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More Exponential Word Problems (page 3 of 3)

Sections: Log-based word problems, exponential-based word problems

• Radio-isotopes of different elements have different half-lives. Magnesium-27 has a half-life of 9.45 minutes. What is the decay constant for Magnesium-27? Round to five decimal places.

Since the decay rate is given in terms of minutes, then time t will be in minutes. However, you may 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 "half-life": 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 half-life 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 
ln(0.5) = 9.45k 
^ln(0.5)/_9.45 = k = –0.073348907996... 

The decay constant for Magnesium-27 is –0.07335/minute.

The constant was negative, because this was a decay problem.

Some people are frightened of certain medical tests because the tests involve the injection of radioactive materials. I recently underwent a hepatobiliary scan of my gallbladder, which involved an injection of 0.5 cc's (or about one-tenth of a teaspoon) of Technetium-99m. Technetium-99m has a half-life of almost exactly 6 hours. While undergoing the test, I heard the technician telling somebody on the phone that "in twenty-four hours, you'll be down to background radiation levels." Figure out just how much radioactive material will remain from my gallbladder scan in twenty-four hours. Show your work below.

For this problem, I need to find the ending amount A of Technetium-99m. 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. But I do not have the decay constant.

Using the half-life information, I can find the decay constant. (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% left:

0.5 = e^6k 
ln(0.5) = 6k 
ln(0.5)/6 = k 

(This evaluates to about –0.1155, but I'll leave the decay constant in exact form to avoid round-off error.) Now that I have the decay constant, I can find out how much Technitium-99m is left in twenty-four hours:   Copyright © Elizabeth Stapel 2006-2008 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 Technitium-99m remaining in twenty-four hours.

By the way: Technetium-99m is one of the most commonly used radioisotope for these medical purposes. Its radiation is extremely low-energy, so the danger of mutation is very low. The half-life is long enough that the doctors have time to take pictures, but not so long as to pose health problems. The dose listed above, 0.5 mL, seems to be about as large as Technetium-99m injections typically get. Technetium-99m is not readily absorbed by the body, so most of the injected material is voided (that is to say, you, um, flush it into the sewer system) long before it ever gets a chance to decay within your body. To check the decay-rates and residual amounts for other isotopes, try here.

• Carbon-14 has a half-life of 5730 years. You are presented with a document which purports to contain the recollections of a Mycenaean soldier during the Trojan War. The city of Troy was finally destroyed in about 1250 BC, or about 3250 years ago. Carbon-dating works by evaluating the ratio of radioactive carbon-14 to the stable isotope carbon-12. Considering the amount of carbon-12 contained a sample cut from the document, there would have been about 1.3 × 10^–12 grams of carbon-14 in the sample when the parchment was new. According to your equipment, there remains 1.0 × 10^–12 grams. Is there a possibility that this is a genuine document? Is this instead a recent forgery? Justify your conclusions.

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 half-life is 5730 years, so:

0.5 = e^5730k 
ln(0.5) = 5730k 
^ln(0.5)/₅₇₃₀ = k 

I'll leave the decay constant in this "exact" form to avoid round-off error.

I have the beginning (expected) amount of C-14 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 
1}/_1.3 = e^{(ln(0.5)/5730)t} 
ln( ¹/_1.3 ) = ( ^ln(0.5)/₅₇₃₀ )t 
5730ln( ¹/_1.3 ) / ln(0.5) = t = 2168.87160124... 

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 much-younger 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 not-quite-old-enough ancient parchment. If possible, the ink should be tested, since a recent forgery would use recently-made ink.

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