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Adopt the lower-bound scale of the Heisenberg uncertainty principle, $\Delta r\Delta p\sim \hslash /2$, with the prescribed radial scale from a. Using $\hslash c=197.327$ MeV fm,For a one-sided exponential distribution of a positive momentum magnitude, $f(p)=p_0^{-1}{e}^{-p/p_0}$, both its mean and its standard deviation are $p_0$. Under the additional identification made in this estimate, set the characteristic momentum to $\Delta p$. This identification concerns a positive magnitude; the mean of a signed momentum component in an isotropic stationary state would instead vanish.

Use the relativistic energy-momentum relation, not a nonrelativistic kinetic-energy approximation, because the momentum-energy scale is much larger than the stipulated quark rest energy:This is a characteristic energy estimated at the characteristic momentum. It is not an exact evaluation of $⟨\sqrt{p^2{c}^2+m_q^2{c}^4}⟩$ over a fully specified three-dimensional wavefunction. An order-of-magnitude convention $\Delta p\sim \hslash /\Delta r$ would give an answer about twice as large; the uncertainty principle by itself specifies a bound, not a unique confinement energy.