U235 and U238 occur in nature in the atomic ratio 140:1.Assuming that at the time of earth's formation the two isotopes were present in equal amounts . Calculate the age of earth.(half life of U238 is 4.5*10^9 and that of U235 is 7.13*10^8 years.

Dear student,

Radiactive decay works along the equation:

n(t) = n_0*2^(-t/h)

where n is the amount of the isotope at time t; n_0 is the original amount, and h is the half-life.

So, now you have a situation where

n_238 = n_0 * 2^(-t/h_238)
n_235 = n_0 * 2^(-t/h_235)

In both equations, n_0 are equal. We also know that (n_235/n_238) = (0.71/99.29) = 0.00715

So, we need to solve for t in the equation

2^(-t/h_235)/2^(-t/h_238) = 0.00715

2^*((t/h_238)-(t/h_235)) = 0.00715

After some not too complicated gyrations, we come up with an age (from these data) of t = 6.0*10^9 years. (This is actually off by about 1.5 bln years from the actual estimate of the earth's age).

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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