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One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a rate of approximately 2.5% per year.

AWrite the ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident.

Verified Answer :

JeanaShupp

Answer: \(P=0.975^t\)


Step-by-step explanation:

We know that the general form of the exponential decay formula is


\(y=A(1-r)^t\), where y is final amount remaining after t time, A is the original amount and r is the rate of decay


Now, the ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident. \(P=\frac{A(1-0.02)^t}{A}=\frac{(0.975)^t}{1}\)

Hence, the ratio of remaining since the nuclear accident is \(P=0.975^t\)

The ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident is \(\boxed{p = {{0.975}^t}}.\)


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 \(2.5\%.\)


Explanation:

Consider the initial amount of the radioactive substance be \(A\).

The final amount of the radioactive substance is \({{Ap}}\)

The ratio of decay can be obtained as follows,

\(\begin{aligned}{\text{A}}p &= {\text{A}}{\left( {1 - r} \right)^t}\\p&= \frac{{{\text{A}}{{\left( {1 - r} \right)}^t}}}{{\text{A}}}\\p&= {\left( {1 - 0.025} \right)^t}\\p&= {0.975^t}\\\end{aligned}\)


The ratio of strontium-90 remaining, \(p\) , as a function of years,\(t\), since the nuclear accident is \(\boxed{p = {{0.975}^t}}.\)


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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Exponential decay


Keywords: 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.

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