Answer :
Answer:
1 day
Explanation:
Let the safe level = x
The current level = x + 0.2x = 1.2 x
Thus,
Half life = 3.8 days
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{3.8}\ days^{-1}[/tex]
The rate constant, k = 0.1824 days⁻¹
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = x / 1.2 x = 0.8333
t = ?
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.8333=e^{-0.1824\times t}[/tex]
t ≅ 1 day
Lab must be vacated in 1 day.
If the resulting radiation level is 20% above the safe level, then in 1 day lab laboratory must be remained vacated.
What is half-lives?
Half lives are the time interval which is need to decay the atomic nuclei of a radioactive sample.
The half life is 3.8 days. Thus, the rate constant can be found out using the following formula.
[tex]k=\dfrac{0.693}{3.8}\\k=0.1824[/tex]
Let concentration at a time t is x. If the resulting radiation level is 20% above the safe level, then the initial concentration will be 1.2x.
The ratio of concentration at time t and initial concentration can be given as,
[tex]\dfrac{x}{1.2x}=e^{-kt}\\0.8333=e^{-0.1824t}\\t \approx 1[/tex]
Hence, if the resulting radiation level is 20% above the safe level, then in 1 day lab laboratory must be remained vacated.
Learn more about the half lives here;
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