Question

Part A How long does it take for the radioactive isotope to decay to 6.0×10−4 μg ? (Assume no excretion of the nuclide from the body.)

Part B A radioactive sample contains 1.70 g of an isotope with a half-life of 3.6 days. What mass of the isotope will remain after 5.8 days? (Assume no excretion of the nuclide from the body.)

Part C A mammoth skeleton has a carbon-14 decay rate of 0.46 disintegrations per minute per gram of carbon (0.46 dis/min⋅gC ). When did the mammoth live? (Assume that living organisms have a carbon-14 decay rate of 15.3 dis/min⋅gC and that carbon-14 has a half-life of 5715 yr.)

Part D A meteor has a Pb-206:U-238 mass ratio of 0.850:1.00. What is the age of the meteor? (Assume that the meteor did not contain any Pb-206 at the time of its formation.)

Answer 1

Part A) Half-life and the amount is not mentioned , so it becomes impossible to get a solution for this.

Part B) Radioactive decay follows first order kinetics and is given by the equation---
A = Ao e^-kt., where A and Ao are the amounts at time t and the initial amount, t is the time and k is the decay constant which is related to the half-life of the radio nuclide.

Given, t_1/2 = 3.6 days

k = ln 2 / t_1/2

=> k = 0.693 / 3.6 days = 0.1925 day^-1

Now, A = Ao e^-kt.

=> A = (1.70g) e-^{(0.1925 day-1​ x 5.8 day)}

=> A = 1.70 g * e^-1.1165

=> A = 1.70 g * 0.327

=> A = 0.5559 g

Part C) We know, K = ln 2 / t_1/2 = 0.693 / 5715 yr = 1.21 x 10^-4 yr^-1

Again, the integrated form of first order rate law is--

lnA = -kt + ln A₀

=> t = 1/k ln A₀ / A

=> t =( 1/1.21 x 10^-4 yr^-1 ) * ln (15.3 / 0.46)

=> t = ( 1/1.21 x 10^-4 yr^-1 ) *ln 33.26

=> t = ( 1/1.21 x 10^-4 yr^-1 ) *3.504

=> t = (3.504 / 1.21 x 10^-4 yr^-1)

=> t = 2.8958 x 10⁴ yr

=> t = 28958 yr

Part D) We can say that the mass of Pb-206 can be 0.850 g.

Now,

0.850 g Pb 206 * (1mol Pb 206 / 207.2g Pb) * (1 mol U 238 / 1 mol Pb 206) * (238.02891g U / 1 mol U 238) =0.9764699493 g U-238

Again, t_1/2 = 4.5 x 10⁹ yr

Now, K = 0.693 / t_1/2 = 0.693 / 4.5 x 10⁹ yr = 1.5 x 10^-10 yr^-1

So initially, there is a total of 1.9764699493 g U-238
The final mass should be 1.00 g U-238

Now, ln A / A₀ = - kt

=> ln (1.00 M U-238 / 1.9764699493 M U-238) = -1.5 x 10^-10 yr^-1 x t

=> ln 0.5059 = -1.5 x 10^-10 yr^-1 x t

=> - 0.68 = -1.5 x 10^-10 yr^-1 x t

=> t = 0.68 / 1.5 x 10^-10 yr

=> t = 0.45 x 10¹⁰ yr

=> t = 4.5 x 10⁹ yr