 Let's just do this problem. We threw alpha and beta negative decay. We go from a radon 220 to lead 208 and We don't know what sequence that is. We just want to know how many alpha and beta particles are we going to have in the end Now let's just write alpha decay again for ourselves alpha decay Remember that's going to start with some parent Atomic number, atomic mass number, and that's going to give us some of the daughter product We're going to lose two from our atomic number and we're going to lose four from our atomic mass number And that means we'll have the alpha particle on this side If we undergo a beta negative decay We're going to go from some parent To the A we're going to have some daughter. The A space is exactly the same and we actually gain one proton As the electron gets Decays into a proton plus an electron and that electron remember is looking like such Now we are going from 220 to 208. That's the difference of 12 That means if you look at it very simply 12 will be four of these If I do this four times, so I'm going to lose four times two This eight I'm going to lose eight protons, but I only need to lose We only need to lose four If I do this three times, I'm going to lose six if I do this three times I'm going to lose 12, which is what I want if I Do it three times. I'm also going to do six, but I only need to lose four That means I've got to make up to and I can make up to by doing this twice It's as simple as that think about it. I'll have four I'll have four mediums for alpha particles Because if I do that Where am I this morning? If I do this three times, I'm going to do this negative four negative four negative four negative 12, which is exactly what I want But that is going to leave me negative two negative four negative six So that will leave me at 80 208 80, but I don't want to be at 80. I want to be at 82 fortunately, I can do this twice It's not going to change my 208 where I'm at, but it won't change my 80 to 81 and then to 82 So we'll have two electrons We have three alpha particles and two beta particles This is an ultra-familiar problem Or the concept to solve it anyway the equations to solve it I have uranium 235 and the decays into Daughter products and we have some gamma-ray release And we noted those gamma-rays those gamma photons have a wavelength of one point one four times ten to the power Decent meters. What is the energy in mega electron volt? You can write our simple equation that we know so well and as much as remember this potentials at the speed of light So I can use HPC divided by the lambda and if I substitute that I get About 1.745 times 10 to the power negative 14 joule I need to convert that to electron volt. One electron volt is 1.60 Times 10 to the power using three significant digits because I can put a 2 in there But more accuracy Depending on how it's important thing actually depending on how you want to handle or how you are taught to handle significant digits Anyway negative 90 joule and that is going to give me 0.1 9 mega electron Simply not Let's do our last problem. We have phosphorus 32 there and it goes beta negative decay to solve for 32 We have one electron and I think in the next section. I think we'll quickly do Neutrinos and that would be an anti-neutrino because of the bar and carbide Positive decay will give you a neutrino. Beta negative decay is going to give you an anti-neutrino And this is the atomic mass Units of phosphorus 32 and the sulfur 32 and the beta negative particle Which is an electron has this 1.20 mega electron volt What is the possible energy of? Anti-neutrino what possible energy can it have well if we look at it this way if we combine If we combine these two if we combine those two It would mean That this electron is included in this which remember is what we have for our units So if I subtract these two from each other, I'll get the units It is the units of this minus this whole lot combined And that gives me if I subtract those two I get 1.837 times 10 to the power negative 3 units There is a difference between those two I can convert that into energy by multiplying by 9 3 1.5 mega electron volt one unit that gives me 1.7 1.7 1.1 Mega electron thought So those two combined should give me 1.7 those two combined minus this gives me 1.7 1.1 mega electron thought Now of that the beta particle because it has velocity. There's no velocity in there So no kinetic energy in there. Let's say it's all in there And that is 1.2 so if I subtract 1.2 mega electron volt from there I get the total of 0.5 1.1 Mega electron volt is the possible energy that this anti-nutriner can carry away