Answer to Question #107152 in Atomic and Nuclear Physics for Abdul

Question

A geological rock sample contains two radioactive isotopes, X and Y.
In the sample, there are N nuclei of X and 5N nuclei of Y. The half-life of isotope X is half of that of isotope Y. Y is not the product of X decaying nor is X the product of Y decaying.
The sample is stored for a time equal to two half-lives of isotope Y.
Determine the magnitude of the ratio

at the end of the storage period.


(3 marks)
b Polonium-210 is an alpha particle emitter with a decay constant of 5.80 × 10−8 s−1.
i Complete the decay equation below.


(2 marks)
ii Calculate the activity of 1.80 mg of polonium-210.


(3 marks)
iii How many nuclei of the polonium remain after 98 days?

Answer 1

a) At the end of two half-lives of isotope Y ($2t_Y$), isotope X will pass through 4 half-lives ($4t_X$). Calculate the number of isotopes X left after this time:

$$N_X(t)=\frac{N}{2^{t/t_{X}}}=\frac{N}{2^{4t_X/t_{X}}}=\frac{N}{2^4},\\ \space\\ N_Y(t)=\frac{5N}{2^{t/t_Y}}=\frac{5N}{2^{2t_Y/t_Y}}=\frac{N}{2^2},\\ \space\\ \chi=\frac{N_Y(t)}{N_X(t)}=20.$$

b) i. The decay equation for polonium-210 is

$$N(t)=N_0e^{-\lambda t}=N_0e^{-5.80 × 10^{−8}t}.$$

ii. Calculate the activity of 1.80 mg of polonium-210:

$$A=\lambda N_0=\lambda\cdot\frac{m}{\Mu}\cdot N_A=\\ =5.8\cdot10^{-8}\cdot\frac{1.8\cdot10^{-3}}{210}\cdot6.022\cdot10^{23}=2.99\text{ Bq}.$$

iii. How many nuclei of the polonium remain after 98 days:

$$N(t)=N_0e^{-\lambda t}=\frac{m}{\Mu}\cdot N_Ae^{-5.80 × 10^{−8}t}=\\ \space\\ =\frac{1.8\cdot10^{-3}}{210}\cdot6.022\cdot10^{23}e^{-5.8\cdot10^{-8}\cdot98\cdot24\cdot3600}=\\=3.16\cdot10^{18}.$$