Respuesta :
Answer:
a). Half life period = 10.31 years
b). Time taken to decay to 90% of the original amount = 1.57 years
Step-by-step explanation:
Decay of a radioactive element is represented by the exponential function.
[tex]A_{t}=A_{0}e^{-kt}[/tex]
Where [tex]A_{t}[/tex] = Amount of the element after decay
[tex]A_{0}[/tex] = Initial amount
k = Decay constant
t = time or duration
Substance decayed 93.5% after one year.
Therefore, 93.5% of [tex](A_{0})[/tex] = [tex]A_{0}e^{-k}[/tex]
[tex]0.935A_{0}=A_{0}e^{-k}[/tex]
[tex]0.935=e^{-k}[/tex]
Take the log on both the sides of the equation
[tex]ln(0.935)=ln(e^{-k})[/tex]
-0.067208 = -k
k = 0.067208
(a). For half life of the substance
[tex]\frac{A_{0}}{2}=A_{0}e^{-0.067208t}[/tex]
\frac{1}{2}=e^{-0.067208t}
Take the log on each side
[tex]-ln2=ln(e^{-0.067208t})[/tex]
0.693147 = 0.067208t
t = [tex]\frac{0.693147}{0.067208}[/tex]
= 10.31 years
(b). Time taken to decay to 90% of the original amount
[tex]0.90A_{0}=A_{0}e^{-0.067208t}[/tex]
[tex]0.90=e^{-0.067208t}[/tex]
Take log on both the sides of the equation
[tex]ln(0.90)=ln(e^{-0.067208t})[/tex]
-0.1053605 = -0.067208t
t = [tex]\frac{0.1053605}{0.067208}[/tex]
= 1.57 years
Using an exponential equation, we have that:
a) The half-life of the substance is of 10.31 years.
b) It would take 1.57 years for the sample to decay to 90% of its original amount.
The equation for the amount of substance after t years is given by:
[tex]A(t) = A(0)e^{-kt}[/tex]
In which:
- A(0) is the initial amount.
- k is the decay rate, as a decimal.
93.5% of its original amount after a year.
This means that [tex]A(1) = 0.935A(0)[/tex], and this is used to find k.
[tex]A(t) = A(0)e^{-kt}[/tex]
[tex]0.935A(0) = A(0)e^{-k}[/tex]
[tex]e^{-k} = 0.935[/tex]
[tex]\ln{e^{-k}} = \ln{0.935}[/tex]
[tex]-k = \ln{0.935}[/tex]
[tex]k = 0.0672[/tex]
Thus:
[tex]A(t) = A(0)e^{-0.0672t}[/tex]
Item a:
The half-life is t for which:
[tex]A(t) = 0.5A(0)[/tex]
Then
[tex]0.5A(0) = A(0)e^{-0.0672t}[/tex]
[tex]e^{-0.0672t} = 0.5[/tex]
[tex]\ln{e^{-0.0672t}} = \ln{0.5}[/tex]
[tex]-0.0672t = \ln{0.5}[/tex]
[tex]t = -\frac{\ln{0.5}}{0.0672}[/tex]
[tex]t = 10.31[/tex]
The half-life of the substance is of 10.31 years.
Item b:
This is t for which:
[tex]A(t) = 0.9A(0)[/tex]
Then
[tex]0.9A(0) = A(0)e^{-0.0672t}[/tex]
[tex]e^{-0.0672t} = 0.9[/tex]
[tex]\ln{e^{-0.0672t}} = \ln{0.9}[/tex]
[tex]-0.0672t = \ln{0.9}[/tex]
[tex]t = -\frac{\ln{0.9}}{0.0672}[/tex]
[tex]t = 1.57[/tex]
It would take 1.57 years for the sample to decay to 90% of its original amount.
A similar problem is given at https://brainly.com/question/16984805
