A rock sample has isotopes X and Y in mole ratio e : 1 (X : Y) now. Initially the ratio X : Y was 1 : 1. How old is the rock sample? [ln 2 = 0.7]
Given : $(t_{1/2})_{X}$ = 6 × $1 0^{9}$ yrs
$(t_{1/2})_{Y}$ = 2 × $1 0^{9}$ yrs

Question

A rock sample has isotopes X and Y in mole ratio e : 1 (X : Y) now. Initially the ratio X : Y was 1 : 1. How old is the rock sample? [ln 2 = 0.7]

Given : $(t_{1/2})_{X}$ = 6 × $1 0^{9}$ yrs

$(t_{1/2})_{Y}$ = 2 × $1 0^{9}$ yrs

TestHub · Accepted Answer

Ratio now / ratio initially = e/1 divided by 1/1 = e.

$N_{X}$ / $N_{Y}$ = ( $N_{X, 0}$ / $N_{Y, 0}$ ) x exp[-( $λ_{X}$ - $λ_{Y}$ )t], so ( $λ_{Y}$ - $λ_{X}$ ) t = ln e = 1.

$λ_{Y}$ = 0.7/(2 × $1 0^{9}$ ) = 0.35 × $1 0^{- 9}$ and $λ_{X}$ = 0.7/(6 × $1 0^{9}$ ) = 0.1167 × $1 0^{- 9}$ . Difference = 0.2333 × $1 0^{- 9}$ per yr.

t = 1/(0.2333 × $1 0^{- 9}$ ) = 4.29 × $1 0^{9}$ , about 4.3 × $1 0^{9}$ yr.

Answer: (C).

Chemistry - Chemical Kinetics Question with Solution | TestHub

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