Homework 6 Answer Key November 4, 2014 1 Q7.7 Why can the angular momentum vector lie on the z axis for two-dimensional rotation in the xy plane but not for rotation in three-dimensional space? For rotation in two dimensions, the angular momentum vector has only one component, and the vector is perpendicular to the plane of rotation. For rotation in three dimensions, the angular momentum vector has three components. Because the corresponding angular momentum operators do not commute, all three components of the angular momentum can't be known simultaneously. Therefore, the angular momentum vector can't lie on the z axis, because this requires that the x and y components are known to be zero, and the z component is known through the value of l2. 2 Q7.20 2 2 For a two-dimensional harmonic oscillator, V (x; y) = kxx + kyy write an expression for the energy levels of such an oscillator in terms of kx and ky. 1 1 E = n + hν + n + hν x 2 x y 2 y s s h kx 1 h ky 1 = nx + + ny + 2π µx 2 2π µy 2 3 P 7.5 1H19F has a force constant of 966 N m−1 and a bond length of 91.68 pm. Calculate the frequency of the light corresponding to the lowest energy pure vibrational and pure rotational transitions. In what regions of the EM spectrum do they lie? v s u −2 1 k 1 u 966 kg s ν = = t −21 2π µ 2π 1:008 amu×18:9984 amu 1:661×10 kg 1:008 amu+18:9984 amu × amu = 1:24 × 1014 s−1 The frequency lies in the infrared region of the EM spectrum. The lowest energy transition is J = 0 ! J = 1. 1:008 amu × 18:9984 amu 1:661 × 10−21 kg I = µr2 = × × (91:68 × 10−12m)2 0 1:008 amu + 18:9984 amu amu = 1:34 × 10−47 kg m2 ¯h2 ¯h2 (1:055 × 10−34 J s)2 ∆E = 2(J + 1) = = = 8:33 × 10−22J rot 2 2 −47 2 2=mur0 µr0 1:34 × 10 kg m ∆E ot ν = r = 1:25 × 1012 s−1 h This frequency lies in the microwave region of the EM spectrum. 1 4 P7.10 Show by carrying out the necessary integration that the eigenfunctions of the Schrodinger¨ equation for rotation in two dimensions, p1 eimlφ and p1 einlφ, m 6= n are orthogonal. 2π 2π l l 1 2 Z 2π 1 Z 2π imlφ inlφ p e e dφ = (cos(ml − nl)φ + i sin(ml − nl)φ) dφ 2π 0 2π 0 1 ([sin 2π(ml − nl) − 0] − i[cos 2π(ml − nl) − 1]) = 0 2π(ml − nl) because (ml − nl) is an integer. 5 P7.30 By substituting in the Schrodinger¨ equation for rotation in three dimensions, show that the rotational wave function (5=16π)1=2(3 cos2 θ − 1) is an eigenfunction of the total energy operator. Determine the energy eigenvalue. ¯h2 1 δ δY (θ; φ) 1 δ2Y (θ; φ) − 2 sin θ + 2 2 = EY (θ; φ) 2µr0 sin θ δθ δθ sin θ δφ 2 0 n 1=2 o1 n 1=2 o3 2 5 2 2 5 2 ¯h 1 δ δ 16π (3 cos θ − 1) 1 δ 16π (3 cos θ − 1) − 2 4 @sin θ A + 2 2 5 2µr0 sin θ δθ δθ sin θ δφ 2 1=2 ¯h 5 1 δ 2 = − 2 ((−6 cos θ sin θ)) 2µr0 16π sin θ δθ 2 1=2 ¯h 5 1 3 2 = − 2 (6 sin θ − 12 cos θ sin θ) 2µr0 16π sin θ 2 1=2 ¯h 5 2 2 = − 2 (6f1 − cos g − 12 cos θ) 2µr0 16π 2 1=2 2 1=2 ¯h 5 2 6¯h 5 2 = − 2 (6 − 18 cos θ) = − 2 (3 cos θ − 1) 2µr0 16π 2µr0 16π 3¯h2 h¯2 The energy eigenvalue is 2 , corresponding to El = l(l + 1) with l = 2. µr0 2I 2.