If the radon concentration has become 1/20th of what it started as, and its half-life is 3.8 days, what amount of time in days has passed?

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If the radon concentration has become 1/20th of what it started as, and its half-life is 3.8 days, what amount of time in days has passed?

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Answer 1 · 2017-12-24T10:28:50

The time is $16.4 days$ $"16.4 days"$

Explanation:

The half life of Radon is $t_{1 / 2} = 3.8 days$ $t_(1//2)="3.8 days"$

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

The equation for radioactive decay is

$\frac{N (t)}{N_{0}} = e^{- λ t}$ $(N(t))/(N_0)=e^(-lambdat)$

Therefore,

$e^{- λ t} = \frac{N (t)}{N_{0}} = \frac{1}{20}$ $e^(-lambdat)=(N(t))/N_0=1/20$ .

Taking natural logs of both sides,

$- λ t = ln (\frac{1}{20}) = - ln 20$ $-lambdat=ln(1/20)=-ln20$

$t = \frac{ln 20}{λ}$ $t= (ln 20)/(lambda)$

$= \frac{ln 20}{ln 2 / t_{1 / 2}}$ $= ln20/(ln2//t_(1//2))$

$= \frac{ln 20}{ln 2} \cdot t_{1 / 2}$ $= ln20/ln2*t_(1//2)$

$= \frac{ln 20}{ln 2} \cdot 3.8 days$ $= ln20/ln2*"3.8 days"$

$=$ $=$ $16.4 days$ $"16.4 days"$

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If the radon concentration has become 1/20th of what it started as, and its half-life is 3.8 days, what amount of time in days has passed?

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