Question

The decay curve of a radioactive nuclide sample is shown. Its initial activity is measured to be X Bq.

14. What is the population at t=8.0

15. After how many half lives will the population be 125 nuclides?

16. What will be the level of activity (in terms of X) after two half lives have elapsed?

17. What would the activity of the above sample be if we tripled its size?

18. If the initial activity of 0.356 grams of the above sample is Y Bq, what would the activity be if we increased the initial sample to 0.950 grams?

The decay curve of a radioactive nuclide 1000 sample is shown. Its initial activity is measured to be X Bo 14, what is the population at t 8.0 hours? 14. 500 15. After how many half lives will the population be 125 nuclides? 15. 16. What will be the level of activity (in terms of X) after two half lives have elapsed? 16 hours 16. 17. What would the activity of the above sample be if we tripled its size? 17 18. If the initial activity of 0.365 grams of the above sample is Y Bq, what would the activity be if we increased the initial sample to 0.950 grams?

Answer 1

It is an exponential decay curve of nuclide. 14) at t = 8 hours, from graph y co - ordinate = population Tilde Equal 325 15) form 1000 to 125 nuclide in how many half lives ? 1 st half life renaming nuclide = 1000/2 = 500 2nd half life renaming nuclide = 500/2 = 250 3rd half life renaming nuclide = 250/2 = 125 and so an or we can by 1000/ 2^n = 15 rightarrow 2^n = 1000/125 = 8 rightarrow 2^n = 2^3 Therefore n = 3 16) activity of a sample is directly proportional to number of atoms of reductive isotope A = k n rightarrow (1) (where k = decay cost) given initial activity A =_i = x Bq \r\n and N_i = 1000 Therefore form eq (1) A = KN rightarrow X = k 1000 rightarrow k = x times 10^-3 s^-1 rightarrow (2) after two half lines N_2 half = N_i/2^2 = 1000/4 = 250 Therefore A = k N_2 half = x times 10^-3 times 250 = 25 tx (unit distraction per second) 17) on triply size N�_i = 3000 thus after two half lives q