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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.

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.

Grade:11

1 Answers

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

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).

 

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

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