Respuesta :
From the statement you can find the function that represents the percent of strontium-90 that remains after t years of the nuclear accident.
Call P the percent:
---> P = e ^ (- 0.025t)
Now you can also estimate the half-life of the isotope strontium-90:
Call Pi the initial percent and P2 the half percent:
P2 = 0.5 * P initial =>
P2 = 0.5 * Pinitial = Pinitial * e ^ (-0.025 t)
=> 0.5 = e ^ (-0.025t)
=> ln 0.5 = -0.025t
=> t = - ln (0.5) / 0.025
=> t = 27.7 years <----- half-time
Call P the percent:
---> P = e ^ (- 0.025t)
Now you can also estimate the half-life of the isotope strontium-90:
Call Pi the initial percent and P2 the half percent:
P2 = 0.5 * P initial =>
P2 = 0.5 * Pinitial = Pinitial * e ^ (-0.025 t)
=> 0.5 = e ^ (-0.025t)
=> ln 0.5 = -0.025t
=> t = - ln (0.5) / 0.025
=> t = 27.7 years <----- half-time
The half-life of the strontium-90, which decays exponentially at an annual rate of approximately [tex]2.8\%[/tex] per year is [tex]\boxed{24.4{\text{ years}}}.[/tex]
Further explanation:
The exponential decay formula can be expressed as follows,
[tex]\boxed{y = a{{\left( {1 - r} \right)}^x}}[/tex]
Here, [tex]a[/tex] represents the initial amount, [tex]y[/tex] is the final amount,r is the rate of decay, and x represents the time.
Given:
The rate of decay is [tex]2.58%.[/tex]
Explanation:
Consider the initial amount of the radioactive substance be [tex]A[/tex].
The final amount of the radioactive substance is [tex]{\text{Ap}}[/tex]
The ratio of decay can be obtained as follows,
[tex]\begin{aligned}{\text{A}}p &= {\text{A}}{\left( {1 - r} \right)^t}\\p&= \frac{{{\text{A}}{{\left( {1 - r} \right)}^t}}}{{\text{A}}}\\p&= {\left( {1 - 0.028} \right)^t}\\p &= {0.972^t}\\\end{aligned}[/tex]
In half-life the final amount is half of initial amount.
[tex]p = 0.5[/tex]
The half-life t can be obtained as follows,
[tex]\begin{aligned}0.5&= {0.972^t}\\\log\left( {0.5} \right) &= t \times \log \left( {0.972} \right)\\t&=\frac{{\log \left( {0.5} \right)}}{{\log \left( {0.972} \right)}}\\t&= 24.4\\\end{aligned}[/tex]
The half-life of the strontium-90, which decays exponentially at an annual rate of approximately [tex]2.8\%[/tex] per year is [tex]\boxed{24.4{\text{ years}}}.[/tex]
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Exponential decay
Keywords: half-life, 2.8%, twenty percent, contaminants, nuclear accident, Chernobyl, strontium-90, exponentially, 2.5% per year, ratio of strontium, p, function of time, radioactive substance, decays, ten years, each year, rate of decay each year, substance.
