Question

Carbon-14 is used to determine the age of ancient objects. If a sample today contains 0.060 g of carbon-14, how much carbon-14 must have been present in the sample 11,430 years ago? The half-life of carbon-14 is 5730 years.

Answer 1

Answer : 0.239 gram of carbon-14 must have been present in the sample 11,430 years ago.

Solution : Given,

As we know that the radioactive decays follow first order kinetics.

First, we have to calculate the rate constant of carbon-14.

Formula used : $t_{1/2}=\frac{0.693}{k}$

Now put the value of half-life, we get the value of rate constant.

$5730years=\frac{0.693}{k}$

$k=1.209\times 10^{-4}year^{-1}$

Now we have to calculate the original amount of carbon-14.

The expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant  = $1.209\times 10^{-4}year^{-1}$

t = time taken for decay process  = 11430 years

a = initial amount of the carbon-14 = ?

a - x = amount left after decay process  = 0.060 g

Putting values in above equation, we get the value of initial amount of carbon-14.

$1.209\times 10^{-4}year^{-1}=\frac{2.303}{11430years}\log\frac{a}{0.060g}$

$a=0.238g$

Therefore, 0.239 gram of carbon-14 must have been present in the sample 11,430 years ago.

Answer 2

Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.

Explanation:

Half-life of sample of carbon -14= 5,730 days

$\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{5,730 days}=0.00012 day^{-1}$

Let the sample present 11,430 years(t) ago = $N_o$

Sample left till today ,N= 0.060 g

$N=N_o\times e^{-\lambda t}$

$ln[N]=ln[N]_o-\lambda t$

$\log[0.060 g]=\log[N_o]-2.303\times 0.00012 day^{-1}\times 11,430 days$

$\log[N_o]=1.9369$

$N_o=86.47 g$

86.47 g of carbon-14 must have been present in the sample 11,430 years ago.