Answer:
At t = 2 minutes, remaining quantity of the radioactive element is 12.5 mg.
Step-by-step explanation:
To get the answer of this question we will solve this further with the help of the equation [tex]A_{t}=A_{0}e^{-kt}[/tex]
where k = decay constant
t = time for decay
[tex]A_{0}[/tex] = Initial quantity taken
From the graph attached we can say that 50 mg of a radioactive element remained half in 1 minute.
So the equation becomes
[tex]50=25e^{-k(1)}[/tex]
Now we take natural log on both the sides of the equation
ln50 = ln[25.[tex]e^{-k}[/tex]
3.912 = ln25 + [tex]ln(e^{-k})[/tex]
3.912 = 3.219 + (-k)lne
3.912 - 3.219 = -k [since lne = 1]
0.693 = -k
k = -0.693
Now we will calculate the remaining quantity of the element after 2 minutes
[tex]A_{t}=50.e^{-(0.693)(2)}[/tex]
= [tex]50.e^{-1.386}[/tex]
= [tex]\frac{50}{e^{1.386}}[/tex]
= [tex]\frac{50}{3.9988}[/tex]
= 12.50 mg
Now we confirm this value from the graph.
At t = 2 minutes, remaining quantity of the radioactive element is 12.5 mg.
