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1 Ci = 3 . 70 × 10 10 Bq, size 12{1" Ci"=3 "." "70" times "10" rSup { size 8{"10"} } " Bq"} {}

or 3 . 70 × 10 10 size 12{3 "." "70" times "10" rSup { size 8{"10"} } } {} decays per second. A curie is a large unit of activity, while a becquerel is a relatively small unit. 1 MBq = 100 microcuries ( μ Ci ) size 12{"1 MBq"="100 microcuries " \( μ"Ci" \) } {} . In countries like Australia and New Zealand that adhere more to SI units, most radioactive sources, such as those used in medical diagnostics or in physics laboratories, are labeled in Bq or megabecquerel (MBq).

Intuitively, you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its half-life. The greater the number of radioactive nuclei present in the sample, the more will decay per unit of time. The shorter the half-life, the more decays per unit time, for a given number of nuclei. So activity R size 12{R} {} should be proportional to the number of radioactive nuclei, N size 12{N} {} , and inversely proportional to their half-life, t 1 / 2 size 12{t rSub { size 8{1/2} } } {} . In fact, your intuition is correct. It can be shown that the activity of a source is

R = 0 . 693 N t 1 / 2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}

where N size 12{N} {} is the number of radioactive nuclei present, having half-life t 1 / 2 size 12{t rSub { size 8{1/2} } } {} . This relationship is useful in a variety of calculations, as the next two examples illustrate.

How great is the 14 C size 12{"" lSup { size 8{"14"} } C} {} Activity in living tissue?

Calculate the activity due to 14 C size 12{"" lSup { size 8{"14"} } C} {} in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.

Strategy

To find the activity R size 12{R} {} using the equation R = 0 . 693 N t 1 / 2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {} , we must know N size 12{N} {} and t 1 / 2 size 12{t rSub { size 8{1/2} } } {} . The half-life of 14 C size 12{"" lSup { size 8{"14"} } C} {} can be found in Appendix B , and was stated above as 5730 y. To find N size 12{N} {} , we first find the number of 12 C size 12{"" lSup { size 8{"12"} } C} {} nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by 1 . 3 × 10 12 size 12{1 "." 3×"10" rSup { size 8{ +- "12"} } } {} (the abundance of 14 C size 12{"" lSup { size 8{"14"} } C} {} in a carbon sample from a living organism) to get the number of 14 C size 12{"" lSup { size 8{"14"} } C} {} nuclei in a living organism.

Solution

One mole of carbon has a mass of 12.0 g, since it is nearly pure 12 C size 12{"" lSup { size 8{"12"} } C} {} . (A mole has a mass in grams equal in magnitude to A size 12{A} {} found in the periodic table.) Thus the number of carbon nuclei in a kilogram is

N ( 12 C ) = 6.02 × 10 23 mol –1 12.0 g/mol × (1000 g) = 5.02 × 10 25 .

So the number of 14 C size 12{"" lSup { size 8{"14"} } C} {} nuclei in 1 kg of carbon is

N ( 14 C ) = ( 5.02 × 10 25 ) ( 1.3 × 10 −12 ) = 6.52 × 10 13 . size 12{N \( rSup { size 8{"14"} } C \) = \( 5 "." "02" times "10" rSup { size 8{"25"} } \) \( 1 "." 3 times "10" rSup { size 8{ - "12"} } \) =6 "." "52" times "10" rSup { size 8{"13"} } } {}

Now the activity R size 12{R} {} is found using the equation R = 0 . 693 N t 1 / 2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {} .

Entering known values gives

R = 0 . 693 ( 6 . 52 × 10 13 ) 5730 y = 7 . 89 × 10 9 y –1 , size 12{R= { {0 "." "693" \( 6 "." "52"´"10" rSup { size 8{"13"} } \) } over {"5730"" y"} } =7 "." "89"´"10" rSup { size 8{9} } /y} {}

or 7 . 89 × 10 9 size 12{7 "." "89" times "10" rSup { size 8{9} } } {} decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,

R = ( 7.89 × 10 9 y –1 ) 1.00 y 3 . 16 × 10 7 s = 250 Bq, size 12{R=7 "." "89"´"10" rSup { size 8{9} } /y cdot { {1 "." "00"" y"} over {3 "." "16"´"10" rSup { size 8{7} } " s"} } ="250"" Bq"} {}

or 250 decays per second. To express R size 12{R} {} in curies, we use the definition of a curie,

R = 250 Bq 3.7 × 10 10 Bq/Ci = 6.76 × 10 9 Ci. size 12{R= { {"250"" Bq"} over {3 "." 7´"10" rSup { size 8{"10"} } " Bq/Ci"} } =6 "." "75"´"10" rSup { size 8{-9} } " Ci"} {}

Thus,

R = 6.76 nCi. size 12{R=6 "." "75" "nCi"} {}

Discussion

Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of 14 C size 12{"" lSup { size 8{"14"} } C} {} decays per second taking place in us. Carbon-14 and other naturally occurring radioactive substances in our bodies contribute to the background radiation we receive. The small number of decays per second found for a kilogram of carbon in this example gives you some idea of how difficult it is to detect 14 C size 12{"" lSup { size 8{"14"} } C} {} in a small sample of material. If there are 250 decays per second in a kilogram, then there are 0.25 decays per second in a gram of carbon in living tissue. To observe this, you must be able to distinguish decays from other forms of radiation, in order to reduce background noise. This becomes more difficult with an old tissue sample, since it contains less 14 C size 12{"" lSup { size 8{"14"} } C} {} , and for samples more than 50 thousand years old, it is impossible.

Questions & Answers

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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