Nuclear Decay help please?

Question

Question: 238-U eventually decays to 206-Pb. How many alpha and beta particles are produced?

Answer: 8 alpha and 6 beta.

My problem is that I don't know how to solve for that.

The book is absolutely no help. It just says that for alpha emission, the number of protons decreases by 2 units (but in this case its 10) and that the mass number decreases by 4 units (in this case its 32). So I'm really confused...any help please?

Answer 1

A big assumption here is that all the nuclear changes are either alpha or beta decay.  With that assumption though...

first find the mass difference between U-238 and Pb-206...32 amu as you stated.  

Next divide that by the mass of an alpha particle (32/4) which give you 8.  Therefore there are 8 alpha particles emitted.

To determine the number of beta particle subtract the number of protons between U and Pb (92-82) which gives you 10 so you need a net loss of 10 protons.  Loosing 8 alpha particles would be 16 (remember an alpha particle is 2 protons and 2 neutrons).  A net of -6.  Therefore you need to "gain" 6 protons which is done by 6 beta particle emmissions..

Answer 2

You're on the right track. You need to find a combination of alpha and beta emissions that will give you 10 fewer atomic units and a 32 unit decrease in mass. The only thing you're missing is what happens with beta decay, which results in no change in mass, but an increase in atomic number by 1.

In order to lose 32 units of mass, we need 32 / 4 or 8 alpha decays. At that point, there will have been 8 * 2 or 16 lost protons, 6 more than the 10 needed, so 6 beta decays are required.

Answer 3

Ok, here it is. An α particle is a helium nucleus - 2p,2n - atomic mass 4. U238 loses 32 atomic mass units, so there's your 8 α particles. The atomic number drops from 92 to 82, so there's your 10 protons. A ß particle is just an electron (or a positron).

Since the proton count is down by 10, the electron count must also be reduced by that amount, so I don't understand the "6 ß particle" answer.

EDIT - OK- I wasn't thinking - see previous excellent answer