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A certain radioactive isotope has leaked into a small stream. Four
hundred days after the​ leak, 13% of the original amount of the substance remained. Determine the​ half-life of this radioactive isotope.

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apologiabiology
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for a thing that is P amount at first, with a half life of k days
after t days, the amount left is A

the equation is
[tex]A=P( \frac{1}{2})^{ \frac{t}{k} } [/tex]

given
13% of P remained=A
t=400

replace A with 0.13P and t with 400


0.13P=[tex]P( \frac{1}{2})^{ \frac{400}{k} } [/tex]
divide both sides by P
[tex]0.13=( \frac{1}{2})^{ \frac{400}{k} } [/tex]
take the ln of both sides
[tex]ln0.13=( \frac{400}{k})ln \frac{1}{2}} [/tex]
multiply both sides by k
[tex]kln0.13=400ln \frac{1}{2}} [/tex]
divide oth sides by ln0.13
k=[tex] \frac{400ln \frac{1}{2} }{ln0.13} [/tex]
evaluate
k=135.897 days


the half life is 135.897 days

The​ half-life of this radioactive isotope is 135.897 days and this can be determined by using the formula of half-life.

Given :

  • A certain radioactive isotope has leaked into a small stream.
  • Four  hundred days after the​ leak, 13% of the original amount of the substance remained.

The formula of the half-life is given by:

[tex]\rm A =P \left(\dfrac{1}{2} \right)^{\dfrac{t}{t_{1/2}}}[/tex]

where A is the final concentration, P is the initial concentration, t is the time period, and [tex]\rm t_{1/2}[/tex] is the half-life.

Now, substitute the known values in the above formula.

[tex]\rm 0.13P =P \left(\dfrac{1}{2} \right)^{\dfrac{400}{t_{1/2}}}[/tex]

Simplify the above expression.

[tex]\rm 0.13 = \left(\dfrac{1}{2} \right)^{\dfrac{400}{t_{1/2}}}[/tex]

Take the log on both sides.

[tex]\rm log(0.13) = {\dfrac{400}{t_{1/2}}} \times log\left(\dfrac{1}{2} \right)[/tex]

[tex]\rm t_{1/2} = \dfrac{400\times log(0.5)}{log(0.13)}[/tex]

Simplify the above expression in order to evaluate the [tex]\rm t_{1/2}[/tex].

[tex]\rm t_{1/2} = 135.897\;days[/tex]

For more information, refer to the link given below:

https://brainly.com/question/22724843

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