An isotope has a half life of 90 days. How many days will it take for a 5 mg sample to decay to 1 mg?

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Answer 1 · 2017-06-08T19:23:51

The number of days is $= 209 d a y s$ $=209days$

The radioactive decay is represented by the equation

$M (t) = M (0) e^{- λ t}$ $M(t)=M(0)e^(-lambdat)$

The radioactive constant is

$λ = \frac{ln 2}{t_{\frac{1}{2}}}$ $lambda=ln2/t_(1/2)$

Here, the half life is

$t_{\frac{1}{2}} = 90 d$ $t_(1/2)=90d$

$λ = \frac{ln 2}{90} = 7.7 \cdot 10^{- 3}$ $lambda=ln2/90=7.7*10^-3$

The initial mass is $M (0) = 5 m g$ $M(0)=5mg$

The final mass is $m (t) = 1 m g$ $m(t)=1mg$

$1 = 5 e^{- {7.710}^{- 3} t}$ $1=5e^(-7.710^(-3)t)$

$e^{- {7.710}^{- 3}} t = \frac{1}{5} = 0.2$ $e^(-7.710^-3)t=1/5=0.2$

$- {7.710}^{- 3} t = ln 0.2 = - 1.609$ $-7.710^-3t=ln0.2=-1.609$

$t = \frac{1.609}{{7.710}^{- 3}}$ $t=1.609/7.710^-3$

$= 209 d$ $=209d$