The half-life of phosphorus-32 is 14.26 days. The decay constant is k=4.86×10−2/days. If you accidentally spill phosphorus-32 onto your shoe, how long would it take before 99.9% of the radioactive material has decayed so that you can safely wear the shoes again?
.001=e^-.692t/th
-.693*t/14.26 = ln .001
t= 142 days? check that.
ln(No/N) = kt
Substitute for No = 100
N = 100-99.9
Solve for t in days.
Yeah, thanks.
To determine how long it would take for 99.9% of the radioactive material to decay, we can use the decay constant, which is denoted by the symbol "k." The decay constant is related to the half-life of a radioactive substance through the equation:
k = ln(2) / T½
Where k is the decay constant and T½ is the half-life.
Given that the half-life of phosphorus-32 is 14.26 days, we can plug this value into the equation to find the decay constant (k):
k = ln(2) / 14.26 days
Using natural logarithm properties, this simplifies to approximately 0.0486 per day.
Now that we have the decay constant, we can determine the time required for 99.9% of the radioactive material to decay.
99.9% decay corresponds to 0.001 remaining, or 0.999 decayed.
The equation for exponential decay is:
A(t) = A₀ * e^(-kt)
Where A(t) is the remaining amount of the substance at time t, A₀ is the initial amount, k is the decay constant, and t is the time.
We want to solve for t when A(t) = 0.001. Rearranging the equation, we have:
0.001 = A₀ * e^(-kt)
Dividing both sides of the equation by A₀, we get:
0.001 / A₀ = e^(-kt)
To isolate t, we can take the natural logarithm of both sides:
ln(0.001 / A₀) = -kt
Finally, solving for t, we have:
t = -ln(0.001 / A₀) / k
Using the given information of A₀ = 1, we can substitute this value into the equation:
t = -ln(0.001 / 1) / 0.0486 per day
Evaluating this expression, we find that the time required for 99.9% of the radioactive material to decay is approximately 119.3 days. Therefore, you would need to wait for approximately 119.3 days before it is safe to wear the shoes again