Temperature Of An Atom, Shocking!

From the previous post "Hot Topic, Less Mumble, Jumble...",

$T=\cfrac{1}{2}(qv^2_{rc}+mv^2_{rg})$

then for an electron, its temperature is given by,

$T_e=\cfrac{1}{2}(qv^2_{rc}+m_ev^2_{rg\,e})$

and for a proton,

$T_p=\cfrac{1}{2}(qv^2_{rc}+m_pv^2_{rg\,p})$

and for a neutron,

$T_n=\cfrac{1}{2}m_nv^2_{rg\,n}$

Temperature associated with an atom is,

$T_{atom}=n(T_e+T_p)+(N-n)T_n$

where  $N$  is the mass number and  $n$  the atomic number,

$T_{atom}=nT_e+nT_p+(N-n)T_n$

Let,

$T_{nucleus}=nT_p+(N-n)T_n$

$T_{atom}=nT_e+T_{nucleus}$

And since,

$T_e<nT_e<nT_p<nT_p+(N-n)T_n$

$T_e<T_{nucleus}$

For the sake of completeness.