Question

Half-life of a radioactive substance is $18$ minutes. The time interval between its $20\%$ decay and $80\%$ decay in minutes is

• A. 6
• B. 9
• C. 18
• D. 36

Answer 1

Answer: (D)

Aftcr $n$ half-life, the numbcrs of atom left undecayed is given by,
$N=N_0{\left(\frac{1}{2}\right)}^n$
$n=\frac{t}{T_{\frac{1}{2}}}$ or $N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}$
For $20\%$ decay of radioactive substance,
$N=N_0-\frac{20}{100}{N}_0=0.8N_0$
or 0.8 $N_0=N_0{\left(\frac{1}{2}\right)}^{t_{1/18}}...(i)$
For $80\%$ decay of radioactive substance,
$N=N_0-\frac{80}{100}{N}_0=0.2N_0$ or $0.2N_0=N_0{\left(\frac{1}{2}\right)}_{\dots }^{t_{2/18}}...(ii)$
From Eqs. (i) and (ii), we get
$\Rightarrow 4={\left(\frac{1}{2}\right)}^{\frac{t_1-t_2}{18}}$
Taking log on the both sides, we get
$\log 4=\log {\left(\frac{1}{2}\right)}^{\frac{t_1-t_2}{18}}$
$\Rightarrow \log 2^2=\log 2^{\left(\frac{t_2-t_1}{18}\right)}$
$\Rightarrow 2=\frac{t_2-t_1}{18}$
$\Rightarrow t_2-t_1=36$ min