The only difference between exponential-growth equations and exponential-decay equations is that the growth constant for decay situations is negative. The equation itself is just the same as for exponential growth, but you should expect a negative value for the constant. If you get a positive value, you should probably go back and check your work.

Content Continues Below

- 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.

Affiliate

Advertisement

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 "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% of that amount (converted to 0.5 for use inside the formula).

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 will 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 with this equation:

0.5 = *e*^{9.45k}

Solving for the decay constant *k*, I get:

0.5 = *e*^{9.45k}

ln(0.5) = 9.45*k*

^{ln(0.5)}/_{9.45} = *k* = −0.073348907996...

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

The constant was negative, as expected, because this was a *decay* problem. If I'd ended up with a positive value, this would have signalled to me that I'd made a mistake somewhere.

Content Continues Below

- Some people are frightened of certain medical tests because the tests involve the injection of radioactive materials. A hepatobiliary scan of my gallbladder involved an injection of 0.5 cc's (or about one-tenth of a teaspoon) of Technetium-99m, which 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 remained from my gallbladder scan after twenty-four hours. Show your work below.

For this exercise, 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. I do not have the decay constant but, by using the half-life 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}

ln(0.5) = 6*k*

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 was left after twenty-four hours:

*A* = 0.5*e*^{(ln(0.5)/6)(24)} = 0.03125

There will be no more than 0.03125 mL (or about 1/160 of a teaspoon) of Technitium-99m remaining in twenty-four hours.

It can be very useful, sometimes in unexpected places, to know that 1 cc (that is, one cubic centimeter) is equal to 1 mL (that is, one milliliter). I would recommend that you add this bit of knowledge to your store.

Affiliate

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 chance of mutation is very low. (Whatever you're being treated for is the greater danger.) The half-life 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", shall we say?, into the sewer system.

- 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 evaluates the ratio of radioactive carbon-14 to stable carbon-12. Given the amount of carbon-12 contained a measured 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, assuming the proposed age is correct. According to your equipment, there remains 1.0 × 10^{−12}grams. Is there a possibility that this is a genuine document? Or 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) = 5730*k*

^{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.

URL: https://www.purplemath.com/modules/expoprob3.htm

© 2022 Purplemath, Inc. All right reserved. Web Design by