Hand in 9 Solutions(07)

# Hand in 9 Solutions(07) - Solutions for Hand-In Set#9 PHYS...

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Unformatted text preview: Solutions for Hand-In Set #9 PHYS 212 – Spring 2007 A62 — Uncertainty The uncertainty in the x position is given in the prob- lem statement: Δ x = 10- 12 m. This implies a minimum uncertainty in the x component of the momentum given by the Heisenberg Uncertainty Principle: Δ x Δ p ≥ ¯ h 2 ⇒ Δ p ≥ ¯ h 2Δ x . The mass is certain, so this can be rewritten as m Δ v ≥ ¯ h 2Δ x , or Δ v ≥ ¯ h 2 m Δ x ≥ 1 . 06 × 10- 34 J · s 2 × . 04 kg × 10- 12 m ≥ 1 . 3 × 10- 21 m/s A65 — Schr¨ odinger equation for classically forbidden situation (a) Rearranging the Schr¨ odinger equation gives d 2 ψ ( x ) dx 2 =- 2 m ¯ h 2 ( E- U ) ψ ( x ) . (b) For the first trial solution we have ψ 1 ( x ) = Ax 2 dψ 1 dx = 2 Ax d 2 ψ 1 dx 2 = 2 A. Plugging these into the rearranged Schr¨ odinger equation gives 2 A ? =- 2 m ¯ h 2 ( E- U ) Ax 2 . where I have used the symbol ? = to emphasize that were are tying to see if the function ψ 1 “works,” i.e., makes the equality true for all values of x . In this case ψ 1 is not a soluton. The lefthand side is a positive constant, and the righthand side is quadratic in x ; these clearly can’t be equal for all x . The function ψ 1 ( x ) is not a solution. (c) For the second trial solution we have ψ 2 ( x ) = B sin kx dψ 2 dx = Bk cos kx d 2 ψ 1 dx 2 =- Bk 2 sin kx....
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## This homework help was uploaded on 04/07/2008 for the course PHYS 212 taught by Professor Ladd during the Spring '08 term at Bucknell.

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Hand in 9 Solutions(07) - Solutions for Hand-In Set#9 PHYS...

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