Statistics of Nuclear DecayThe ”radioactivity” of a radioactive sample decreases in time. By radioactivity wemean the number of times a radioactive decay occu r s in the sample per unit time. Astandard unit of activity is decays per secon d . In our exp er i m ents in the laboratorywe will observe that the radioactivity decreases in a very simple manner, by a ”half-life” formula. In this chapter, we will present the half-life formula, discuss the physicsbehind the formula, and derive a n expression for the statistical uncertainty for ourmeasurements of activity. These three ideas derive from one principle.”Half-life decay formula”The number of radioactive nuclei and the activity of a sample decrease in time ina very simple way, via a ” h al f- l ife formula”. Let N(t) be the number of radioactivenuclei in a sample. There exists a particular time τ, called th e half-life, such thatN(t) decreases by a factor of 1/2 after a time τ , for any t i m e t. In other words:N(t + τ ) = N(t)/2 for any ti m e t. For example, suppose initially we start with 64radioactive nuclei. After a time τ we have 32 ra d i oac ti ve nuclei. After a tim e 2τ wehave 16 radioactive nuclei, etc. Th e table below summarizes the results:Time N(t)0 64τ 32 = 64(1/2)22τ 16 = 64(1/2)23τ 8 = 64(1/2)34τ 4 = 64(1/2)4From the ri ght column, one can deduce a simple formula for N(t):N(t) = N0(12)t/τ(1)where N0is the number of radioactive nuclei at time t = 0. The formula holds forour simple ex am ple, and al so satisfies N(t + τ) = N(t)/2 for any time t. In fact theformula is valid for any time t, even for tim es between multiples of τ.As previously mentioned, the activity of a sample is the number of decays/time.Units for activity that we will use in this class are:1 Curie = 3.7 x 1010decays/sec1 milli-Cur ie = 1 mCi = 3.7 x 107decays/sec1 micro-Curi e = 1µCi = 37,000 decays/sec1 Bequerel = 1 Bq = 1 decay/sec1Samples with an activity of a Curie and a mil li - Cu r i e is quite hazardous. In ourlaboratory, our samples will be a few micro-Curies. Since the activity A(t) equals−dNdt, the activity of a sam p l e also follows the simple half-life formula:A(t) = A0(12)t/τ(2)where A0is the initial activity, A(t = 0), of the sample.The half-life τ depends on the particular radioact i ve nucleus. For example, theradioisotope137Cs has a half-l ife of 30 years. The radioisotope127Ag h as a h a lf- l ifeof 200 seconds, and238U a half-l i fe of around 4.47 x 109years. Half-lives can be a slong as a billion years, or as short as a fraction of a secon d . To find the half-life ofa particular isotope, you need to look up the value in an appropriate table. I havelinks on my home page to sites tha t have extensive data on nuclei.The ”Physics” behind the half-life formulaWhy does the activi ty and th e number of radioactive nuclei in a sample decrease asthe ”half-life” formula? The answer can be understood from the physics of quantummechanics:A radioactive nucleus has a certain probability per unit time to decay.The probability to decay/time is termed the ”decay constant”, and is given t h esymbol λ. The value of the decay constant depen ds on the nature of the particulardecay pro cess.λ ≡ the probabilty to decay per unit time (units of 1/time)From this assumption, one can ”derive” the half-life decay rule as follows. LettingN(t) stand for the number of radioa ct ive nuclei in the sample at time t, the numberof nuclei that decay in a time ∆t equals N(t) times the pr o b ability that one will decayin the time ∆t. Stated in an equat i on , we have:N(t) − N(t + ∆t) = N(t)λ(∆t) (3)Dividing by ∆t, and multiplying by - 1 gives:N(t + ∆t) − N(t)∆t= −λ∆t (4)If we take the limit as ∆ t goes to zero, we obtain a simple differential equation forN(t):2dNdt= −λN (5)This is a fairly simple differential equation, whose solution is a decaying exponentialfunction:N(t) = N0e−λt(6)where N0is the number of radioacti ve nuclei at time t = 0. This exponential decayis the same as the half-life decay formula given above. If we change from base e tobase 2, we can see this. Since 2 = eln2, we have e = 21/(ln2). If we substitute thisexpression for e into the above equation, we obtain:N(t) = N02−(λt)/(ln2)= N0(12)(λt)/ln2)(7)Comparing this expression with the half-life for mula, we see that they are identicaland that the half-life is related to the decay co n st ant by:τ =ln2λ(8)This is an important relationship. It allows us to d et er m i n e the decay constant bymeasuring t he half-life.Lets get a feeling for the numbers involved for one of our classroom isotopes.Consider137Cs, which has a half-life of 30 years. To calculate the decay constant, λ,in units of 1/min we just substitute into the above equation:λ =ln2τ=ln230years(3 65d ays/year)(24hrs/day)(60min/hr)= 4.4x10−81min(9)This m eans, that if you wait one minute, t he prob ab i l i ty that a particular radioactiveCs137nucleus will decay i s only 4.4 × 10−8, or 1 out of 23 million. That is about thesame chance as winning the state lottery! If t he chance is so small that a nucleus willdecay, then how come we have so many decays in one minute? Our Geiger counterreads a few thousand counts per minute. The reason is that there are many manyradioactive nuclei in t h e sample. Although each one has a tiny chance to decay,there are so many nuclei in the sample that a relative large number actually decayeach minute. To determine how many radioactive nuclei are in the sample, we needthe relationship between activity and the decay constant. This relationship can bedetermined as follows:3The activity, A(t), is th e number of decays per unit time, or decays/sec. Considermeasuring for a time ∆t. In terms of the number of radioactive nuclei at time t, N(t),we have:A(t) =N(t) − N(t + ∆t)∆t(10)Takin g the limit as ∆t goes to zero, gives:A(t) = −dNdt= −d(N0e−λt)dt= λN(t) (11)This relationship is often written as:A = λN =Nln2τ(12)This is a very i m portant relationship, and we will use it in our laboratory ex-periments. With our knowledge of t h e hal f-l i fe of an isot o pe, we can measure theactivity of a sample and determine the number of radioactive nuclei. For example,consider137Cs. For a one microcurie137Cs sampl e, the number of radi oact i ve nucleican be calculated. One microcu r i e i s an activity of 37,000 decays per second, whichmeans that A = 37,000 (1/sec). We need to express the half-life,