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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.

Verified Answer :

apologiabiology

ok so

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
\(A=P( \frac{1}{2})^{ \frac{t}{k} }\)

given
13% of P remained=A
t=400

replace A with 0.13P and t with 400


0.13P=\(P( \frac{1}{2})^{ \frac{400}{k} }\)
divide both sides by P
\(0.13=( \frac{1}{2})^{ \frac{400}{k} }\)
take the ln of both sides
\(ln0.13=( \frac{400}{k})ln \frac{1}{2}}\)
multiply both sides by k
\(kln0.13=400ln \frac{1}{2}}\)
divide oth sides by ln0.13
k=\(\frac{400ln \frac{1}{2} }{ln0.13}\)
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:

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

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

Now, substitute the known values in the above formula.

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

Simplify the above expression.

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

Take the log on both sides.

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

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

Simplify the above expression in order to evaluate the \(\rm t_{1/2}\).

\(\rm t_{1/2} = 135.897\;days\)

For more information, refer to the link given below:

https://brainly.com/question/22724843

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