How much time would it take for 5.2 x 105 atoms of fermium-253 (half-life 3 days) to decay to 6.5 x 104 atoms?
{[log(5.2E5 / 6.5E4)] / [log(2)]} * 3 d
the fraction left after t days is (1/2)^(t/3)
So, solve for t in
(1/2)^(t/3) = (6.5 * 10^4)/(5.2 * 10^5) = 0.125 = (1/2)^3
so, t/3 = 3
t = 9 days, or 3 half-lives
5.2/.65 = 8
1/8 left
(1/2)(1/2)(1/2) =1/8
three days :)
I mean 3 half lives, 9 days
To determine the time it would take for the given number of fermium-253 atoms to decay, we can use the radioactive decay formula:
N = N₀ * (1/2)^(t / T)
Where:
- N is the final number of atoms (6.5 x 10^4 atoms)
- N₀ is the initial number of atoms (5.2 x 10^5 atoms)
- t is the time
- T is the half-life of fermium-253 (3 days)
We can rearrange the formula to solve for t:
t = T * log₂(N / N₀)
Substituting the values we have:
t = 3 days * log₂(6.5 x 10^4 / 5.2 x 10^5)
To calculate the logarithm, we use the fact that log₂(a / b) = log₂(a) - log₂(b):
t = 3 days * (log₂(6.5 x 10^4) - log₂(5.2 x 10^5))
Using a calculator or logarithmic table, we can find these logarithmic values:
t ≈ 3 days * (-1.6383 - (-2.2840))
t ≈ 3 days * (0.6457)
t ≈ 1.9371 days
So, it would take approximately 1.9371 days for 5.2 x 10^5 atoms of fermium-253 to decay to 6.5 x 10^4 atoms.