90Sr is one of the harmful radionuclides resulting from nuclear fission explosions. It decays by a first-order process with a half-life of 28 years. How many years would it take for 99.99% of a given sample released in the atmosphere to disintegrate? Answer in units of years.
i solved for k and got .02476 from the half life equation for first order and then i know i need to use this equation: ln(No/N) = kt but i don't know what to use for No/N...
I would just make up a convenient number, like 100 for example for No. That would make 100 x 0.9999 = 99.99 to disappear which leaves 100-99.99 = 0.01 = N. Check my math
372.12 yrs
To find how many years it would take for 99.99% of a sample of 90Sr to disintegrate, let's set up the equation using the exponential decay formula for first-order processes:
N(t) = N₀ * e^(-kt)
Where:
- N(t) is the remaining quantity of 90Sr after time t
- N₀ is the initial quantity of 90Sr
- k is the decay constant
- t is the time (in years)
First, let's find the value of k using the half-life equation for first-order processes:
t₁/₂ = (ln 2) / k
Given that the half-life of 90Sr is 28 years, we substitute this value into the equation:
28 = (ln 2) / k
Rearranging the equation to solve for k:
k = (ln 2) / 28
Using the value of k, we can now find the time it takes for 99.99% of the sample to disintegrate. Let N(t) be 0.0001 times N₀ (99.99% reduction):
0.0001N₀ = N₀ * e^(-kt)
Dividing both sides by N₀ and taking the natural logarithm of both sides:
ln(0.0001) = -kt
Substituting the value of k we found earlier:
ln(0.0001) = (-((ln 2) / 28)) * t
Now, solve for t:
t = (ln(0.0001)) / (-((ln 2) / 28))
Calculating this expression, we find the number of years it would take for 99.99% of the 90Sr sample to disintegrate.