``^238U`decays to ``^206(Pb)`with a half-life of `4.47xx10^9 y.This happens in a number of steps .Can you justify a single half-life for this chain of processes? A sample of rock is found to contain 2.00mg of ``238U`and 0.600 mg of ``206(Pb)`.assuming that all the lead has come form uranium ,find the life of the rock.

Answer

A, B

Explanation

Half life period can be a single for all the
process. It is the time taken for (1)/(2) of the uranium to convert to lead
No of atons of
` U^(238) = (6 xx 10^(23 ) xx 2 xx 10^(-3))/(238) `
` = (12)/(238) xx 10^(20) `
` = 0.05042 xx 10^(20)`
Initially total no. of uranium atoms
` =((12)/(238) + (36)/(206)) xx 10^(20) `
`=0.06789`
` N= N_(0) e^(-lambda t) `
`rArr N = e^(-0.093/(t_1//2)`
`rArr 0.05042 = 0.06789 = e^ (-0.6931)/(4.47 xx10^(9)) `
` rArr log ((0.05042)/(0.06789)) = (-0.6931)/(4.47 xx10^(9)) `
` t = 1.92 xx 10^(9) year `