In moon rock sample the ratio of the number of stable argon-`40` atoms present to the number of radioactive potassium`-40` atoms is `7:1`. Assume that all the argon atoms were produced by the decay of potassium atoms, with a half-life of `2.5xx10^(9)yr`. The age of the rock is

Options

A. `2.95 xx 10^(11)yr`
B. `2.95xx10^(9)yr`
C. `4.18xx10^(9)yr`
D. `4.18xx10^(11)yr`

Answer

C

Explanation

If No potassium atoms were present at the time the rock was formed by solidification from a molten form, the number of potassium atoms remaining at the time of analysis is :-
`N_(k) =N_(0)e^(-lambda t) " " `…(i)
in which t is the age of the rock.
For every potassium atom that decays, an argon atom is produced. Thus, the number of argon atoms present at the line of the analysis is
`N_(Ar)=N_(0) - N_(K) " " `...(ii)
We cannot measure `N_(0)` , so let's eliminate it from Eqs. (i) and (ii).
We find, after some algebra, that
`lambda t= "In" (1+(N_(Ar))/(N_(K)))`,
in which `N_(Ar)//N_(k)` can be measured. Solving for t
`t=(T_(1//2) "In"(1+N_(Ar)//N_(K)))/("In"2) `
`=((1.25 xx10^(9)y)["In"(1+9.1)])/("In"2) =4.18 xx 10^(9)yr`.