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3063_Exam2_solutions_sp06

3063_Exam2_solutions_sp06 - PHY3063 Spring 2006 R D Field...

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PHY3063 Spring 2006 R. D. Field Exam 2 Solutions Page 1 of 7 March 9, 2006 PHY 3063 Exam 2 Solutions Problem 1 (25 points): Consider an “atom” consisting of a proton and a muon. A proton has an electric charge +e and rest mass energy M p c 2 940 MeV. A muon is an elementary particle with charge –e similar to an electron but with a mass that is 207 times the mass of an electron (m e c 2 0.511 MeV). Use Bohr’s model of an atom to calculate the following: (a) (8 points) The radius of the first Bohr orbit ( i.e. the ground state) of the muonic “atom” ( in fm ). Answer: 284 fm Solution: Bohr’s model tell us that fm A A r m m r m Z r e e e 284 10 84 . 2 ) 529 . 0 ( 186 1 186 1 3 0 0 1 = × = = o o µ , where I used e p e e p e p e m M m m M m M m M m mM 186 / 207 1 1 207 207 207 + = + = + µ and o D A r e 529 . 0 0 = α . (b) (8 points) The ground state energy of the muonic “atom” ( in keV ). Answer: -2.53 keV Solution: Bohr’s model tell us that keV eV eV E m Z E e 53 . 2 530 , 2 ) 6 . 13 ( 186 0 2 1 = = = = µ , where eV c m E e 6 . 13 ) ( 2 1 2 2 0 = = α and 137 1 2 = c Ke h α . (c) (9 points) The ground state energy of the hydrogen atom can be estimated using the uncertainty relation. For a hydrogen atom in the ground state the uncertainty r in the position of the electron is approximately the 1 st Bohr radius, r 0 , consequently the elecrron’s momentum p r is uncertain by about r / h . Compute the ground state energy of the hydrogen atom ( in eV ), r Ke m p E e r 2 2 2 = , by setting r = r 0 and 0 / r p p r r h = = , where r 0 is the Bohr radius of the ground state. Answer: -13.6 eV Solution: We see that eV c m c m c m c m Ke c m m r Ke r m r Ke m p E e e e e e e e e r 6 . 13 ) ( 2 1 ) ( ) ( 2 1 ) /( )) /( ( 2 2 2 2 2 2 2 2 2 2 2 2 2 0 2 2 0 2 2 2 = = = = = α α α α α h h h h
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PHY3063 Spring 2006 R. D. Field Exam 2 Solutions Page 2 of 7 March 9, 2006 Problem 2 (25 points): The parity operator, P op , is defined by ) ( ) ( x x P op = ψ ψ . (a) (10 points) Prove that the parity operator is hermitian. Remember hermitian operators have the property that O op = O op where [ ] = dx x O x dx x x O op op ) ( ) ( ) ( ) ( 1 * 2 1 * 2 ψ ψ ψ ψ . Solution: We see that [ ] +∞ +∞ = dx x x dx x x P op ) ( ) ( ) ( ) ( 1 2 1 * 2 ψ ψ ψ ψ , but +∞ +∞ −∞ + +∞ = = = dx x x dy y y dy y y dx x x ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 1 2 1 2 1 2 1 2 ψ ψ ψ ψ ψ ψ ψ ψ , where I changed variables with y = -x. Thus, [ ] +∞ +∞ +∞ +∞ = = = dx x P x dx x x dx x x dx x x P op op ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 1 2 1 2 1 2 1 * 2 ψ ψ ψ ψ ψ ψ ψ ψ and hence P op = P op which means that P op is hermitian. (b) (5 points) Prove that the eigenvalues of the parity operator are real.
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