Home
Class 12
CHEMISTRY
The analysis of a uranium reveals that r...

The analysis of a uranium reveals that ratio of mole of `.^(206)Pb`and `.^(238) U` in sample is 0.2 . If effective decay constant of process `.^(238)U rarr .^(206) Pb` is `lambda` then age of rock is

A

`(1)/(lambda) In (5)/(4)`

B

`(1)/(lambda) In ((5)/(1))`

C

`(1)/(lambda) ln(4)/(1)`

D

`(1)/(lambda) In ((6)/(5))`

Text Solution

Verified by Experts

The correct Answer is:
D
`t=(1)/(lambda)In[(n_(Pb+n_(U))/(n_(U)))] " "...(i)`
given `(n_(Pb))/(n_(U))=0.2=(1)/(5) rArr (n_(Pb+n_(U))/(n_(U)))=(1)/(5)+1=((6)/(5)) " " ...(ii)`
Substituting (ii)in(i)
`t=(1)/(lambda) In ((6)/(5))`
Promotional Banner

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

The analysis of a rock shows that relative no. of ""^(206)Pb and ""^(238)U atoms is Pb/U =0.25. If t_(0.5) ""^(238)U rarr ""^(206)Pb is 4 xx 10^(9) years, the age of the rock is

.^(238)U decays with a half-life of 4.5 xx10^(9) years, the decay series eventaully ending at .^(206)Pb , which is stable. A rock sample analysis shows that the ratio of the number of atoms of .^(206)Pb to .^(238)U is 0.0058. Assuming that all the .^(206)Pb is prodduced by the decay of .^(238)U and that all other half-lives on the chain are negligilbe, the age of the rock sample is (1n 1.0058 =5.78 xx10^(-3)) .

A closed vessel with rigid walls contains 1 mole of ._(92)^(238)U and 1 mole of air at 298K . Considering complete decay of ._(92)^(238)U to ._(82)^(206)Pb the ratio of the final pressure to the initial pressure of the system at 298K is

A sample of U^(238) ("half life" = 4.5 xx 10^(9)yr) ore is found to contain 23.8 g" of " U^(238) and 20.6g of Pb^(206) . Calculate the age of the ore

How many moles of helium are produced when one mole of ._(92)^(238)U disintegrates into ._(82)^(206)Pb ?

How many moles of helium are produced when 1 mole of ._(92)U^(238) disintegrate into ._(82)Pb^(206) ?