Uranium ores on the earth at the present time typically have a composition consisting of `99.3%` of the isotope `._92U^238` and `0.7%` of the isotope `._92U^235`. The half-lives of these isotopes are `4.47xx10^9yr` and `7.04xx10^8yr`, respectively. If these isotopes were equally abundant when the earth was formed, estimate the age of the earth.

Explanation

Let `N_(0)` be number of atoms of each isotope at the time of formation of the earth `(t=0)` and `N_(1)` and `N_(2)` the number of atoms at present `(t=t)`. Then, according to law of radioactivity,
` N_(1)=N_(0)e^(-lamda_(1) t)" "`…(i)
and `" "N_(2)=N_(0)e^(-lamda_(2)t)" "...(ii)`
`therefore" "(N_(1))/(N_(2))=e^((lamda_(2)-lamda_(1))t )" "`...(iii)
Further it is given that, `(N_(1))/(N_(2))=(99.3)/(0.7)" "`...(iv)
Equating Eqs. (iii) and (iv) and taking log on both sides, we have
`" "(lamda_(2)-lamda_(1))t=In((99.3)/(0.7))`
`therefore" "t=((1)/(lamda_(2)-lamda_(1)))In ((99.3)/(0.7))`
Where `lamda_(1) and lamda_(2)` are the decay constant.
Substituting the values, we have
`" "t=(1)/((0.693)/(7.04xx10^(8))-(0.693)/(4.47xx10^(9)))In((99.3)/(0.7))`
`" "(becauset_(1//2)(0.693)/(lamda)rArrlamda=(0.693)/(t_(1//2)))`
or age of the earth, `t=5.97xx10^(9)` yr