Question

A radioactive element following first order kinetics reduces to half in two years. How much time would it take for the radioactive element to decay such that only ${\left(\frac{1}{4}\right)}^{th}$ of its original concentration remains?

• A. $1.5\times t_{\frac{1}{2}}$
• B. $2\times t_{\frac{1}{2}}$
• C. $t_{\frac{1}{2}}$
• D. $0.5\times t_{\frac{1}{2}}$

Answer 1

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

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

$\left[A\right]=[A{]}_0{e}^{-kt}$
Time at which $\left[A\right]=\frac{[A{]}_0}{4}$
$\frac{[A{]}_0}{4}=[A{]}_0{e}^{-kt}$
$\ln 4=kt$
Putting value of k
$\ln 4=\frac{\ln 2}{t_{\frac{1}{2}}}\times t$

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

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