A radioactive isotope has a half life of 25 years. There was 100 mg of the isotope in 2005.
(a) Find an expression for the amount of the isotope, A(t), that is still active using time t, measured in years since 2005. Your expression must be of the form A(t)=Pe^ct (base "e").
You are modelling
amount = P e^(kt), where P is the original amount
so when t = 0 , P = 100
when t = 25, P = 50
50 = 100 e^(25k)
.5 = e^(25k)
ln .5= 25k
k = ln .5/25 = appr -0.027725... (I stored this in may calculator's memory)
amount = 100 e^(-.027725...(t))
Of course, if we had used a base of 1/2 it simply would have been
amount = 100 (1/2)^(t/25)
check: let t = 25 in both equations, both yield an amount of 50
Thank you!!
To find the expression for the amount of the isotope that is still active, we need to use the formula for exponential decay, which is given by:
A(t) = A₀ * e^(kt)
Where:
A(t) is the amount of the isotope at time t
A₀ is the initial amount of the isotope
k is the decay constant
t is the time
In this case, the half-life of the isotope is given as 25 years. This means that the amount of the isotope is halved every 25 years. We can use this information to find the decay constant, k.
The decay constant (k) can be calculated using the formula:
k = ln(2) / half-life
where ln denotes the natural logarithm.
In this case, the half-life is 25 years, so:
k = ln(2) / 25
Now, we can substitute the values of A₀ = 100 mg and k into the exponential decay formula:
A(t) = 100 * e^((ln(2) / 25) * t)
Therefore, the expression for the amount of the isotope that is still active using time t, measured in years since 2005, is:
A(t) = 100 * e^(0.02773 * t)
where t is the time in years since 2005.