\(p_x=\cfrac{n_{\lambda\,e\_x}.\Delta Q}{mc^2}\)
from the post "Ding Dong And A Small Peck" dated 10 May 2016.
When,
\(\cfrac{\Delta Q}{mc^2}=1\)
\(p_x=n_{\lambda\,e\_x}\)
Which means, from the Poisson Distribution Plot with \(\lambda=1.6678205\),
\(Probability=0.30\)
of one encounter in \(10^9\) seconds. This translates to one entanglement event in \(11574.07\,\, days\)!
Rare, but given the number of particles around, entanglement happens all the time.
Ouch!
