Question

A radioactive element following first order kinetics reduces to half in one year. How much time would it take for the radioactive element to decay such that only ${\left(\frac{1}{8}\right)}^{th}$ of its original concentration remains?

• A. $3\times t_{\frac{1}{2}}$
• B. $8\times t_{\frac{1}{2}}$
• C. $\frac{1}{3}\times t_{\frac{1}{2}}$
• D. $\frac{1}{8}\times t_{\frac{1}{2}}$

Answer 1

The correct option is A $3\times t_{\frac{1}{2}}$

Since, radioactive element following first order kinetics reduces to half in two years.
$t_{\frac{1}{2}}=1yr$
$k=\frac{\ln 2}{t_{\frac{1}{2}}}$

$\left[A\right]=[A{]}_0{e}^{-kt}$
Time at which $\left[A\right]=\frac{[A{]}_0}{8}$
$\frac{[A{]}_0}{8}=[A{]}_0{e}^{-kt}$
$\ln 8=kt$
Putting value of k
$\ln 8=\frac{\ln 2}{t_{\frac{1}{2}}}\times t$

$3\ln 2=\frac{\ln 2}{t_{\frac{1}{2}}}\times t$

$t=3\times t_{\frac{1}{2}}$