Alpha and Beta decay....PLEASE HELP!?

Question

Radon 220/86 Rn produces a daughter nucleus that is radioactive. The daughter in turn produces its own radioactive daughter and so on. This process continues until Lead 208/82 Pb is reached. What is the total number of alpha particles? and the total number of Beta particles that are generated in this series of radioactive decays?

Answer 1

an alpha particle consists of 2 protons and 2 neutrons, therefore if radon emmitted an alpha paticle it would take four (2+2) away from the mass number (the big number) and two away from the atomic number ( the little number).

an beta paticle is the emmission of an electron, which would add one to the atomic number.

You can work the number of stages out from this.

this means it is the emitting, in this case of 3 alpha particles and two beta particles

Answer 2

Alpha particles are basically a helium nucleus, and so have an atomic number of 4, proton number of 2.

Assuming it's Beta minus decay where it is an electron that is released, not a positron, the symbol we write has an atomic number of 0, proton number of -1. This is because Beta minus decay occurs when an neutron splits into a proton and electron.

Alpha: Rn 220/86 => A 4/2 + X 216/84 (A = alpha, X = new element)

Beta: Rn 220/86 => B 0/-1 + Y 220/87 (B = beta, Y = new element)

So, only alpha decay affects the atomic number, so from 220 => 208, we need to lose 12 nucleons (neutrons or protons).

This means we need 3 alpha decays, as we lose 4 nucleons per decay.

3 alpha decays means we lose 6 protons, which will only give us 80 protons.

Each beta minus decay gives us 1 proton, so to get back to 82, 2 beta minus decays are required.

Soooo:

3 alpha decays, 2 beta minus decays