From key@physics.utoronto.ca Tue Mar 11 08:45:44 2008
Date: Tue, 11 Mar 2008 08:45:39 -0400
From: Tony Key <key@physics.utoronto.ca>
To: xiao.ou@utoronto.ca, key@physics.utoronto.ca
Subject: Re: MP problem

Hi Xiao - all of these questions use the same ideas and formulae.  the total
energy available for a decay is the Q value.  Q/c^2 is the difference in
NUCLEAR masses between the initial state and the final state.  End of story.
To do the calculation, you  need to calculate the NUCLEAR masses (m) from
the ATOMIC masses that are given.  This is done by subtracting the masses of
the Z electrons in the atoms from the atomic mass - m = M -Zm_e.  In all
cases excpet that for beta+plus (positron) decay, the Q value reduces to the
difference in atomic masses, as the electron masses cancel out from the
equation.   You can prove this for yourself.

I find the hint in the last question confusing -  it is a MasteringPhysics
question - but if you follow my recipe above you will get the correct
answer.  The hint simply says that the Q/c^2 value can be calculated by
ignoring the electron masses - apply my method to prove it!  i.e. Q/c^2 =
m(I) - {m(X) + m_e + m(neutrino)}  = {M(I) - 53m_e} - {M(X) - 54m_e + m_e +
m(neutrino)} = M(I) - M(X)

The 3H decays to 3He + the electron + an antineutrino (yes, one of the
neutrons in the 3H tranforms into a proton - the nucleus with two protons is
then 3He) .  Use the formalism above.

The nuclear masses are used because we are considering nuclear decays (!) -
what else would  we use?  Unfortunately all that we are given in the tables
are the  atomic masses, so we have to calculate the former from the latter.

TonyK

PS  The question you mention does not  ask for "the kinetic energy of a
negative beta decay of 3H"  - it asks for the "total kinetic energy released
in the beta-minus decay of  3H" -  make  sure you  understand the
difference!

xiao.ou@utoronto.ca wrote:
> sir I have a question. One of the mastering physics questions asks us to
> find the kinetic energy of a negative beta decay of 3H. And the question
> before that took into account the mass of the electron in the nucleus, or
> so it said for my hint. So my question is how would you find the energy of
> the 3H? does it still have 1 proton? And when it decays to 3/1H, is one of
> the neutrons from the isotope transformed into a proton?
>
> Also why do we only take the nuclear masses into account for calculating
> kinetic energy?
>