Question

I need the answer for 4 yr,6yr,15yr & find the half life

Answer 1

We have a mass that follows the given exponential model:

`A(t)=500e^(-0.032t)`

a) We have to calculate how much mass is present after 4 years. This correspond to the value of A when t = 4, as t is expressed in years.

We can replace t with 4 and calculate A(4) as:

`\begin{gathered} A(4)=500e^(-0.032(4)) \\ A(4)=500e^(-0.128) \\ A(4)\approx500\cdot0.8799 \\ A(4)\approx440 \end{gathered}`

b) We have to calculate the remaining mass after 6 years. This is similar to the previous point, but with t = 6 years.

We can calculate A(6) as:

`\begin{gathered} A(6)=500e^(-0.032(6)) \\ A(6)=500e^(-0.192) \\ A(6)\approx500\cdot0.8253 \\ A(6)\approx413 \end{gathered}`

c) We have to calaculate the mass after 15 years (t = 15):

`\begin{gathered} A(15)=500\cdot e^(-0.032(15)) \\ A(15)=500\cdot e^(-0.48) \\ A(15)\approx500\cdot0.6188 \\ A(15)\approx309 \end{gathered}`

d) We have to calculate the half-life for this element.

This represents the interval of time t for which the mass is halved. This value is constant for an exponential decay model and we can find this value t as:

`\begin{gathered} (A(x+t))/(A(x))=0.5 \\ \\ (500\cdot e^(-0.032(x+t)))/(500\cdot e^(-0.032x))=0.5 \\ \\ e^(-0.032t)=0.5 \\ \\ \ln(e^(-0.032t))=\ln(0.5) \\ \\ -0.032t=\ln(0.5) \\ \\ t=-(\ln(0.5))/(0.032) \\ \\ t\approx21.66 \end{gathered}`

Answer:

a) 440 g

b) 413 g

c) 309 g

d) 21.66 years