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Home Modern Physics Rules of Randomness Exponential Decay and HalfLife  
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Exponential Decay and HalfLifeMost people know that radioactivity "lasts a certain amount of time," but that simple statement leaves out a lot. As an example, consider the following medical procedure used to diagnose thyroid function. A very small quantity of the isotope ^{131}I, produced in a nuclear reactor, is fed to or injected into the patient. The body's biochemical systems treat this artificial, radioactive isotope exactly the same as ^{127}I, which is the only naturally occurring type. (Nutritionally, iodine is a necessary trace element. Iodine taken into the body is partly excreted, but the rest becomes concentrated in the thyroid gland. Iodized salt has had iodine added to it to prevent the nutritional deficiency known as goiters, in which the iodinestarved thyroid becomes swollen.) As the ^{131}I undergoes beta decay, it emits electrons, neutrinos, and gamma rays. The gamma rays can be measured by a detector passed over the patient's body. As the radioactive iodine becomes concentrated in the thyroid, the amount of gamma radiation coming from the thyroid becomes greater, and that emitted by the rest of the body is reduced. The rate at which the iodine concentrates in the thyroid tells the doctor about the health of the thyroid. If you ever undergo this procedure, someone will presumably explain a little about radioactivity to you, to allay your fears that you will turn into the Incredible Hulk, or that your next child will have an unusual number of limbs. Since iodine stays in your thyroid for a long time once it gets there, one thing you'll want to know is whether your thyroid is going to become radioactive forever. They may just tell you that the radioactivity "only lasts a certain amount of time," but we can now carry out a quantitative derivation of how the radioactivity really will die out. Let P_{surv}(t) be the probability that an iodine atom will survive without decaying for a period of at least t. It has been experimentally measured that half all ^{131}I atoms decay in 8 hours, so we have P_{surv}(8 hr) = 0.5 . Now using the law of independent probabilities, the probability of surviving for 16 hours equals the probability of surviving for the first 8 hours multiplied by the probability of surviving for the second 8 hours, P_{surv}(16 hr) = 0.50 × 0.50 = 0.25 . Similarly we have P_{surv}(24 hr) = 0.50 × 0.5 × 0.5 = 0.125 . Generalizing from this pattern, the probability of surviving for any time t that is a multiple of 8 hours is P_{surv}(t) = 0.5^{t/8 hr} . We now know how to find the probability of survival at intervals of 8 hours, but what about the points in time in between? What would be the probability of surviving for 4 hours? Well, using the law of independent probabilities again, we have P_{surv}(8 hr) = P_{surv}(4 hr) × P_{surv}(4 hr) , which can be rearranged to give
This is exactly what we would have found simply by plugging in P_{surv}(t) = 0.5^{t/8 hr} and ignoring the restriction to multiples of 8 hours. Since 8 hours is the amount of time required for half of the atoms to decay, it is known as the halflife, written t_{1/2}. The general rule is as follows: exponential decay equation P_{surv}(t) = 0.5^{t/t1/2} Using the rule for calculating averages, we can also find the number of atoms, N(t), remaining in a sample at time t: N(t) = N(0) × 0.5^{t/t1/2} Both of these equations have graphs that look like dyingout exponentials, as in the example below.
Rate of decayIf you want to find how many radioactive decays occur within a time interval lasting from time t to time t + Δt, the most straightforward approach is to calculate it like this:
A problem arises when Δt is small compared to t_{1/2}. For instance, suppose you have a hunk of 10^{22} atoms of ^{235}U, with a halflife of 700 million years, which is 2.2×10^{16} s. You want to know how many decays will occur in t = 1 s. Since we're specifying the current number of atoms, t = 0. As you plug in to the formula above on your calculator, the quantity 0.5 ^{Δt/t1/2} comes out on your calculator to equal one, so the final result is zero. That's incorrect, though. In reality, 0.5^{Δt/t1/2} should equal 0.999999999999999968, but your calculator only gives eight digits of precision, so it rounded it off to one. In other words, the probability that a ^{235}U atom will survive for 1 s is very close to one, but not equal to one. The number of decays in one second is therefore 3.2 × 10^{5}, not zero. Well, my calculator only does eight digits of precision, just like yours, so how did I know the right answer? The way to do it is to use the following approximation:
(The symbol << means "is much less than.") Using it, we can find the following approximation:
This also gives us a way to calculate the rate of decay, i.e., the number of decays per unit time. Dividing by Δt on both sides, we have
By the way, none of the equations we've derived so far was the actual probability distribution for the time at which a particular radioactive atom will decay. That probability distribution would be found by substituting N(0) = 1 into the equation for the rate of decay. If the sheer number of equations is starting to seem formidable, let's pause and think for a second. The simple equation for P_{surv} is something you can derive easily from the law of independent probabilities any time you need it. From that, you can quickly find the exact equation for the rate of decay. The derivation of the approximate equations for Δt << t is a little hairier, but note that except for the factors of ln 2, everything in these equations can be found simply from considerations of logic and units. For instance, a longer halflife will obviously lead to a slower rate of decays, so it makes sense that we divide by it. As for the ln 2 factors, they are exactly the kind of thing that one looks up in a book when one needs to know them. Discussion Questions


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