Measurements of a certain isotope tell you that the decay rate decreases from 9354 decays/minute to 3185 decays/minute over a period of 3.00 days. The half-life of this isotope in units of days and to one decimal place is:
Question
Measurements of a certain isotope tell you that the decay rate decreases from 9354 decays/minute to 3185 decays/minute over a period of 3.00 days. The half-life of this isotope in units of days and to one decimal place is:
Solution
To solve this problem, we need to use the formula for exponential decay, which is:
N = N0 * (1/2)^(t/T)
where: N is the final amount of the substance, N0 is the initial amount of the substance, t is the time that has passed, and T is the half-life of the substance.
We are given that N0 = 9354 decays/minute, N = 3185 decays/minute, and t = 3 days. We want to find T.
First, we need to convert the time from days to minutes, since the decay rate is given in decays/minute. There are 60 minutes in an hour and 24 hours in a day, so t = 3 days * 24 hours/day * 60 minutes/hour = 4320 minutes.
Next, we can plug the given values into the formula and solve for T:
3185 = 9354 * (1/2)^(4320/T)
Divide both sides by 9354 to isolate the exponential term:
0.3404 = (1/2)^(4320/T)
Take the natural logarithm of both sides to get rid of the exponential:
ln(0.3404) = 4320/T * ln(1/2)
Solve for T:
T = 4320 * ln(1/2) / ln(0.3404)
Using a calculator, we find that T ≈ 2.6 days. So, the half-life of the isotope is approximately 2.6 days.
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