Quantum Engi Q and A_Part_16

# Quantum Engi Q and A_Part_16 - 46 CHAPTER 4 SINGLE-PARTICLE...

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46 CHAPTER 4. SINGLE-PARTICLE SYSTEMS Answer: The square wave function is | ψ 100 ( r ) | 2 = 1 πa 3 0 e 2 r/a 0 and the value at the nucleus r = 0 is then | ψ 100 (0) | 2 = 1 πa 3 0 For the value at r above to be one percent of this, you must have e 2 r/a 0 = 0 . 01 or taking logarithm, r = 2 . 3 a 0 . Expressed in ˚ A, a 0 = 0 . 53 ˚ A, so r = 1 . 22 ˚ A. 4.2.4.2 Solution hydd-b Question: Check from the conditions n > l ≥ | m | that ψ 200 , ψ 211 , ψ 210 , and ψ 21 1 are the only states of the form ψ nlm that have energy E 2 . (Of course, all their combinations, like 2p x and 2p y , have energy E 2 too, but they are not simply of the form ψ nlm , but combinations of the “basic” solutions ψ 200 , ψ 211 , ψ 210 , and ψ 21 1 .) Answer: Since the energy is given to be E n = E 2 , you have n = 2. The azimuthal quantum number l must be a smaller nonnegative integer, so it can only be 0 or 1. In case l = 0, the absolute value of the magnetic quantum number m cannot be more than zero, allowing only

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## This note was uploaded on 11/13/2011 for the course PHY 4458 taught by Professor Garvin during the Fall '11 term at University of Florida.

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Quantum Engi Q and A_Part_16 - 46 CHAPTER 4 SINGLE-PARTICLE...

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