s11f1019

s11f1019 - Solution for assignment 19: Effective...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution for assignment 19: Effective Potential. Ueff (r) = − α M2 + r 2mr2 with M angular momentum and m reduced mass. (A) Solve now for r = r0 : Ueff (r) = + 2M 2 M2 α M2 − ⇒ r0 = . = 0 ⇒ α r0 = r 2mr3 m αm (B) The minimum value of the effective potential: min Ueff = Ueff (r0 ) = − α2 m α2 m α2 m + =− . M2 2M 2 2M 2 (C) Calculation of the turning points for E < 0: We are looking of solutions of the quadratic equation M2 =0 2m E r2 + α r − α M2 , q=− . E 2m E M2 α2 + , 2E 2mE r2 + p r − q = 0 with p = p rmin,max = − ± 2 p 2 2 −q = − α ± 2E (D) When do we have real solutions? The upper bound is E < 0, the lower follows from α 2E 2 + M2 α2 M2 α2 m = 0⇒ + =0⇒E=− . 2mE 4E 2m 2M 2 So, we find − α2 m <E<0 2M 2 and the lower value is the minimum of the effective potential as calculated under (B). 1 ...
View Full Document

This note was uploaded on 12/11/2011 for the course PHY 4936 taught by Professor Berg during the Fall '11 term at FSU.

Ask a homework question - tutors are online