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Question
238U decays to 206Pb with a half-life of 4.47 × 109 y. This happens in a number of steps. Can you justify a single half for this chain of processes? A sample of rock is found to contain 2.00 mg of 238U and 0.600 mg of 206Pb. Assuming that all the lead has come from uranium, find the life of the rock.
Solution
Given:
Half-life of 238U, t1/2 = 4.47 × 109 years
Total Number of atoms present in the rock initially , `N_0 = ((6.023 xx 10^23 xx 2)/238 + (6.023 xx 10^23 xx 0.6)/206)`
= `(12.046/238 + 3.62/206) xx 10^20`
= `(0.0505 + 0.0175) xx 10^20`
= `0.0680 xx 10^20`
Now , `N = N_0e^(-lambdat)`
Here, `lambda` = Disintegration constant
t = Life of the rock
⇒ `N = N_0e^((-0.693)/(t/"1/2") xx t)`
⇒ `0.0505 = 0.0680e^((-0.693)/(4.47 xx 10^9) xx t)`
⇒ `"In" (0.0505/0.0680) = (-06931)/(4.47 xx 10^9) xx t`
⇒ `t = 1.92 xx 10^9` years
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