More Exponential Word Problems (page 3 of 3) Sections: Logbased word problems, exponentialbased word problems
Since the decay rate is given in terms of minutes, then time t will be in minutes. However, I note that there is no beginning or ending amount given. How am I supposed to figure out what the decay constant is? I can do this by working from the definition of "halflife": in the given amount of time (in this case, 9.45 minutes), half of the initial amount will be gone. That is, from t = 0 to t = 9.45, I will have gone from 100% ("1") of however much I started with to 50% ("0.5") of that amount. Since the halflife does not depend on how much I started with, I can either pick an arbitrary beginning amount (such as 100 grams) and then calculate the decay constant after 9.45 minutes, at which point only 50 grams will remain (the other 50 grams have mutated into some other isotope or element). Or else I can just deal with the 50% that is left. Either way, I will end up dealing with this equation: 0.5 = e^{9.45k} Solving for the decay constant, I get: 0.5 = e^{9.45k} The decay constant for Magnesium27 is –0.07335/minute. The constant was negative, because this was a decay problem.
For this exercise, I need to find the ending amount A of Technetium99m. Recalling that 1 cc (cubic centimeter) equals 1 mL (milliliter), I know that the beginning amount is P = 0.5 mL. The ending time is 24 hours. I do not have the decay constant but, by using the halflife information, I can find it. (Since this is a decay problem, I expect the constant to be negative. If I end up with a positive value, I'll know that I should go back and check my work.) In 6 hours, there will be 50% of the original amount left: 0.5
= e^{6k} (This evaluates to about –0.1155, but I'll leave the decay constant in exact form to avoid roundoff error.) Now that I have the decay constant, I can find out how much Technitium99m was left after twentyfour hours: Copyright © Elizabeth Stapel 20022011 All Rights Reserved A = 0.5e^{(ln(0.5)/6)(24)} = 0.03125 There will be no more than 0.03125 mL (about 1/160 of a teaspoon) of Technitium99m remaining in twentyfour hours. By the way: Technetium99m is one of the most commonly used radioisotope for these medical purposes. Its radiation is extremely lowenergy, so the chance of mutation is very low. (Whatever you're being treated for is the greater danger.) The halflife is just long enough for the doctors to have time to take their pictures. The dose I was given is about as large as these injections typically get. Your body does not easily absorb this chemical, so most of the injection is voided into the sewer system.
First, as usual, I have to find the decay rate. (In "real life", you'd look this up on a table, or have it programmed into your equipment, but this is math, not "real life".) The halflife is 5730 years, so: 0.5 = e^{5730k} I'll leave the decay constant in this "exact" form to avoid roundoff error. I have the beginning (expected) amount of C14 and the present (ending) amount; from this information, I can calculate the age of the parchment: 1.0 ×
10^{–12} = (1.3
× 10^{–12})e^{(ln(0.5)/5730)t} Then the parchment is about 2170 years old, much less than the necessary 3250 years ago that the Trojan War took place. But the parchment is indeed old, so this isn't a total fake. Since the parchment is genuinely old (2170 years), but clearly not old enough to be the actual writings of a soldier in the Trojan War (3250 years), either this is a muchyounger copy of an earlier document (in which case it is odd that there are no references to it in other documents, since only famous works tended to be copied), or, which is more likely, this is a recent forgery written on a notquiteoldenough ancient parchment. If possible, the ink should be tested, since a recent forgery would use recentlymade ink. << Previous Top  1  2  3  Return to Index



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