# Homework Help: Alpha decay

1. Feb 16, 2016

### Drizzy

1. The problem statement, all variables and given/known data

Th-232 ---> He-4 + Ra-228 + energy

How much is the energy?

2. Relevant equations

3. The attempt at a solution

my solution is this:

[m(Th-232)-90m(electron)] - m(He-4) - [8m(Ra-228)-88m(electron)]

My book skipped the electrons, why?

2. Feb 16, 2016

### SteamKing

Staff Emeritus
What's the mass of an electron compared to the mass of say, a proton?

3. Feb 16, 2016

### Drizzy

veeeeeeery small

4. Feb 16, 2016

### SteamKing

Staff Emeritus
So if you took 88 veeery small electrons and put them together, would you have something which approached the mass of even 1 proton?

5. Feb 16, 2016

### Drizzy

no :P but my friend is saying that they skipped the electrons because there are 88 electrons before the decay and equally as much after. but if that was the case then I should be able to write the mass of the electrons and then cross them off because they cancel out each other

6. Feb 17, 2016

### Drizzy

Can someone tell me if my friend is right?

7. Feb 17, 2016

### Burner1

Firstly, I think there should not be a 8(the blue one)here.

Secondly, I doubt if the red 88 is true. In alpha decay, a positive helium nucleus without electrons is emitted and the 90 electrons in Th-232 should be retained in the Ra-228 formed. Thus, they are fully cancelled out.

By the way, from what I have learnt, we focus on the mass defect inside the nucleus when calculating the energy released from nuclear reactions, so we don't consider the electrons.

Correct me if I was mistaken though.