Example from Physics, 22.3 Half Life and Radiometric Dating

Carbon-11 Decay

Carbon-11 has a half-life of 20.334 min. (a) What is the decay constant for carbon-11?

If 1 kg of carbon-11 sample exists at the beginning of an hour, (b) how much material will remain at the end of the hour and (c) what will be the decay activity at that time?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
Since $N_{O}$ refers to the amount of carbon-11 at the start, then after one half-life, the amount of carbon-11 remaining will be $N_{O} / 2 .$ The decay constant is equivalent to the probability that a nucleus will decay each second. As a result, the half-life will need to be converted to seconds.
Solution
(a)

$N = N_{O} e^{- λ t}$
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Since half of the carbon-11 remains after one half-life, $N / N_{O} = 0.5$ .

$0.5 = e^{- λ t}$
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Take the natural logarithm of each side to isolate the decay constant.

$\ln (0.5) = - λ t$
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Convert the 20.334 min to seconds.

$\begin{array}{l} - 0.693 = (- λ) (20.334 \min) (\frac{60 s}{1 \min}) \\ - 0.693 = (- λ) (1,220.04 s) \\ \frac{- 0.693}{1,220.04 s} = - λ \\ λ = 5.68 \times 10^{- 4} s^{- 1} \end{array}$
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(b) The amount of material after one hour can be found by using the equation

$N = N_{O} e^{- λ t},$
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with t converted into seconds and N_O written as 1,000 g

$\begin{array}{l} N = (1,000 g) e - (0.000568) (60.60) \\ N = 129.4 g \end{array}$
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(c) The decay activity after one hour can be found by using the equation

$R = λ N$
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for the mass value after one hour.

$R = λ N = (0.000568 \frac{decays}{second}) (129.4 grams) = 0.0735 Bq$
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0 of 2 steps open

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Strategy

Solution

Strategy

Solution