years
Final Answer : 5.65 T years