s11f1019

# s11f1019 - Solution for assignment 19: Eﬀective...

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

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