Question 68: Atoms & Nuclei

Question

A piece of wood  from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of C
14
is 5730 years, then age of the wooden piece placed in the museum is approximately :

Accepted Answer

Fresh wood:
20 decays/min
(current activity if alive / just cut).
Old museum sample:
2 decays/min
.
Half-life of CÃ¢ÂÂ14 =
5730 years
.
Find the age of wooden piece in museum.
Step 1: Ratio of activities
$ \frac{A}{A_0} = \frac{2}{20} = 0.1 $
So, the museum sampleÃ¢ÂÂs activity is 10% of that in fresh wood.
Step 2: Radioactive decay law
$ A = A_0 e^{-\lambda t} $
$ \frac{A}{A_0} = e^{-\lambda t} $
So,
$ 0.1 = e^{-\lambda t} $
$ t = \frac{\ln(0.1)}{-\lambda} $
Step 3: Decay constant
$ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5730 \,	ext{yr}} \approx 1.2097 	imes 10^{-4} \,	ext{yr}^{-1} $
Step 4: Solve for $t$
$ t = \frac{\ln(0.1)}{-1.2097 	imes 10^{-4}} $
$ \ln(0.1) = -2.3026 $
$ t = \frac{2.3026}{1.2097 	imes 10^{-4}} $
$ t \approx 19039 \,	ext{years} $
Ã¢ÂÂ Final Answer:
Option C: 19039 years

Answer

10439 years

Answer

13094 years

Answer

19039 years

Answer

39049 years

Atoms & Nuclei

A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of C 14 is 5730 years, then age of the wooden piece placed in the museum is approximately :

Choose the correct answer: