Excitation levels and Nucleus Size

[genloz]

How would you use excitation levels (or the lowest excitation level) to find the radius of a nucleus? And how would it differ to using the r=r0A^(1/3) formula?

[genloz]

Is it possible to use heisenberg again?
eg.
\Delta x \Delta p \approx \hbar
\Delta x \approx \frac{\hbar}{\sqrt{2mE}}

so subbing in the excitation level for E and the A value for m?

[malawi_glenn]

Have you tried?

[genloz]

Well, for:
^{17}O (oxygen)
r=r_{0}A^{1/3}
r=1.3*17^{1/3}=3.34

and using the other way:
\Delta x=\frac{\hbar}{\sqrt{2mE}}
Knowing that the first excitation level of 17-oxygen is 3055keV
\Delta x=\frac{0.197}{\sqrt{2*17*3055}}
\Delta x=6.11*10^{-4} units?? m I guess?

Clearly not the same...

[malawi_glenn]

Strange, I got 6.34*10^(-16) m for \Delta x .....

The thing by using excitation energy is that you must know the energy of the ground state, 3055keV is just the delta E, not E of that level.

[genloz]

Okay, so if the base level is 870keV, then the total excitation energy is 3055+870=3925... If we say 17u = 6*mp+8*mn+6*me then we arrive at:
\Delta x = \frac{197 MeV fm}{\sqrt{2*(6*938.8+8*939.7+6*0.511)*3.925}}
\Delta x = \frac{197 MeV fm}{\sqrt{2*(6*938.8MeV+8*939.7MeV+6*0.511MeV)*3.9 25MeV}}
\Delta x = 0.613 fm
Far too small...

[malawi_glenn]

Sounds very strange that the ground energy is smaller than its excitation energy.
But any way, the higher the energy, the smaller deltaX.

[genloz]

Okay, I think I get it.. the mass is only 938.8 or 939.7 because only 1 nucleon is excited, yeah? and the E is the actual value of the excitation, not the difference between excitation and ground state... so that:
\Delta x = \frac{197 MeV fm}{\sqrt{2*938.8*3.055}}
\Delta x = 2.6fm
Which is still much smaller than 3.34... any reason for that?

[malawi_glenn]

HUP gives lower limit.

And also the deltaX should be the smallest cube-length that a particle with momenta (2mE)^½ can be contained in.

If we could get exact numbers with HUP, then we would use it instead of scattering and so on.