Question

A radioactive isotope has a half-life of \[T\] years. How long will it take the activity to reduce to (a) \[3.125\% \] (b) \[1\% \] of its original value?

Answer 1

Hint: You can start by mentioning the equation for radioactive decay, i.e. \[\dfrac{N}{{{N_0}}} = {e^{ - \lambda t}}\]. Then you can use the equation \[\lambda = \dfrac{{0.693}}{T}\] to simplify the former equation. Then use that equation to calculate the time taken for both the cases.

Complete step-by-step solution:
We know that for the radioactive decay of a radioactive substance
\[\dfrac{N}{{{N_0}}} = {e^{ - \lambda t}}\] (Equation 1)
Here, \[{N_0} = \] The initial amount of the substance
\[N = \] Amount of the substance at the time \[t\]
\[\lambda = \] Disintegration constant
\[t = \] Time

(a). For the activity of the given radioactive isotope to decrease to \[3.125\% \] of its initial amount, the amount of substance also decreases to \[3.125\% \] of its initial amount.
So, \[\dfrac{N}{{{N_0}}} = \dfrac{{3.125}}{{100}}\]
\[\dfrac{N}{{{N_0}}} = \dfrac{1}{{32}}\]
Substituting this value in equation 1, we get
\[\dfrac{1}{{32}} = {e^{ - \lambda t}}\]
\[ \Rightarrow \ln \left( 1 \right) - \ln \left( {32} \right) = - \lambda t\]
\[ \Rightarrow - \lambda t = 0 - 3.4657\] ( \[\because \] \[\ln 1 = 0\] and \[\ln \left( {32} \right) = 3.4657\] )
\[t = \dfrac{{3.4657}}{\lambda }\]
Now, we know that
\[\lambda = \dfrac{{0.693}}{T}\]
Here, \[T = \] The half-life of the radioactive isotope
So, \[t = \dfrac{{3.4657}}{{\dfrac{{0.693}}{T}}}\]
\[t \approx 5Tyears\]
Hence, the radioactive isotope will take \[5T\] years to reduce to about \[3.125\% \] its initial value.

(b). For the activity of the given radioactive isotope to decrease to \[1\% \] its initial amount, the amount of substance also decreases to \[1\% \] its initial amount.
So, \[\dfrac{N}{{{N_0}}} = \dfrac{1}{{100}}\]
\[\dfrac{N}{{{N_0}}} = \dfrac{1}{{32}}\]
Substituting this value in equation 1, we get
\[\dfrac{1}{{100}} = {e^{ - \lambda t}}\]
\[ \Rightarrow \ln \left( 1 \right) - \ln \left( {100} \right) = - \lambda t\]
\[ \Rightarrow - \lambda t = 0 - 4.6052\] ( \[\because \] \[\ln 1 = 0\] and \[\ln \left( {100} \right) = 4.6052\] )
\[t = \dfrac{{4.6052}}{\lambda }\]
Now, we know that
\[\lambda = \dfrac{{0.693}}{T}\]
Here, \[T = \] The half-life of the radioactive isotope
So, \[t = \dfrac{{4.6052}}{{\dfrac{{0.693}}{T}}}\]
\[t \approx 6.645Tyears\]

Hence, the radioactive isotope will take \[6.645T\] years to reduce to about \[1\% \] its initial value.

Note: To solve such types of problems in which we use the equation for radioactive decay, i.e. \[\dfrac{N}{{{N_0}}} = {e^{ - \lambda t}}\], it is very important to know how to calculate the log values ( like in this question we used the value of \[\ln 0\], \[\ln 32\] and \[\ln 100\] ). Most of the time it is already given in the problem, but it was not in the given problem.