Question

One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at an annual rate of approximately 2.8% per year.

Answer 1

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

Answer 2

The half-life of the strontium-90, which decays exponentially at an annual rate of approximately $2.8\%$ per year is $\boxed{24.4{\text{ years}}}.$

Further explanation:

The exponential decay formula can be expressed as follows,

$\boxed{y = a{{\left( {1 - r} \right)}^x}}$

Here, $a$ represents the initial amount, $y$ is the final amount,r is the rate of decay, and x represents the time.

Given:

The rate of decay is $2.58%.$

Explanation:

Consider the initial amount of the radioactive substance be $A$.

The final amount of the radioactive substance is ${\text{Ap}}$

The ratio of decay can be obtained as follows,

$\begin{aligned}{\text{A}}p &= {\text{A}}{\left( {1 - r} \right)^t}\\p&= \frac{{{\text{A}}{{\left( {1 - r} \right)}^t}}}{{\text{A}}}\\p&= {\left( {1 - 0.028} \right)^t}\\p &= {0.972^t}\\\end{aligned}$

In half-life the final amount is half of initial amount.

$p = 0.5$

The half-life t can be obtained as follows,

$\begin{aligned}0.5&= {0.972^t}\\\log\left( {0.5} \right) &= t \times \log \left( {0.972} \right)\\t&=\frac{{\log \left( {0.5} \right)}}{{\log \left( {0.972} \right)}}\\t&= 24.4\\\end{aligned}$

The half-life of the strontium-90, which decays exponentially at an annual rate of approximately $2.8\%$ per year is $\boxed{24.4{\text{ years}}}.$

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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.