Angular momentum & Energy using Yukawa's potential

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Nigsia
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Hello there!
I was doing my Gravitation problems and I found this problem that I'm unable to solve.

Yukawa's theory for nuclear forces states that the potential energy corresponding to the attraction force produced by a proton and a neutron is:
[tex]U(r) = \frac{k}{r}e^{-\alpha r},\ k<0,\ \alpha > 0[/tex]
From the expression of it's effective potential, find the module of it's angular momentum and it's energy, for which it's possible a circular movement with a radius r0

I've tried several things, none of them leading to something meaningful. In fact, I know that expression for effective potential is:
[tex]U_{ef}(r)=U(r)+\frac{L}{2r^2}[/tex]
So I imagine I would need to find L fist in order to get the expression for Uef, but I'm not able to remember nor find any kind of formula linking U and L. Would you please help me out?

PS: Once I know how to find L I know how to end it, since:
[tex]\frac{dU_{ef}}{dr} = 0 \Leftrightarrow r = r_0[/tex]
is the expression of the energy of a circular movement with a radius r0

Thanks in advance.
 
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Okay, I have it differentiated. How do I find angular momentum? As far as I know, I only find the energy with what I've done.
 
Do you mean that the Potential is twice the kinetic?
 
I don't really know that formula. The only thing that I can think of is [tex]E=\frac{U}{2}[/tex]
 
I really don't know. Can you give me a hint?
 
The potential is twice the kinetic, so [tex]\left| E\right| = \left| \frac{U}{2} \right| = \left| K \right|[/tex]
 
Nigsia said:
The potential is twice the kinetic, so [tex]\left| E\right| = \left| \frac{U}{2} \right| = \left| K \right|[/tex]
No. This is simply wrong. You have to be aware when certain theorems hold and when they do not. The answer is much simpler and does not require anything else than very basic mechanics.
 
Could you please give me another hint? I'm really struggling to get anything clear.
 
In theory [tex] L=(-mkr_0(1+\alpha r_0)e^{-\alpha r_0})^{1/2}[/tex]