This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY4604 Fall 2007 R. D. Field Department of Physics Page 1 of 8 Exam 2 Solutions PHY 4604 Exam 2 Solutions (Total Points = 100) Problem 1 (20 points): Circle the correct answer for following (4 points each). (1a) An electron is in a onedimensional infinite square well with zero potential energy in the interior and infinite potential energy at the walls. A graph of its wave function ψ (x) versus x is shown. What is the value of the quantum number n ? ( circle the correct answer ) (a) 0 (b) 2 (c) 4 (d) 6 (e) 8 (1b) An atom is in a state with orbital quantum number l = 2 . What are the possible values of the magnetic quantum number m . ( circle the correct answer ) (a) 1, 2 (b) 0, 1, 2 (c) 0, 1 (d) 1, 0, 1 (e) 2, 1, 0, 1, 2 (1c) An electron is in a quantum state for which the magnitude of the orbital momentum is h 10 3 . How many allowed values of the zcomponent of the angular momentum are there? ( circle the correct answer ) (a) 9 (b) 10 (c) 17 (d) 18 (e) 19 (1d) An electron is in a quantum state for which there are six allowed values of the zcomponent of the angular momentum. What is the magnitude of the angular momentum vector? ( circle the correct answer ) (a) h 6 (b) 2 / 5 h (c) h 6 (d) h 10 3 (e) 2 / 35 h (1e) What is 2 × 2 × 2 × 2 in SU(2)? ( circle the correct answer ) (a) 5 + 3+ 3+ 3 + 1 +1 (b) 16 (c) 9 + 5 + 2 (d) 6 + 5 + 3 + 2 (e) 10 + 6 Problem 2 (40 points): Consider a spin ½ system described by the Hamiltonian: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = 1 1 ε ε ε ε i i H where ε and ε 1 are real positive constants. (a) (6 points) Find the energy levels of the system. How many energy levels are there? What is the ground state energy, E , and the first excited state energy, E 1 ? Answer: There are two energy levels: E = ε – ε 1 , E 1 = ε + ε 1 . Solution: The energy levels are the solution of 1 1 = − − − λ ε ε ε λ ε i i PHY4604 Fall 2007 R. D. Field Department of Physics Page 2 of 8 Exam 2 Solutions which yields ) ( 2 1 2 = − − ε λ ε , which implies that 1 ε λ ε ± = − and hence 1 ε ε λ ± = . There are two energy levels, E = ε – ε 1 and E 1 = ε + ε 1 . (b) (6 points) What are the ( normalized ) eigenkets corresponding to the ground state,  E > , and the first excited state , E 1 > ? Answer: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − >= 1 2 1  i E ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ >= 1 2 1  1 i E . Solution: For the ground state we see that ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − b a b a i b i a b a i i ) ( 1 1 1 1 1 ε ε ε ε ε ε ε ε ε ε and hence ε a +i ε 1 b = ( ε ε 1 )a, which implies that a = ib and hence if b = 1 then a = i and ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − >= 1 2 1  i E ....
View
Full
Document
This note was uploaded on 05/29/2011 for the course PHY 4064 taught by Professor Fields during the Spring '07 term at University of Florida.
 Spring '07
 FIELDS
 mechanics, Energy, Potential Energy

Click to edit the document details