Minimum Kinetic Energy in a Deuteron: Which Formula Is Correct?

[ZedCar]

Homework Statement

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 answer is correct?

Homework Equations

The Attempt at a Solution

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

 

[bossman27]

I believe it's the former. Where did you get the idea to drop the 2?

 

[ZedCar]

I believe it's the former. Where did you get the idea to drop the 2?

Because that 2nd version was given to us by the lecturer.

The former attempt is mine.

 

[bossman27]

Because that 2nd version was given to us by the lecturer.

After further reading, $\sigma x \sigma p \geq \frac{\hbar}{2}$ is the modern form, which is most accurate.

$\sigma x \sigma p \geq \hbar$ was Heisenbergs original formula. It is obviously still a true mathematical statement since the minimum uncertainty is greater than in the above formula, but it is not quite as accurate.

I think your professor probably wants the original Heisenberg equation, since he decided to teach it that way. Either way, you can always write a note in next to the problem if you're worried about using the incorrect one.

 

[ZedCar]

Thanks very much bossman. That's appreciated.

$\sigma x \sigma p = \hbar$ was Heisenbergs original formula.

I was unaware of that. I'll look into that now.

At least now I know where the "problem" lies.