1201_Physics ProblemsTechnical Physics

# 1201_Physics ProblemsTechnical Physics - 542 P42.72 Atomic...

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

542 Atomic Physics P42.72 ψπ 25 12 0 32 0 2 0 2 1 4 2 1 22 00 = F H G I K J F H G I K J =− F H G I K J −− a f a r a eA r a e ra d dr Ae a r a ψ + F H G I K J 2 0 0 2 0 2 2 d dr Ae a r a 2 2 2 0 2 0 0 3 24 = F H G I K J F H G I K J Substituting into Schrödinger’s equation and dividing by Ae 2 0 , we will have a solution if −+ + + = 5 4 8 2 2 2 2 0 2 2 0 2 0 3 2 0 2 0 == = ma ke a r mar r E Er a e e e e e . Now with a mek ee 0 2 2 = = , this reduces to F H G I K J F H G I K J r a E r a 42 2 8 = . This is true, so 25 is a solution to the Schrödinger equation, provided EE 1 4 340 1 . e V . P42.73 (a) Suppose the atoms move in the + x direction. The absorption of a photon by an atom is a completely inelastic collision, described by mv h mv if ±± ± ii i +−= λ ej so vv h m fi −= . This happens promptly every time an atom has fallen back into the ground state, so it happens every 10 8 = s t . Then, a t h mt = ×⋅ × ∆∆ ~ . ~ 6626 10 500 10
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online