Nuclear Chemistry/Physics And you 1. Radiation What exactly is this radiation stuff you hear about? Radiation is the release of Energy (and matter) that.

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Presentation transcript:

Nuclear Chemistry/Physics And you

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

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

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

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!!!!

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!!!!!!!

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.

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

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!

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

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

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

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

Fun and Exciting example: 238 U  4 α Th

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

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

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

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

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 U  4 2 α Th +

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

Now how about some β decay of Pa Pa  0 -1 β U WOW!!! That makes this guy happy

What about α decay of U-234? U  4 2 α Th Fantastic!!!

α decay of Th – 230?? Th  4 2 α Ra Wonderful!!! That one is a bit scary!

α decay of Ra Ra  4 2 α Rn NICE!!!

α decay of Rn – 222? Rn  4 2 α Po Well Done!

α decay of Po – 218 ? Po  4 2 α Pb Keep up the great work!!!!

β decay of Pb Pb  0 -1 β Bi Super dooper!!!

β decay of Bi Bi  0 -1 β Po Fantastico!!!!!

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!

Just a few more here Am – 241 undergoes α and gamma decay Am  4 2 α Np

Extension problem Pu – 242 undergoes α decay to form Pb – 218 Think about how many alpha particles are needed Pu  α Pb