1. Unit: Nuclear Chemistry Half-Life Day 2 – Notes

2. After today you will be able to… Identify the factor that nuclear stability is dependent on. Calculate the half-life for a given radioisotope. Calculate how much of a radioisotope remains after a given amount of time.

3. Nuclear Stability Close to 2,000 different nuclei are known. Approximately 260 are stable and do not decay or change with time. The stability (resistance to change) depends on its neutron:proton ratio.

4. Nuclear Stability Plotting a graph of number of neutrons vs. number of protons for each element results in a region called the band of stability.

5. Nuclear Stability For elements with low atomic numbers (below 20) the ratio of neutrons:protons is about 1. – Example: (6n/6p = 1) C 12 6

6. Nuclear Stability For elements with higher atomic numbers, stable nuclei have more neutrons than protons. The ratio of n:p is closer to 1.5 for these heavier elements. – Example: 124n/82p = approx. 1.5 Pb 206 82

7. Nuclear Stability The neutron:proton ratio is important because it determines the type of decay that occurs. All nuclei that have an atomic number greater than 83 are radioactive. Ever wonder why some atomic masses listed on the Periodic Table have ( ) around them? Because their atomic masses are estimated due to radioactive decay!

8. Half-Life Half-life: (t 1/2 ) the time required for half of the nuclei of a radioisotope sample to decay to products. Example: If you have 20 atoms of Radon- 222, the half life is ~4 days. How many atoms remain at the end of two half lives? 0 t 1/2 1 t 1/2 2 t 1/2 4 days8 days 20 atoms initially 10 atoms 5 atoms remain

9. Half-Life We can represent half- life graphically as well. —Example: Carbon-14

10. However, very seldom do we count atoms. Therefore it is more appropriate to calculate amount that remains in terms of mass.

11. Half-Life Example: Carbon-14 emits beta radiation and decays with a half-life (t 1/2 ) of 5730 years. Assume you start with a mass of 2.00x10 -12 g of carbon-14. a.How long is three half-lives? b.How many grams of the isotope remain at the end of three half-lives? a.3(5730) = b. 2.00x10 -12 g x 1 / 2 x 1 / 2 x 1 / 2 = 17,190 years 2.5x10 -13 g

12. Half-Life b. How many grams of the isotope remain at the end of three half-lives? Alternatively, part b can also be calculated like this: 0 t 1/2 2.00x10 -12 g initially 1.00x10 -12 g 2.50x10 -13 g remains 1 t 1/2 2 t 1/2 3 t 1/2 5.00x10 -13 g x 1 / 2

13. Half-Life Example: Manganese-56 is a beta emitter with a half-life of 2.6 hours. a.How many half-lives did the sample go through at the end of 10.4 hours? b.What is the mass of maganese-56 in a 1.0mg sample of the isotope at the end of 10.4 hours? a.10.4 h/2.6 h = b. 1.0mg x 1 / 2 x 1 / 2 x 1 / 2 x 1 / 2 = 4 half-lives 0.063 mg

14. Questions? Complete and turn in the exit ticket.