Natural RadiationThere are a few radioisotopes that exist in our environment. Isotopes that werepresent when the earth was formed and isotopes that are continuously produced bycosmic rays can exist today if they have long enough half-lives. Here we will discuss5 of these isotopes. Four were produced when the earth was formed:40K (half-lifeof 1.277 × 109years),238U (half-life of 4.51 × 109years,232T h (half-life of 1.4 × 1010years), and235U (half-life of 7.1× 108years).14C (half-life 5280 years) is continuouslyproduced in the upper atmosphere by cosmic rays entering the earth.40KMost of the potassium found on earth is the stable39K. However, a small fraction,0.0117%, of all potassium is40K.40K is radioactive with a half-life of 1.277 × 109years.40K decays in the following way:40K has an 89.28% chance to undergo betadecay to the ground state of40Ca, and a 10.72% chance to undergo electron captureto40Ar. When electron capture occurs it is almost always to the excited state of40Ar, 99.53% of the time. From the excited state,40Ar decays to the ground stateemitting a gamma particle with energy 1460.83 KeV. Thus when40K decays, thereis a 89.28% chance a beta particle is emitted and a 10.72(0.9953) = 10.67% chance agamma is emitted.Although 0.0117% seems like a small amount, it means that one out of every 8500potassium atoms is radioactive. This is a large amount! The average 70 Kg man hasaround 140 grams of potassium in his body. So the number of radioactive40K nucleiin his body is:Number of40K nuclei =140g40g(6.02 × 1023)(0.000117) = 2.47 × 1020(1)which is a lot of nuclei. The activity due of40K decays in the average person can becalculated using the relation: activity A = Nλ, or A = N(ln2)/τ:A =2.47 × 1020ln(2)1.277 × 109(365)(24)(3600)sec= 4240decayssec(2)This amount of activity is equal to 4240/37000 = 0.11 µCi. Perhaps we should weara radioactive sign around our necks! Fruits and vegetables can have as much as 0.4%potassium, and an average soil sample contains 2% potassium. Thus, one Kg ofsoil has an activity of 0.016 µCi due to the potassium content alone. By measuringthe gamma spectrum of soil and food samples for long times, one or two hours, the1potassium content can be measured to an accuracy of as good as 10%.40K is thelargest contributor to our natural background radiation.238UAnother isotope found in the earth is uranium 238. Due to its long half-life somestill remains since the formation of the earth.238U has a long decay series, undergoingalpha, beta and gamma decays until it finally becomes stable as lead 206,206P b. Welist the complete238U decay series:2Isotope half-life gamma energies238U 4.468 × 109years —-234T h 24.1 days 63.3 (3.67%)92.38 (2.13%)92.80 (2.10%)112.8 (0.21%)234mP a 1.17 minutes 765 (0.317%)1001 (0.842%)234U 2.47 × 105years 53.2 (0.123%)230T h 7.54 × 104years 67.7 (0.377%)226Ra 1602 years 186.2 (3.59%)222Rn 3.823 days 510 (0.076%)218P o 3.05 minutes ——214P b (99.98%) 26.8 minutes 53.2 (1.1%)242.0 (7.25%)295.2 (18.4%)351.9 (35.6%)785.9 (1.06%)218At (0.02%) 2 seconds —–214Bi 19.7 minutes 609.3 (45.5%)768.4 (4.89%)806.2 (1.26%)934.1 (3.16%)1120.3 (14.9%)1238.1 (5.83%)1377.7 (3.99%)1408.0 (2.39%)1509.2 (2.13%)1764.5 (15.3%)214P o (99.98%) 164 microsec 799 (0.014%)210T l (0.02%) 1.3 minutes 296 (80%)795 (100%)1310 (21%)210P b 21 years 46.5 (4.05%)210Bi 5.01 days ——210P o 138.4 days 803 (0.0011%)206P b Stable3In the first column we list the isotopes in the decay series. If the decay is Alphaemission, the atomic number is lowered by 2 and the atomic mass is lowered by 4. Fora Beta decay process, the atomic number increases by 1 and the atomic mass remainsunchanged. In the middle column we list the half-life of the radioisotope shown inthe first column.In the last column we list the energy of the gamma rays emitted by the isotope inthe first column. The number in parentheses is the yield percentage of the gamma.The isotopes in the series are called the daughter isotopes of238U. From the table, onecan see that as238U decays via its daughter isotopes many alpha, beta and gammaparticles are emitted.Consider the following question: During the long decay series, how many of thevarious daughter isotopes are present at any one time? The half-lives are listed inthe center column. Are there a smaller number of isotopes with short half-lives thanisotopes with long half-lives at any particular moment? Any particular isotope willhave a certain rate at which it is produced, and a rate at which it decays. Considerthe situation in which isotope A decays to isotope B, which decays to isotope C. LetNAbe the number of nuclei of isotope A, NBbe the number of nuclei of isotope B, andNCbe the number of nuclei of isotope C. Let λA, λB, and λCbe the correspondingdecay constants for the decays. The rate at which the number of nuclei of isotope Bdecreases is λBNB, which is just the decays/sec or the activity. The rate of formationof isotope B is just the decay rate of isotope A, λANA. So the change in the numberof nuclei of isotope B per unit time is given by:Change of NBper unit time = +λANA− λBNB(3)dNBdt= λANA− λBNB(4)This rate-of-change equation applies to every radioactive nucleus in the decay series.The first nucleus,238U, will only have the decay term −λN, and the final nucleus,206P b, will only have the rate of formation term, +λN. The solution to the series ofdifferential equations is complicated. However, if we observe the decay series after along time, long enough for the series to come into equilibrium, then the solution issimple. After a long time, the number of radioactive nuclei of a particular isotoperemains constant in time, dNB/dt = 0. The rate of formation, +λANA, is equal tothe rate of decay, −λBNB:λANA= λBNB(5)4for every isotope in the decay series. This equilibrium condition is referred to as secu-lar equilibrium. Since λN is equal to the activity of an isotope, for secular equilibriumthe activity of each isotope in the series is the same. In terms of the half-lives of theisotopes, we have:NAτA=NBτB=NCτC(6)Isotopes with longer half-lives have more nuclei at any particular time in the decayseries than isotopes with shorter half-lives.In the experiment where we measure the spectrum of Brazil nuts, the decay seriesmight not be in secular equilibrium. In fact, by measuring the counts under theappropriate photopeaks we can determine the age of the nuts. As before, supposethere are three isotopes A, B, and C,