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

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.