Question

where Ais the initial amount present and A is the amount presenteStrontium-90 is a radioactive material that decays according to the function A(t) = Age-0.02441at time t (in years). Assume that a scientist has a sample of 800 grams of strontium-90.ce(a) What is the decay rate of strontium-90?(b) How much strontium-90 is left after 30 years?(c) When will only 200 grams of strontium-90 be left?(d) What is the half-life of strontium-90?-stilCOREcio (a) The decay rate of strontium-90 is %.(Type an integer or a decimal. Include the negative sign for the decay rate.)LiolewMat

Answer 1

The given information is:

- The function of decay is:

`A(t)=A_0e^(-0.0244)`

Where A0 is the initial amount of strontium-90, A is the amount present at time t (in years).

- The initial amount is 800 grams.

a. What is the decay rate of strontium-90?

The given formula is written in the form:

`A(t)=A_0e^(rt)`

Where r is the decay rate in decimal form, so:

`r=-0.0244*100\%=-2.44\%`

The decay rate is -2.44%.

b. How much strontium-90 is left after 30 years?

Replace t=30 and solve:

`\begin{gathered} A(30)=800g*e^(-0.0244*30) \\ A(30)=800g*e^(-0.732) \\ A(30)=800g*0.481 \\ A(30)=384.8g \end{gathered}`

There is 384.8 grams after 30 years.

c. When will only 200 g of strontium-90 be left?

A(t)=200g, then replace it and solve for t:

`\begin{gathered} 200g=800g*e^(-0.0244t) \\ (200g)/(800g)=e^(-0.0244t) \\ \ln((200g)/(800g))=\ln e^(-0.0244t) \\ -1.386=-0.0244t \\ t=(-1.386)/(-0.0244) \\ t=56.8 \end{gathered}`

There will be 200 g left after 56.8 years.

d. What is the half-life of strontium-90?

The half-life is when A(t)=A0/2, then if we replace this into the decay formula we obtain:

`\begin{gathered} (A_0)/(2)=A_0*e^(-0.0244t) \\ Simplify\text{ A0 on both sides} \\ (1)/(2)=e^(-0.0244t) \\ \ln(0.5)=\ln e^(-0.0244t) \\ -0.693=-0.0244t \\ t=(-0.693)/(-0.0244) \\ t=28.4 \end{gathered}`

The half-life of strontium-90 is 28.4 years.