Problem Set 3_Sol

# Problem Set 3_Sol - Answer Key Side 1 of 4 Problem One...

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Answer Key Ch 1a, Problem Set Three Side 1 of 4 5 Points 5 Points 5 Points 5 Points 5 Points Problem One – Waves – 25 Points, 5 Each Nodes are the points where the wavefunction amplitude crosses zero at all times (not including our boundary conditions). Therefore: a. Final Answer: 2 Nodes b. Final Answer: 3 Nodes c. The correspondence principle states that the more nodes a wavefunction has, the higher in energy it is (for identical sytems, of course). Therefore: Final Answer: B has more nodes, therefore B is higher in energy. d. For any wave of the system to have higher energy than B, it must have more nodes then B. Therefore the minimum number of nodes that wave C may have is: Final Answer: 4 Nodes e. The principle quantum number, n , is related to the number of nodes through a simple relation: n = number of nodes + 1 Therefore, wave A with 2 nodes has n =3, and wave B with 3 nodes has n =4. Final Answer: n A =3, n B =4

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Answer Key Ch 1a, Problem Set Three Side 2 of 4 2 Points 8 Points 3 Points 2 Points Problem Two – Heisenberg Uncertainty principle – 15 Points We cannot change the mass of the proton, so the uncertainty in the momemtum is wholly in the velocity component: " p x " x # h 4 \$ " p x = " mv x ( )= m " v x + v x " m = m " v x with Δ m going to zero. The uncertainty in velocity (
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## This note was uploaded on 03/30/2010 for the course CH 1a taught by Professor Lewis during the Fall '08 term at Caltech.

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Problem Set 3_Sol - Answer Key Side 1 of 4 Problem One...

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