Respuesta :
Answer:
(1) 3.467 grams of a 4-gram sample of silver-108m would remain after 27 years.
(2) 2.632 grams of a 4-gram sample of silver-108m would remain after 79 years.
Step-by-step explanation:
We are given that the half-life of silver-108m is approximately 130 years.
We can assume this situation as an exponential decay where A(t) be the amount of sample of silver-108m that remain after t years.
Let [tex]A_0[/tex] be the initial amount and r will be the decay constant (which will be a negative number)
Now, it is stated that the half-life of silver-108m is approximately 130 years, this means any amount after this time period will reduce to ([tex]A_0[/tex]/2), that is;
[tex]A(t)=A_0\times e ^{rt}[/tex]
[tex]\frac{A_0}{2} =A_0\times e ^{(r\times 130)}[/tex]
[tex]\frac{1}{2} = e ^{(r\times 130)}[/tex]
Taking log on both sides, we get;
[tex]ln(\frac{1}{2}) = ln(e ^{(r\times 130)})[/tex]
[tex]ln(\frac{1}{2}) = 130 \times r[/tex]
[tex]r=\frac{ln(0.5)}{130}[/tex]
r = -0.0053
(1) We have to find how much of a 4-gram sample of silver-108m would remain after 27 years, that is;
Here, t = 27 years, [tex]A_0[/tex] = 4 grams and r we have calculated above.
So, [tex]A(t)=A_0\times e ^{rt}[/tex]
[tex]A(27)=4 \times e ^{-0.0053 \times 27}[/tex]
A(27) = 3.467 grams
Hence, 3.467 grams of a 4-gram sample of silver-108m would remain after 27 years.
(2) We have to find how much of a 4-gram sample of silver-108m would remain after 79 years, that is;
Here, t = 79 years, [tex]A_0[/tex] = 4 grams and r we have calculated above.
So, [tex]A(t)=A_0\times e ^{rt}[/tex]
[tex]A(79)=4 \times e ^{-0.0053 \times 79}[/tex]
A(79) = 2.632 grams
Hence, 2.632 grams of a 4-gram sample of silver-108m would remain after 79 years.
