Question

Phosphorus-32 (p-32) has a half-life of 14.2 days. if 250 g of this substance are present initially, find the amount q(t) present after t days. (round your growth constant to four decimal places.)

Answer 1

f(t) = 250g * 0.9524^t Since we have a half-life of 14.2 days we want to find an X such that X^14.2 = 0.5 We can do that easily using logarithms. Just take the logarithm of 0.5, divide by 14.2, and the get the anti-log. So 10^(log(0.5)/14.2) = 10^(-0.301029996/14.2) =10^(-0.021199295) = 0.952359031 = 0.9524 So the function now becomes f(t) = 250g * 0.9524^t Let's check the function with a few test points at t = 14.2 and t = 28.4 which should return a result of 125 and 62.5 grams respectively. It will actually return slightly larger values since the growth constant was rounded up. But should be close enough to verify the formula f(14.2) = 250g * 0.9524^14.2 = 250 * 0.500305517 = 125.0763791 f(28.4) = 250g * 0.9524^28.4 = 250 * 0.25030561 = 62.57 As predicted, the values are close, but slightly high. To demonstrate the effects of the rounding, consider this function using 0.952359031 as the growth factor. 250g * 0.952359031^ 28.4 = 250 * 0.25 = 62.5

Answer 2

The amount of substance left out of a given amount with the given half-life and the number of days can be calculated through the equation,

                          q(t) = qo(0.5)^(t/h)

where q(t) is the amount of substance after t days, qo is the initial amount of substance. q(t) and q(o) should have the same units. t is the number of days and h is the half-life.