You are looking for a function of time, t, that reduces itself by half every 72 hours. If t is in hours, then t/72 increases by 1 every 72 hours. So the function
f(t) = 2-t/72does that. Make sure you see why.
Trouble is, at
What's more, if you multiply f(x) by any constant, it still reduces itself by half every 72 hours. So the function
g(t) = 6 × 1017 × 2-t/72meets all the requirements for part a of the problem.
Part b asks you to express this in terms of an exponential based upon ex. In this section we discussed the identity
bx = ex ln(b)So simply substitute 2 for b and -t/72 for x and you get
g(t) = 6 × 1017 × 2-t/72 = 6 × 1017 × e(-t/72) ln(2)Using a calculator, you can compute that this is the same as
g(t) = 6 × 1017 × e-t/103.8740429Part c wants to know the rate at which the material decays. For each scintillation there will be the loss of one atom of radioactive material. And the term, rate, suggests that what we need to do here is take the derivative with respect to time, t. Using the
g'(t) = (-6 × 1017 / 103.8740429) × e-t/103.8740429and using a calculator, you can see that this is the same as
g'(t) = -5.776226505 × 1015 × e-t/103.8740429Note that this number is negative. That indicates that the number of radioactive atoms is decreasing. The rate of scintillations per hour, then, must be the opposite of that:
scintillations/hour = 5.776226505 × 1015 × e-t/103.8740429
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