a dispersion which is compatible with the energy of the ground state of the hydrogen atom, i.e., such that ${\Delta P_0^2\over 2M}\approx E_I$. Considering that the total mass is: $M \approx M_p \approx 1.67 \times 10^{-27}$ kg, we have $\Delta P_0 \approx 8.6 \times 10^{-23}$ Js/m, which is compatible with Heisenberg’s principle, as is clear that: $\Delta Q_0\Delta P_0 \approx 45.6 \times 10^{-34}~{\rm Js}\approx 43\, \hbar > \hbar / 2$ ($\hbar \approx 1.05 \times 10^{-34}$ Js).

Inserting the above values in (\[eq1\]), and observing that the second term in the square root is negligible compared to the first, we thus obtain $\Delta Q_t \approx 5.1 \times 10^4\, t$ m/s, that is: $$t\approx 0.2\times 10^{−4}\,\, \Delta Q_t ~{\rm s/m}. 
\label{eq1bis}$$ This last expression tells us the time we roughly need to wait for the center of mass of the wave packet of the hydrogen atom to reach the spatial spreading $\Delta Q_t$. Let us consider some specific values. To obtain a spreading of $5$ km, that is, of $5 \times 10^3$ m, we have to wait about $10^{-1}$ s, i.e., a tenth of a second. In 1 s, instead, the packet will have reached a width of about $50$ km, while 