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.

Answer 1

Given:
Half-life of ²³⁸U, t_1/2 = 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