When Uranium 238 decays, it emits 8 alpha particles and 6 beta particles before finally becoming the stable daughter product; What is the atomic number and mass number of the daughter product? Show your work.

Answer 1

each α-particle is composed of two protons and two neutrons
... a helium nucleus

a beta particle is an electron from the decay of a neutron
... this results in a proton and the beta electron

32 nucleons are emitted ... 16 protons and 16 neutrons

6 of the remaining neutrons decay into protons

Answer 2

92U238 ==> 2He4 + 90X234
90X234 ==> 2He4 + 88Y230
88Y230 ==> 2He4 + 86Z226
etc. Subtract 2 for atomic number and 4 for mass number.
For beta particle, you increase atomic number by 1 and leave mass number the same. Follow the series down the chain until you get to the end with 8 alpha and 6 beta emissions.
OR you can do it this way.
For alpha. 8 x -2 protons = -16 protons
................8 x -2 neutrons = -16 neutrons or -32 for mass number.
For beta. 6 x +1 proton = 6 protons.
So 92U238
238-32 = 206 for mass number.
92 -16 + 6 = 82 for the atomic number and you're left with 82X206. What's X? Look on the periodic chart and find atomic number of 82 is Pb so the chain ends at 82Pb206