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