Radioactivity and radioisotopes

1. Radioactivity and radioisotopes • Decay Constant and Half-life • Exponential law of decay

2. Decay constant and half-life It is possible to relate the decay constant l to the half-life T½. We already said that N0 is the number of radioactive atoms for t = 0. So, ½N0 will be the number of radioactive atoms after one half-life t = T½.

3. Decay constant and half-life Using logarithms to base e, solve the last equation to find the relationship between l and T½.

4. Half-life We measure the activity of a radioactive isotope in BEQUERELS (Bq), i.e. no of disintegrations per sec. Use the table to draw the half-life graph of Iodine-128. Show on the graph the activity after 40 min Half-life Half-life Half-life

5. Exponential law of decay Using a similar reasoning to the discussion on half-life and number of undecayed atoms N, we can conclude that: Where A is the activity (in Bq) at time t, A0 is the activity for t = 0 and x the number of half-lives elapsed.

6. Activity and number of radioactive atoms A very useful equation is the relationship between the activity of a radioactive nuclide and the number of radioactive atoms:

7. Questions • Polonium 210 has a half-life of approximately 138 days. At the beginning of an investigation the sample contains 2 g of Po-210. How long will it take to have 1/32 g of Po-210? • Show answer • Plot the graph of N = N0e-lt for values of t between 0 – 10 s. What is the half-life if N0 = 6.2 x 1012 and l = 0.25 s-1? • Show answer

8. Answer to question 1 After 1 half-life the mass of Po-210 has reduced of 1/2, i.e. to 1g. After 2 half-lives to 1/2 g After 3 half-lives to 1/4 g After 4 half-lives to 1/8 g After 5 half-lives to 1/16 g After 6 half-lives to 1/32 g So, 6 half-lives must pass before the mass of Po-210 left is 1/32 g  6 x 138 (days) = 828 days ~ 2.3 yr

9. Answer to question 2 T½ = log2/l = 2.77 s