The half-life of a radioactive isotope is 140d. How many days would it take for the decay rate of a sample of this isotope to fall to one-fourth of its initial value?

1. The half-life of the radioactive isotope,$T_{1/2}=140d$
2. The decay rate falls to one-fourth of its initial value.

The radioactive decay is due to the loss of the elementary particles from an unstable nucleus to convert them into a more stable one. Thus, from the given formula for the rate of decay, consider the decay rate is inversely proportional to the half-life. Thus, using the condition that the rate of decay would fall to one-fourth, get the number of days of the substance for the new decay rate.

Consider the formula for the rate of decay as:

$R=\frac{N\ln 2}{T_{1/2}}$

By the definition of half-life and equation (i), the decay rate is reduced; then, the half-life of the substance has doubled. Thus, the same has reduced to$\frac{1}{2}$its initial amount after 140 d. So, reducing it to$\left(\frac{1}{4}\right)={\left(\frac{1}{2}\right)}^2$ of its initial number requires that two half-lives have passed:

$\begin{array}{l}\mathrm{t}=2{\mathrm{T}}_{1/2}\\ =2\left(140\mathrm{d}\right)\\ =280\mathrm{d}\\ \end{array}$

Hence, the number of days is 280d.