2. Nuclear Chemistry/Physics And you

3. 1. Radiation What exactly is this radiation stuff you hear about? Radiation is the release of Energy (and matter) that results from changes in the nucleus of an atom

4. Say What???? A nucleus of an atom can change? That is right – so far in chemistry we have focused mainly on the interaction of electrons – the basis of most chemistry The field of nuclear chemistry focuses on changes within the nucleus of an atom

5. Exactly what kind of changes are we talking about here? There are 3 main types of changes that we will discuss here today: 1) Alpha decay 2) Beta decay 3) Gamma radiation

6. Let’s start with Alpha decay shall we? Alpha (α) radiation results from a loss of 2 protons and 2 neutrons from the nucleus of an atom Wait, 2P+ and 2 n O ? That sounds like a helium nucleus That is correct – a helium atom without the electrons!!!!

7. Do you recall When we represented the helium atom like this? 4 2 He Well we can represent an alpha particle like this: 4 2 α That is like so totally cool!!!!!!!

8. When would alpha decay happen? Well it only happens when the atomic number of an element is greater than 83 That is to say it needs to be so large that it has at least 83 protons in the nucleus.

9. Can you give us an example please??? You bet – 238 U  4 α + 234 Th Here a Uranium atom is changing into a Thorium atom This is called a nuclear decay reaction 92290

10. Say What??? One type of atom is changing into another type of atom?? That is correct! Smart scientist type people (like yourself) refer to the changing of one element into another element as “transmutation” You can use this term at home and impress your mom!

11. That is like – so SICK The second type: Beta (β) radiation is basically the loss of an electron (beta particle) from the nucleus

12. An electron in the nucleus?? Yes – a neutral neutron looses an electron (neg. charge) and becomes a…. Proton!!!! 1 n  1 P + 0 e But we will represent the 0 e as 0 β NO WAY!! WAY!!!! 01- 1

13. β emission example 14 C  0 β + 14 N This only happens when the n 0 to P+ ratio is too large 6 7

14. The third type is: Gamma ( ) radiation release Light E with short wavelength leaves nucleus This happens as a by product of other radioactive decay processes You can not predict when it will happen

15. Fun and Exciting example: 238 U  4 α + 234 Th + 0 92 2 90 0

16. Are there other types? Why yes – I am glad you asked! You do not have to write these in your notes

17. There is Positron emission – when an electron with a positive charge is released (electron antimatter) 11 C  0 e + 11 B A P+ changes into a n 0 6 +15

18. There is K capture – when an inner shell electron is captured by the nucleus (P+ changes into a neutron) 7 Be + 0 e  7 Li When the n 0 to P+ ratio is too small 43

19. Boy that was a lot of content! But not too much for you!!!! Hey I have a great idea, why don’t we take a bit of class time to write out some nuclear equations Now that sounds like fun! Even if it makes my brain hurt

20. What would be the equation for U- 238 that undergoes α decay??? Write out what you are starting with Draw in a yields arrow Write in the alpha particle product Write in the other product of the reaction 238 92 U  4 2 α 234 90 Th +

21. What about some β decay of Thorium 234 234 90 Th  0 -1 β + 234 91 Pa How did you do??? Great!!!!!! That makes me happy

22. Now how about some β decay of Pa - 234 234 91 Pa  0 -1 β + 234 92 U WOW!!! That makes this guy happy

23. What about α decay of U-234? 234 92 U  4 2 α + 230 90 Th Fantastic!!!

24. α decay of Th – 230?? 230 90 Th  4 2 α + 226 88 Ra Wonderful!!! That one is a bit scary!

25. α decay of Ra-226 226 88 Ra  4 2 α + 222 86 Rn NICE!!!

26. α decay of Rn – 222? 222 86 Rn  4 2 α + 218 84 Po Well Done!

27. α decay of Po – 218 ? 218 84 Po  4 2 α + 214 82 Pb Keep up the great work!!!!

28. β decay of Pb - 214 214 82 Pb  0 -1 β + 214 83 Bi Super dooper!!!

29. β decay of Bi - 214 214 83 Bi  0 -1 β + 214 84 Po Fantastico!!!!!

30. I am thinking that you are getting the pattern here! Just think about all you have learned about nuclear chemistry today – and how enriched your life is!

31. Just a few more here Am – 241 undergoes α and gamma decay. 241 95 Am  4 2 α + 0 0 + 237 93 Np

32. Extension problem Pu – 242 undergoes α decay to form Pb – 218 Think about how many alpha particles are needed 242 94 Pu  6 4 2 α + 218 82 Pb