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Unformatted text preview: CHAPTER 7 1. (a) The system under consideration has rotational degrees of freedom, allowing it to rotate about two orthogonal axes perpendicular to the rigid rod connecting the two masses. If we define the z axis as represented by the rod, then the Hamiltonian has the form H = L x 2 + L y 2 2 I = L 2 L z 2 2 I where I is the moment of inertia of the dumbbell. (b) Since there are no rotations about the z axis, the eigenvalue of L z is zero, so that the eigenvalues of the Hamiltonian are E = h 2 l ( l + 1) 2 I with l = 0,1,2,3,… (c) To get the energy spectrum we need an expression for the moment of inertia. We use the fact that I = M red a 2 where the reduced mass is given by M red = M C M N M C + M N = 12 × 14 26 M nucleon = 6.46 M nucleon If we express the separation a in Angstroms, we get I = 6.46 × (1.67 × 10 27 kg )(10 10 m / A ) 2 a A 2 = 1.08 × 10 46 a A 2 The energy difference between the ground state and the first excited state is 2 h 2 / 2 I which leads to the numerical result ∆ E = (1.05 × 10 34 J . s ) 2 1.08 × 10 46 a A 2 kg . m 2 × 1 (1.6 × 10 19 J / eV ) = 6.4 × 10 4 a A 2 eV 2. We use the connection x r = sin θ cos φ ; y r = sin θ sin φ ; z r = cos θ to write Y 1,1 =  3 8 π e i φ sin θ =  3 8 π ( x + iy r ) Y 1,0 = 3 4 π cos θ = 3 4 π ( z r ) Y 1, 1 = ( 1) Y 1,1 * = 3 8 π e i φ sin θ = 3 8 π ( x iy r ) Next we have Y 2,2 = 15 32 π e 2 i φ sin 2 θ = 15 32 π (cos 2 φ + i sin2 φ )sin 2 θ = 15 32 π (cos 2 φ sin 2 φ + 2 i sin φ cos φ )sin 2 θ = 15 32 π x 2 y 2 + 2 ixy r 2 Y 2,1 =  15 8 π e i φ sin θ cos θ =  15 8 π ( x + iy ) z r 2 and Y 2,0 = 5 16 π (3cos 2 θ 1) = 5 16 π 2 z 2 x 2 y 2 r 2 We may use Eq. (746) to obtain the form for We may use Eq....
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This note was uploaded on 03/09/2008 for the course PHYS 401 taught by Professor Mokhtari during the Spring '05 term at UCLA.
 Spring '05
 mokhtari
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