If 20% of a radioactive substance disappears in 70 days, what is its half-life?
y = C*e^(k*t)
where t is time in days
k is the constant of proportionality?
y is the current amount of substance
The Attempt at a Solution
20% disappears, so 100% - 20% = 80% remaining = 0.80 after 70 days
.8*C = C*e^(70*k)
Solve for k
k = -3.188 * 10^-3
Plug k value back into equation to find t at the substance's half-life (.5)
.5*C = C*e^((-3.188 * 10^-3)*t)
Solve for t
t = 217.44
217.44 days is the substance's half-life.
Is that correct?