Age of wood and fossil

source : newatlas.com

## To calculate the age of wood and fossil or bone-

*The age of wood and fossil or bone is determined by Carbon dating method. Activity of C ^{14} is determined in the leaving plant or animal and then activity of C^{14} is determined in old wood or fossil. the following formula is used to determine the age of dead animal or plant-*

#### K = (2.303 /t) [log ( activity of C^{14} in fresh wood or animal / activity of C^{14} in dead wood or animal )

#### K = (2.303 / t) [log ( N_{0} / N )]

**Number of atoms of C ^{14} or gram atoms of C^{14} may be taken in place of**

**activity.**## Numerical-

### Question 1) The activity of C^{14} in an old piece of wood is 7 particles per minute. Half life of C^{14} is 5770 years . The activity of C^{14} in fresh piece of wood is 14 particles per minute. Calculate the age of wood ?

### Solution )

#### K = (2.303 /t [log) ( activity of C^{14} in fresh wood or animal / activity of C^{14} in dead wood or animal )

*activity of C ^{14} in fresh wood = 14 particles*

*activity of C ^{14} in old wood = 7 particles*

*Age of wood ( t) = ?*

*t _{1/2} = 5770 years*

*K = 0.693 / t _{1/2} = 0.693/ 5770 = 0.00012= 1.2 x 10^{-4} years ^{-1}*

*K = 2.303 / t [ log 14 / 7 ]*

*1.2 x 10 ^{-4} =( 2.303 / t )[ log 2]*

*t = (2.303 / 1.2 x 10 ^{-4} )[0.3010 ] = 2.303 x 0.3010 / 1.2 x 10^{-4}*

*t ( age of wood) = 5776.7 years Ans.*

### Question 2) The mass of C^{14} in wood piece remains 1/16 of the C^{14} in a living plant. Half life of C^{14} is 5577 years. Calculate the age of wood?

### Solution )

*Mass of C ^{14} in living plant / Mass of C^{14} in wood piece remain = 16 / 1*

*Means N=1, N _{0} = 16*

*t _{1/2} = 5577 years*

*K = 0.693 / t _{1/2} = 0.693/ 5577 = 0.000124 = 1.24 x 10^{-4} years^{-1}*

*K = 2.303 /t [log ( N*_{0} / N )]

_{0}/ N )]

*t = 2.303 / K [ log 16 / 1]*

*t = 2.303 / K [ log 2 ^{4 }]*

*t = (2.303 / 1.24 x 10 ^{-4} )[4 log 2 ]*

*t = (2.303 / 1.24 x 10 ^{-4} )[4 x 0.3010 ] = 2.303 x 0.3010 x 4 / 1.24 x 10^{-4}*

*t ( age of wood) = 22361 years Ans.*

### Question 3) One of the hazards of nuclear explosion is the generation of Sr^{90} and its subsequent incorporation in bones . This element has a half life of 28.1 years. If one microgram was absorbed by a new born baby, how much Sr^{90} will remain in his bones after 20 years?

### Solution )

*t _{1/2} = 28.1 years , initial amount ‘N_{0}‘ = one microgram = 10-6 gm , N =? , t = 20 years*

*K = 0.693 / t _{1/2} = 0.693/ 28.1 = 0.0246 years^{-1}*

*K = 2.303 /t [log ( N*_{0} / N )]

_{0}/ N )]

*0.0246 = (2.303 / 20)[log 10 ^{-6} / N]*

*0.0246 x 20 / 2.303 = log 10 ^{-6} / N*

*0.2136 = log 10 ^{-6} / N*

*log 10 ^{-6} / N =Antilog 0.2136 = 1.635*

*N = 10 ^{-6} / 1.635 = 0.611 x 10^{-6}*

*Sr*^{90} will remain in his bones after 20 years (N )= 6.11 x 10^{-7} gm Ans.

^{90}will remain in his bones after 20 years (N )= 6.11 x 10

^{-7}gm Ans.

### Question 4) The half life period of C^{14} is 5760 years. A piece of wood when buried in the earth had 1 % C^{14} .Now as charcoal it has only 0.25 % C^{14} . How long has the piece of wood been buried ?

### Solution )

*t _{1/2} = 5760 years , N_{0} = 1 gm , N = 0.25 gm , t =?*

*K = 0.693 / t _{1/2} = 0.693/ 5760 = 0.00012 = 1.2 x 10^{-4} years^{-1}*

*K = (2.303 /t) [log ( N*_{0} / N )]

_{0}/ N )]

*t = 2.303 / K [ log 1 /0.25]*

*= 2.303 / K [ log 4]*

*t= 2.303 / K [ log 2 ^{2 }]*

*= (2.303 / 1.2 x 10 ^{-4} )[2 log 2 ]*

*t = (2.303 / 1.2 x 10 ^{-4} )[2 x 0.3010 ] = 2.303 x 0.3010 x 2 / 1.2 x 10^{-4}*

*t = 1.1553 x 10*^{4} = 11553 years ans.

^{4}= 11553 years ans.