Cold Temperature Requires Thick Skin, Very Thick Too

From the post "Critical Matter/Anti-Matter Fusion Temperature, $T_c$", at double root,

$A=\cfrac{m_e}{2r_e}$

And the y-intercept becomes,

$v^{ 2 }Ar^{ 2 }_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } }$

$\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { 1 }{ 2 } m_{ e }v^{ 2 }r_{ e }$

$T_{ e }T_{ n }=\cfrac { m_{ e }m_{ p } }{ (m_{ e }+m_{ p })^{ 2 } } T^{ 2 }_{ c }$

$T^{ 2 }_{ c }=4\pi \tau _{ o }\cfrac { (m_{ e }+m_{ p })^{ 2 } } { m_{ e }m_{ p } }(\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -\cfrac { 1 }{ 2 } m_{ e }v^{ 2 }r_{ e })$

$T^2_c\approx$4*pi*9.632e42*( 9.10938e-31+1.67262e-27)^2/( 9.10938e-31*1.67262e-27)*((1.602176565e-19)^2/(4*pi*8.8541878176e-12)-1/2*9.10938291e-31*(2^(1/2)*299792458)^2*5.2917721092e-11)

$T^2_c\approx$-9.612e23

$T^2\lt 0$

This implies that it is not possible for hydrogen's electron orbital radius to have a double root.

Let examine the the individual terms of the expression at double root,

$\cfrac { 1 }{ 2 } m_{ e }v^{ 2 }r_{ e }$=1/2*9.10938291e-31*(2^(1/2)*299792458)^2*5.2917721092e-11=4.332e-24

and

$\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } }$=(1.602176565e-19)^2/(4*pi*8.8541878176e-12)=2.307e-28

$r_e$ can increase by an order of 1 for heavier atoms with  $n$  positive charges at the nucleus.  If the mass of electron is estimated correctly it is not possible for a double root unless the nucleus has  $n\gt\cfrac{10^{-23}}{10^{-28}}\gt10\,000$  positive charge.

An increase in conductivity at low temperature is not due to the higher orbit physically moving closer to the kink point but, because of the decrease in band gap and the readiness to lose a packet of energy at lower temperature that increases occupancy at the higher orbit.  This higher valid value of  $r_e$  is the electron's orbit in the conduction band.

High temperature may leave a single root value for  $r_e$  above the kink point; the other root being negative.

$r_e=\cfrac { m_{ e }\pm \sqrt { m^{ 2 }_{ e }-\cfrac { A }{ \pi v^{ 2 } } (\cfrac { q^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \tau _{ o } } ) }  }{ 2A }$

Conductivity may increase with increasing temperature for some material.  High conductivity is associated with high  $r_e$, with other factors in consideration.