10.3 Radioactive decay (Page 2/15)

University physics volume 3 Page 2 / 15

\frac{N_{0}}{2} = N_{0} e^{− λ T_{1 / 2}} .

Dividing both sides by $N_{0}$ and taking the natural logarithm yields

ln \frac{1}{2} = ln e^{− λ T_{1 / 2}}

which reduces to

λ = \frac{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 $\bar{T}$ 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

\bar{T} = \frac{1}{λ} .

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

A = - \frac{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

The decay constant is found to be

$λ = \frac{0.693}{T_{1 / 2}} = (\frac{0.693}{T_{1 / 2}}) (\frac{1 yr}{3.16 \times 10^{7} s}) = 7.61 \times 10^{−10} s^{− 1} .$
The atomic mass of $_{38}^{90} S r$ is 89.91 g. Using Avogadro’s number $N_{A} = 6.022 \times 10^{23}$ atoms/mol, we find the initial number of nuclei in 1.00 g of the material:

$N_{0} = \frac{1.00 g}{89.91 g} (N_{A}) = 6.70 \times 10^{21} nuclei .$

From this, we find that the activity $A_{0}$ at $t = 0$ for 1.00 g of strontium-90 is

$\begin{matrix} A_{0} & = λ N_{0} \\ = (7.61 \times 10^{−10} s^{−1}) (6.70 \times 10^{21} nuclei) \\ = 5.10 \times 10^{12} decays/s . \end{matrix}$

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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 \times 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 \times 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) = \frac{6.02 \times 10^{23} {mol}^{−1}}{12.0 g/mol} \times (1000 g) = 5.02 \times 10^{25} .

The number of $^{14} C$ nuclei in 1 kg of carbon is therefore

N (^{14} C) = (5.02 \times 10^{25}) (1.3 \times 10^{−12}) = 6.52 \times 10^{13} .

Now we can find the activity A by using the equation $A = \frac{0.693 N}{t_{1 / 2}} .$ Entering known values gives us

A = \frac{0.693 (6.52 \times 10^{13})}{5730 y} = 7.89 \times 10^{9} y^{− 1}

or $7.89 \times 10^{9}$ decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,

A = (7.89 \times 10^{9} y^{− 1}) \frac{1.00 y}{3.16 \times 10^{7} s} = 250 Bq,

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

A = \frac{250 Bq}{3.7 \times 10^{10} Bq/Ci} = 6.76 \times 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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10.3 Radioactive decay (Page 2/15)

Decay constant and activity of strontium-90

Strategy

Solution

What is 14 C Activity in living tissue?

Strategy

Solution

Significance

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What is $^{14} C$ Activity in living tissue?