\sect{Equilibrium of Fermions and condensed Bosons F==A+B}
${N}$ fermions of type ${F}$ that have spin $1/2$ are placed in a box of volume ${V}$.
Each fermion can decay into a boson of type ${B}$ that have spin ${0}$,
and a fermion of type ${A}$ that has spin $1/2$.
The reaction is ${F\rightarrow A+B}$, and it has an energy gain ${\epsilon_0}$.
This means that ${A+B}$ has a lower binding energy than ${F}$.
The masses of the particles are $m_{F},m_{A},m_{B}$ respectively.
\Dn
(1) Assuming ideal gases at temperature ${T}$, write the
chemical equilibrium condition
that determine the densities ${n_{F}, n_{A}, n_{B}}$ at equilibrium.
\Dn
(2) Write the chemical equilibrium condition at ${T=0}$.
Describe the dependece of the densities on ${\epsilon_{0}}$.
Find $\epsilon_c$ such that for $\epsilon_0>\epsilon_c$
the number of F fermions vanishes.
\Dn
(3) Assume that the condensation of bosons B occurs at~$T_c$
such that ${T_c \ll p_F^2/(2m_A) < \epsilon_0}$,
where $p_F$ is the Fermi momentum of fermions A.
Evaluate $T_c$ and rewrite the condition on $T_c$ in terms of the given parameters.
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