{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1155_Physics ProblemsTechnical Physics

1155_Physics ProblemsTechnical Physics - 496*P41.12 Quantum...

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

View Full Document Right Arrow Icon
496 Quantum Mechanics *P41.12 (a) The energies of the confined electron are E h m L n n e = 2 2 2 8 . Its energy gain in the quantum jump from state 1 to state 4 is h m L e 2 2 2 2 8 4 1 e j and this is the photon energy: h m L hf hc e 2 2 15 8 = = λ . Then 8 15 2 m cL h e = λ and L h m c e = F H G I K J 15 8 1 2 λ . (b) Let λ represent the wavelength of the photon emitted: hc h m L h m L h m L e e e = = λ 2 2 2 2 2 2 2 2 8 4 8 2 12 8 . Then hc hc h m L m L h e e λ λ = = 2 2 2 2 15 8 8 12 5 4 e j and ′ = λ λ 1 25 . . Section 41.3 The Particle Under Boundary Conditions Section 41.4 The Schrödinger Equation P41.13 We have ψ ω = Ae i kx t b g and = − 2 2 2 ψ ψ x k . Schrödinger’s equation: = − = − 2 2 2 2 2 ψ ψ ψ x k m E U = a f . Since k p h p 2 2 2 2 2 2 2 2 2 = = = π λ π a f b g = and E U p m = 2 2 . Thus this equation balances. P41.14 ψ x A kx B kx a f = + cos sin = − + ψ x kA kx kB kx sin cos = − 2 2 2 2 ψ x k A kx k B kx cos sin = − + 2 2 2 m E U mE A kx B kx = = a f a f ψ cos sin Therefore the Schrödinger equation is satisfied if
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}