Question

Flourine is found to undergo 10% radioactivity decay in 366 minutes determine its halflife​

Answer 1

Answer:

$\boxed{\text{2408 min}}$

Explanation:

The integrated rate law for radioactive decay is

$\ln\dfrac{N_{0}}{N_{t}} = kt$

1. Calculate the decay constant

$\begin{array}{rcl}\ln \dfrac{100}{90} & = & k \times 366\\\\1.054 & = & 366k\\\\k & = & \dfrac{1.054 }{366}\\\\k & = & 2.879 \times 10^{-4} \text{ min}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k}\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{2.879 \times 10^{-4} \text{ min}^{-1}} = \text{2408 min}\\\\\text{The half-life for decay is } \boxed{\textbf{2408 min}}$