Question

strontium - 90 is a radioactive material that decays according to the function ( a(t)=a_{0}e^{-0.0244t} ), where ( a_{0} ) is the initial amount present and ( a ) is assume that a scientist has a sample of 800 grams of strontium - 90
(a) what is the decay rate of strontium - 90?
(b) how much strontium - 90 is left after 40 years?
(c) when will only 600 grams of strontium - 90 be left?
(d) what is the half - life of strontium - 90?
(a) the decay rate of strontium - 90 is ( -2.44% ).
(type an integer or a decimal. include the negative sign for the decay rate.)
(b) approximately 301 grams of strontium - 90 is left after 40 years.
(do not round until the final answer. then round to the nearest whole number as needed.)
(c) only 600 grams of strontium - 90 will be left in about 11.79 years.
(do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Analyze the decay function

The general form of exponential decay is \(A(t)=A_0e^{rt}\), where \(r\) is the decay rate. In the given function \(A(t) = A_0e^{- 0.0244t}\), comparing with the general form, we can directly identify the decay rate \(r=-0.0244\). To convert this to a percentage, we multiply by \(100\), so the decay rate is \(-2.44\%\).

Step2: Calculate the amount left after 40 years

We are given \(A_0 = 800\) grams and \(t = 40\) years. Substitute these values into the function \(A(t)=A_0e^{-0.0244t}\).

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Using a calculator, \(e^{-0.976}\approx0.376\), then \(A(40)=800\times0.376 = 300.8\approx301\) grams.

Step3: Find the time when 600 grams are left

We are given \(A(t)=600\) and \(A_0 = 800\). Substitute into the function \(A(t)=A_0e^{-0.0244t}\), so \(600 = 800e^{-0.0244t}\).
First, divide both sides by \(800\): \(\frac{600}{800}=e^{-0.0244t}\), which simplifies to \(0.75=e^{-0.0244t}\).
Take the natural logarithm of both sides: \(\ln(0.75)=- 0.0244t\).
Since \(\ln(0.75)\approx - 0.2877\), then \(t=\frac{-0.2877}{-0.0244}\approx11.79\) years.

Step4: Calculate the half - life

The half - life occurs when \(A(t)=\frac{A_0}{2}\). Substitute into the function \(\frac{A_0}{2}=A_0e^{-0.0244t}\).
Divide both sides by \(A_0\): \(\frac{1}{2}=e^{-0.0244t}\).
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=-0.0244t\).
Since \(\ln(\frac{1}{2})\approx - 0.6931\), then \(t=\frac{-0.6931}{-0.0244}\approx28.4\) years.

Answer:

(a) The decay rate of strontium - 90 is \(-2.44\%\).
(b) Approximately \(301\) grams of strontium - 90 is left after 40 years.
(c) Only 600 grams of strontium - 90 will be left in about \(11.79\) years.
(d) The half - life of strontium - 90 is approximately \(28.4\) years.