The model for radioactive decay is y=y0e−kt. A radioactive substance has a half-life of 230 years. If 35 grams are present today, in how many years will 14 grams be present?
less than half takes longer than a half life
m = M e^-kt
m/M = 1/2 when t = 230
0.5 = e^-230 k
ln 0.5 = -230 k = -0.693
k = 0.003
so
14 = 35e^-0.003 t
0.4 = e^-0.003 t
ln 0.4 = -0.916 = -0.003 t
t = 305
well duh. I tried to cut too many corners, and even then didn't catch my error!
To solve this problem, we can use the model for radioactive decay, which is given by the formula y = y0 * e^(-kt), where:
- y is the amount of the radioactive substance at a given time
- y0 is the initial amount of the radioactive substance
- k is the decay constant
- t is the time in years
Given that the half-life of the radioactive substance is 230 years, we can use this information to find the decay constant (k).
Since the half-life is the time it takes for the substance to decay to half of its original amount, we can set up the following equation:
1/2 = e^(-k * 230)
To solve for k, we can take the natural logarithm of both sides of the equation:
ln(1/2) = -k * 230
Using the fact that ln(1/2) is equal to -ln(2), we can rewrite the equation as:
-ln(2) = -k * 230
Finally, we can solve for k:
k = ln(2) / 230
Now that we have the decay constant (k), we can use it to determine the time it takes for 14 grams to be present, starting from an initial amount of 35 grams.
Using the formula y = y0 * e^(-kt), we can substitute the known values:
14 grams = 35 grams * e^(-k * t)
Divide both sides of the equation by 35 grams:
14/35 = e^(-k * t)
Now, we can solve for t by taking the natural logarithm of both sides:
ln(14/35) = -k * t
Substituting the value of k we found earlier:
ln(14/35) = -(ln(2) / 230) * t
Finally, we can solve for t by dividing both sides of the equation by -(ln(2) / 230):
t = ln(14/35) / (ln(2) / 230)
Calculating t using a calculator, we find that t ≈ 120.58 years.
Therefore, it will take approximately 120.58 years for 14 grams of the radioactive substance to be present.
e^(-t/230) = 14/35
-t/230 = ln(14/35)
t = 230 ln(35/14) = 210.747
makes sense, since it will take 230 years to drop by half to 17.5 g