A defective nuclear reactor is leaking radioactive iodine-131. The rate of decay of iodine-131is given by the function defined by A(t)=A_0 e^(-0.087t), where t is time in days. How long will it take any quantity of iodine-131 to decay to 25% of its initial amount?
just solve for t:
e^(-.087t) = .25
t = 16.00 days
Alternatively, if you express the function as a power of 1/2, then since
1/2 = e^(ln 1/2) = e^(-ln2)
(1/2)^x = e^(-x/ln2) = e^(-x/.693)
e^(-.087t) = (1/2)^(.125t) = (1/2)^(t/8)
So, the half-life is 8 days
25% left means 2 half-lives, or 16 days
Thank you, Steve!
To find the time it takes for any quantity of iodine-131 to decay to 25% of its initial amount, we need to solve the function A(t) = 0.25 * A₀, where A₀ represents the initial amount of iodine-131.
Given that A(t) = A₀ * e^(-0.087t), we substitute A(t) with 0.25 * A₀:
0.25 * A₀ = A₀ * e^(-0.087t)
Dividing both sides by A₀, we have:
0.25 = e^(-0.087t)
To solve for t, we can take the natural logarithm of both sides since ln(e^x) cancels out the exponential function:
ln(0.25) = ln(e^(-0.087t))
Using the property ln(e^x) = x, we have:
ln(0.25) = -0.087t
Now, we divide both sides by -0.087:
t = ln(0.25) / -0.087
Using a calculator or mathematical software, we can evaluate this expression to find the value of t.
Note: When evaluating the expression ln(0.25) / -0.087, make sure to use a negative value for t since the time cannot be negative.