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iVinArrow [24]
3 years ago
15

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 m

ust have been present in the sample 11,430 years ago? The half-life of carbon-14 is 5730 years.
Physics
2 answers:
taurus [48]3 years ago
8 0

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.

MakcuM [25]3 years ago
5 0

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.

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