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N 0 2 = N 0 e λ T 1 / 2 .

Dividing both sides by N 0 and taking the natural logarithm yields

ln 1 2 = ln e λ T 1 / 2

which reduces to

λ = 0.693 T 1 / 2 .

Thus, if we know the half-life T 1/2 of a radioactive substance, we can find its decay constant. The lifetime     T of a radioactive substance is defined as the average amount of time that a nucleus exists before decaying. The lifetime of a substance is just the reciprocal of the decay constant, written as

T = 1 λ .

The activity     A is defined as the magnitude of the decay rate, or

A = d N d t = λ N = λ N 0 e λ t .

The infinitesimal change dN in the time interval dt is negative because the number of parent (undecayed) particles is decreasing, so the activity ( A ) is positive. Defining the initial activity as A 0 = λ N 0 , we have

A = A 0 e λ t .

Thus, the activity A of a radioactive substance decreases exponentially with time ( [link] ).

Figure a shows a graph of A versus t. It starts at point A subscript 0 and reduces with time. The rate of reduction decreases slowly till A is very close to 0, making a curved plot on the graph. The plot is labeled A = A subscript 0 e to the power minus lambda t. Figure b shows a graph of ln A versus t. It starts at ln A subscript 0 and slopes downwards in a straight line. The slope of the line is labeled minus lambda t.
(a) A plot of the activity as a function of time (b) If we measure the activity at different times, we can plot ln A versus t , and obtain a straight line.

Decay constant and activity of strontium-90

The half-life of strontium-90, 38 90 Sr , is 28.8 y. Find (a) its decay constant and (b) the initial activity of 1.00 g of the material.

Strategy

We can find the decay constant directly from [link] . To determine the activity, we first need to find the number of nuclei present.

Solution

  1. The decay constant is found to be
    λ = 0.693 T 1 / 2 = ( 0.693 T 1 / 2 ) ( 1 yr 3.16 × 10 7 s ) = 7.61 × 10 −10 s 1 .
  2. The atomic mass of 38 90 S r is 89.91 g. Using Avogadro’s number N A = 6.022 × 10 23 atoms/mol, we find the initial number of nuclei in 1.00 g of the material:
    N 0 = 1.00 g 89.91 g ( N A ) = 6.70 × 10 21 nuclei .

    From this, we find that the activity A 0 at t = 0 for 1.00 g of strontium-90 is
    A 0 = λ N 0 = ( 7.61 × 10 −10 s −1 ) ( 6.70 × 10 21 nuclei ) = 5.10 × 10 12 decays/s .
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Expressing λ in terms of the half-life of the substance, we get

A = A 0 e ( 0.693 / T 1 / 2 ) T 1 / 2 = A 0 e −0.693 = A 0 / 2 .

Therefore, the activity is halved after one half-life. We can determine the decay constant λ by measuring the activity as a function of time. Taking the natural logarithm of the left and right sides of [link] , we get

ln A = λ t + ln A 0 .

This equation follows the linear form y = m x + b . If we plot ln A versus t , we expect a straight line with slope λ and y -intercept ln A 0 ( [link] (b)). Activity A is expressed in units of becquerels (Bq), where one 1 Bq = 1 decay per second . This quantity can also be expressed in decays per minute or decays per year. One of the most common units for activity is the curie (Ci)    , defined to be the activity of 1 g of 226 Ra . The relationship between the Bq and Ci is

1 Ci = 3.70 × 10 10 Bq .

What is 14 C Activity in living tissue?

Approximately 20 % of the human body by mass is carbon. Calculate the activity due to 14 C in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.

Strategy

The activity of 14 C is determined using the equation A 0 = λ N 0 , where λ is the decay constant and N 0 is the number of radioactive nuclei. The number of 14 C nuclei in a 1.00-kg sample is determined in two steps. First, we determine the number of 12 C nuclei using the concept of a mole. Second, we multiply this value by 1.3 × 10 −12 (the known abundance of 14 C in a carbon sample from a living organism) to determine the number of 14 C nuclei in a living organism. The decay constant is determined from the known half-life of 14 C (available from [link] ).

Solution

One mole of carbon has a mass of 12.0 g, since it is nearly pure 12 C . 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 .

The number of 14 C nuclei in 1 kg of carbon is therefore

N ( 14 C ) = ( 5.02 × 10 25 ) ( 1.3 × 10 −12 ) = 6.52 × 10 13 .

Now we can find the activity A by using the equation A = 0.693 N t 1 / 2 . Entering known values gives us

A = 0.693 ( 6.52 × 10 13 ) 5730 y = 7.89 × 10 9 y 1

or 7.89 × 10 9 decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,

A = ( 7.89 × 10 9 y 1 ) 1.00 y 3.16 × 10 7 s = 250 Bq ,

or 250 decays per second. To express A in curies, we use the definition of a curie,

A = 250 Bq 3.7 × 10 10 Bq/Ci = 6.76 × 10 −9 Ci .

Thus,

A = 6.76 nCi .

Significance

Approximately 20 % of the human body by weight is carbon. Hundreds of 14 C decays take place in the human body every second. Carbon-14 and other naturally occurring radioactive substances in the body compose a person’s background exposure to nuclear radiation. As we will see later in this chapter, this activity level is well below the maximum recommended dosages.

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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