Question

If ${\frac{7}{8}}^{th}$ of the initial mass of a radioactive substance decays in $15$ days, then what is the half-life period of the radioactive substance?

Answer 1

Let the initial number of radioactive nuclei be $N_0$.
Then, after a half-life period, the number of nuclei present would be $N_0\times \frac{1}{2}$.
After two half-lives, the number of nuclei present would be $\frac{1}{2}\times (N_0\times \left(\frac{1}{2}\right))=N_0{\left(\frac{1}{2}\right)}^2$
Similarly, after ${}^{\prime }{n}^{\prime }$ half-lives, the number of nuclei present would be $N=N_0{\left(\frac{1}{2}\right)}^n$ $\to$(1)
If ${}^{\prime }{t}^{\prime }$ is the given time period and $T$ is the half-life period of the radioactive substance,
then $n=\frac{t}{T}$ $\to$(2)
Also, if ${}^{\prime }{m}_0^{\prime }$ is the initial mass and ${}^{\prime }{m}^{\prime }$ the mass of undecayed nuclei of the substance,
then $\frac{N}{N_0}=\frac{m}{m_0}$ $\to$(3)
$\therefore$ Equation (1) can be written as
$m=m_0{\left(\frac{1}{2}\right)}^n$ or $m=m_0{\left(\frac{1}{2}\right)}^{\frac{t}{T}}$ $\to$(4)
In the given problem, undecayed mass
$m=m_0-\frac{7}{8}{m}_0=\frac{1}{8}{m}_0$ ; $t=15$ days, $T=$ ?
Substituting the values of ${}^{\prime }{m}^{\prime }{,}^{\prime }{m}_0^{\prime }$ and ${}^{\prime }{t}^{\prime }$ in equation (4), we get
$\frac{1}{8}{m}_0=m_0{\left(\frac{1}{2}\right)}^{\frac{15}{T}}\Rightarrow \frac{1}{8}={\left(\frac{1}{2}\right)}^{\frac{15}{T}}\Rightarrow {\left(\frac{1}{2}\right)}^3={\left(\frac{1}{2}\right)}^{\frac{15}{T}}$
$\Rightarrow \frac{15}{T}=3\Rightarrow T=\frac{15}{3}$ days $=5$ days