Radiocarbon Dating A skull believed to belong to an ancient human being is found to have a carbon-14 decay rate of 4.50 disintegrations per minute per gram of carbon (4.50 dis/min ·g C). If living organisms have a decay rate of 15.3 dis/min · g C, how old is the skull? (The decay rate is directly

Question

Radiocarbon Dating A skull believed to belong to an ancient human being is found to have a carbon-14 decay rate of 4.50 disintegrations per minute per gram of carbon (4.50 dis/min ·g C). If living organisms have a decay rate of 15.3 dis/min · g C, how old is the skull? (The decay rate is directly proportional to the amount of carbon-14 present.)

SORT You are given the current rate of decay for the skull and the assumed initial rate. You are asked to find the age of the skull, which is the time that passed in order for the rate to have reached its current value.

GIVEN rate[latex]_{t}[/latex] = 4.50 dis/min · g C;
rate[latex]_{0}[/latex] = 15.3 dis/min · g C;
FIND t

STRATEGIZE Use the expression for half-life (Equation 19.1) to find the rate constant (k) from the half-life for C-14, which is 5730 yr (Table 19.3).

t[latex]_{1/}2 = \frac{0.693}{k}[/latex] [19.1]

Use the value of the rate constant and the initial and current rates to find t from the integrated rate law (Equation 19.4).

ln[latex]\frac{ rate_{t}}{rate_{0}}[/latex]= -kt [19.4]

CONCEPTUAL PLAN

t[latex]_{1/2}[/latex] → k

. t[latex]_{1/2} = \frac{0.693}{k}[/latex]

k, rate[latex]_{t}[/latex] , rate[latex]_{0}[/latex] → t

. ln[latex]\frac{ rate_{t}}{rate_{0}}[/latex]= -kt

SOLVE Follow your plan. Begin by finding the rate constant from the half-life.
Substitute the rate constant and the initial and current rates into the inte-grated rate law and solve for

TABLE 19.3 Selected Nuclides and Their Half-Lives
Nuclide	Half-life	Type of Decay
[latex]^{232}_{90}[/latex]Th	1.4×10[latex]^{10}[/latex]yr	Alpha
[latex]^{238}_{92}[/latex]U	4.5×10[latex]^{9}[/latex]yr	Alpha
[latex]^{14}_{6}[/latex]C	5730yr	Beta
[latex]^{220}_{86}[/latex]Rn	55.6s	Alpha
[latex]^{219}_{90}[/latex]Th	1.05×10[latex]^{-6}[/latex]s	Alpha

Accepted Answer

t[latex]_{1/2} = \frac{0.693}{k}[/latex]
k = [latex]\frac{0.693}{t_{1/2}}= \frac{0.693}{4.5 imes 10^{9} yr}[/latex]
= [latex]1.2\underline{0}9 imes 10^{-4}[/latex]/yr
ln[latex]\frac{ rate_{t}}{rate_{0}}[/latex]= -kt
t = -[latex]\frac{In \frac{rate_{t}}{rate_{0}}}{k} = -\frac{In \frac{4.50 dis/min · g C}{15.3 dis/min · g C}}{1.2\underline{0}9 imes 10^{-4}/yr}[/latex]
= 1.01 × 10[latex]^{4}[/latex] yr

CHECK The units of the answer (yr) are correct. The magnitude of the answer is about 10,000 years, which is a little less than two half-lives. This value is reasonable given that two half-lives would result in a decay rate of about 3.8 dis/min · g C.

Question 19.5: Radiocarbon Dating A skull believed to belong to an ancient ...

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