An isotope of 238,92-Uranium decays into a stable isotope of 206,82-Lead through a series of alpha and beta decays, how many of each does it go through?

This is a good example of where simplification can make an otherwise tricky question a lot easier. It is easy to get confused trying to work out different isotopes and the extra products of the decays, but all this can be ignored, and the question is just a simple maths problem. Firstly, we must consider what each decay does to the atomic number and proton number of the isotope. Alpha decay emits a helium nucleus, and so reduces the atomic number by 4 and the proton number by 2. Beta decay turns a proton into an electron, and increases the proton number by 1, leaving the atomic number unaffected. Using this information, we can work out that in order to reduce the atomic number from 238 to 206 we must use alpha decay, because beta doesn’t affect the proton number. Therefore 8 alpha decays are needed, this brings the atomic number to the required 206, but leaves the proton number at 76. To bring this up to the required 82 for lead, 6 beta decays are necessary. And so, the answer is 8 alpha decays and 6 beta decays.