Question

How much 206 Pb will be in a rock sample that is 4.4 x 10^8 years old and contains 3.16 mg 238 U?

Answer 1

half life of 238-U = 4.468*10^9 years

K = 0.693/t1/2

   = 0.693/(4.468*10^9)

   = 1.55*10^-10 s-1.

k = 1/tln(a0/a0-x)

(1.55*10^-10) = (1/(4.4*10^8))ln(a0/3.16)

a0-x = amount of 238-U in the sample now = 3.16 mg

a0 = initial concentration of 238-U = 3.383 mg

1 mole 238-U = 1 mole 206-Pb

no of mole of 238-U decayed = (3.383-3.16)/238 = 0.00094 mmol

no of mole of 206-Pb formed = 0.00094 mmol

amount of 206-Pb present in sample = 0.00094*206 = 0.194 mg