in this video we're looking at the alpha decay of radon 222 and we're given a little bit of
background about radon like where does it come from it's a daughter nucleus of uranium-238 decay
so there's trace amounts of uranium-238 in rock and soil and those decays will constantly produce
trace amounts of radon 222 interestingly enough this actually causes about 21 000 lung cancer
deaths per year in the united states but what we're here for is the nuclear decay so it emits an
alpha particle and then the daughter nuclei that come after this they also decay by alpha decay so
if you're inhaling this gas you end up ingesting a source of alpha radiation which is very bad
because 100 of that alpha radiation is going to be absorbed in your tissues alpha radiation doesn't
penetrate very far we're asked to write down the nuclear reaction equation for the alpha decay of
radon 222 and compute the energy in mev that's released in the decay we have a reference table
here we're given the atomic mass unit written in mev per c-squared the reason we need that
is because the standard tables for these masses the mass of radon the mass of polonium the mass
of helium the standard tables give these masses in terms of atomic mass units we're going to want
to convert those to mev per c-squared at some point in order to get our energy calculation
done so just a real quick overview of what's happening in alpha radiation that means a helium
nucleus flies out of this very large nucleus and a helium nucleus has two protons and two neutrons
so when my radon alpha decays i'm going to bounce two spaces to the left on the periodic table
because i've lost two protons this is why i get polonium out of the reaction putting things in
nuclide notation i start with radon 222 nucleons i can see from the periodic table 86 protons
this leaves me with 136 neutrons when i do the subtraction and the result of this decay is that i
get a helium nucleus also called an alpha particle and helium has two protons two neutrons four total
nucleons so there's that guy and then again i've lost two protons so now i'm talking about polonium
for my remaining large nucleus and polonium has 84 protons that's two less than i had before i
also lost two neutrons which means there should be 134 of those now and of course i lost four total
nucleons so i'm talking about polonium-218 now there's a subtle point about these masses given
in the table another unfortunate thing about those tables is they include the mass of the electrons
fortunately that's not going to mess up our calculation here but you should be aware of it
so this is the mass of a radon 222 with all of its 86 electrons this is the mass of a polonium-218
with all of its 84 electrons and then the helium with its two electrons fortunately this doesn't
cause a problem because the electrons just cancel out of the calculation here
so the idea for finding the total energy released is just compare the mass that we started with
compare the mass that we ended up with you're going to find that there's some mass missing
there and then we use e equals m c squared to compute the energy equivalent of that mass
so written in math it looks like that and this time i'm going to go ahead and compute the missing
mass in atomic mass units and then i'll convert to mev for c-squared at the end and again notice
in this first term that includes 86 electrons but then in these parenthesis that includes
two electrons for the helium and 84 for the polonium so the electron masses just cancel out
and the missing mass here really is a reduction in the mass of the nuclei so i end up with a
reduction in mass of .006 atomic mass units i convert that to mev per squared and i get 5.59
mev per c squared now i think most nuclear physicists would just ignore the c-squared
and say okay 5.59 mev was released in the reaction but to be totally complete about
this what we're doing is finding the energy equivalent of that mass that disappeared and
for this we use e equals m c squared and our mass was 5.59 mev per c squared and we multiply by c
squared and this is why you can just kind of ignore the c squared when you get to this point
in the calculation it is the same number i get 5.59 mev released in the reaction
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