Minimum KE in a proton/neutron

Smythe

In a deuteron, a proton and a neutron are very weakly bound by the strong nuclear force with an average distance between the particles of about 5fm.
Due to the uncertainty principle, the particles have a minimum momentum and hence a minimum kinetic energy of ...?

Which solution below is correct?


ΔpΔx ~ (h-bar)/2
Δp ~ (h-bar)/(2Δx)

Then using:
ΔE = (Δp)^2 / 2m
ΔE = [ {(h-bar) / (2Δx)}^2 / 2m ]
Using values of Δx=5x10^-15, m=1.67x10^-27
ΔE = 3.3 x 10^-14 J


Or should it be:
Δp ~ (h-bar)/(Δx)

Then using:
ΔE = (Δp)^2 / 2m
ΔE = [ {(h-bar) / (Δx)}^2 / 2m ]
Using values of Δx=5x10^-15, m=1.67x10^-27
ΔE = 1.3 x 10^-13 J

krd

Smythe wrote: »

In a deuteron, a proton and a neutron are very weakly bound by the strong nuclear force with an average distance between the particles of about 5fm.
Due to the uncertainty principle, the particles have a minimum momentum and hence a minimum kinetic energy of ...?

Which solution below is correct?


ΔpΔx ~ (h-bar)/2
Δp ~ (h-bar)/(2Δx)

Then using:
ΔE = (Δp)^2 / 2m
ΔE = [ {(h-bar) / (2Δx)}^2 / 2m ]
Using values of Δx=5x10^-15, m=1.67x10^-27
ΔE = 3.3 x 10^-14 J


Or should it be:
Δp ~ (h-bar)/(Δx)

Then using:
ΔE = (Δp)^2 / 2m
ΔE = [ {(h-bar) / (Δx)}^2 / 2m ]
Using values of Δx=5x10^-15, m=1.67x10^-27
ΔE = 1.3 x 10^-13 J

A deuteron, a proton and a neutron walk into a bar.......And the barman says...

Δp ~ (h-bar)/(2Δx) looks correct. Because even though there is uncertainty, losing that 2 would be pushing it.

Δp ~ (h-bar)/(Δx) looks a bit much.