Age of earth or mineral

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## Age of earth or mineral-

**Uranium – Lead dating method** is used to determine the** age of earth or a mineral**. One gram atom of U^{238} disintegrates to give one gram atom of Pb ^{206}. Gram atoms of U^{238} and Pb ^{206} are determined in the given sample. To use the following formula , it is assumed that no lead was present originally in the mineral-

### K = (2.303 / t) [log( gram atom of U ^{238} + gram atom of Pb ^{206} )/ gram atom of U ^{238} ]

### t = (2.303 / K) [log( gram atom of U ^{238} + gram atom of Pb ^{206} )/ gram atom of U ^{238} ]

### t = Age of earth or mineral

### Question 1) A sample of U – 238 is found to contain 23.8 gm of U – 238 and 20.6 gm Pb- 206. Calculate the age of uranium ore. t_{1/2} for uranium is 4.5 x 10^{9} years.

### Solution )

*Half life of uranium (t _{1/2}) = 4.5 x 10^{9} years*

*K = 0.693 / t _{1/2} = 0.693 / 4.5 x 10^{9}*

*K = 0.154 x 10 ^{-9} = 1.54 x 10^{-10} years^{-1}*

*Weight of uranium = 23.8 gm*

*Gram atoms of U ^{238} = w/m = 23.8 / 238 = 0.1*

*Gram atoms of Pb ^{206} =w/m = 20.6 / 206 = 0.1*

*t = (2.303 / K) [log( gram atom of U *^{238} + gram atom of Pb ^{206} )/ gram atom of U ^{238} ]

^{238}+ gram atom of Pb

^{206})/ gram atom of U

^{238}]

*= (2.303 / 1.54 x 10 ^{-10} ) [log (0.1 + 0.1)/ 0.1]*

*= 1.495 x 10 ^{10}( log 2)*

*Because ,log 2 = 0.3010*

*= 1.495 x 10 ^{10} x 0.3010 = 0.4499 x 1010*

*t = 4.5 x 10*^{9} years

^{9}years

### Question 2) 92 U 238 disintegrates to give _{82} Pb ^{206}. A sample of uranium contains 1.0 gram of U ^{238 } and 0.1 gram of Pb ^{206}. Half life of uranium is 4.5 x 10^{9} years . Calculate the age of uranium sample?

### Solution )

*Half life of uranium (t _{1/2}) = 4.5 x 10^{9} years*

*K = 0.693 / t _{1/2} = 0.693 / 4.5 x 10^{9}*

*K = 0.154 x 10 ^{-9} = 1.54 x 10^{-10} years^{-1}*

*Weight of uranium = 1.0 gm*

*Gram atoms of U ^{238} = w/m = 1.0 / 238 = 0.0042*

*Gram atoms of Pb ^{206} =w/m = 0.1 / 206 = 0.000485*

*t = (2.303 / K) [log( gram atom of U *^{238} + gram atom of Pb ^{206} )/ gram atom of U ^{238} ]

^{238}+ gram atom of Pb

^{206})/ gram atom of U

^{238}]

*= (2.303 / 1.54 x 10 ^{-10} ) [log (0.0042 + 0.000485)/ 0.0042]*

*= 1.495 x 10 ^{10}[log ( 0.004685)/ 0.0042]*

*= 1.495 x 10 ^{10}( log 1.1154)*

*Because ,log 1.1154 = 0.0474*

*= 1.495 x 10 ^{10} x 0.0474 = 0.0708 x 10^{10}*

*t = 7.08 x 10*^{8} years

^{8}years

### Question 3) On analysis a sample of uranium ore was found to contain 0.277 gm of _{82} Pb^{206} and 1.667 gm of _{92}U ^{238}. The half life period of U^{238} is 4.51 x 10^{9} years. If all the lead was assumed to have come from decay of _{92} U ^{238}, What is the age of earth?

### Solution )

*Let , age of earth = t years*

*Weight of uranium = 1.667 gm*

*Gram atoms of U ^{238} = w/m = 1.667 / 238 = 0.0070*

*Gram atoms of Pb ^{206} =w/m = 0.277 / 206 = 0.001345*

*half life of uranium (t _{1/2}) = 4.51 x 10^{9} years*

*K = 0.693 / t _{1/2} = 0.693 / 4.51 x 10^{9}*

*K = 0.154 x 10 ^{-9} = 1.54 x 10^{-10} years ^{-1}*

*t = (2.303 / K) [log( gram atom of U *^{238} + gram atom of Pb ^{206} )/ gram atom of U ^{238} ]

^{238}+ gram atom of Pb

^{206})/ gram atom of U

^{238}]

*= (2.303 / 1.54 x 10 ^{-10} ) [log (0.0070 + 0.001345)/ 0.0070]*

*= 1.495 x 10 ^{10} [log ( 0.00835)/ 0.0070]*

*= 1.495 x 10 ^{10}( log 1.192)*

Because , log 1.192 = 0.07658

*t = 1.495 x 10 ^{10} x 0.07658 = 0.1144 x 10^{10}*

*Age of earth ‘t ‘= 1.144 x 10*^{9} years

^{9}years