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Phys 197 Homework Solution 41A

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Phys 197 Homework Solution 41A Phys 197 Homework Solution 41A Q3. For a body orbiting the sun, such as a planet, comet, or asteroid, is there any restriction on the z-component of its orbital angular momentum such as there is with the z-component of the electron’s orbital angular momentum in hydrogen? Explain. ————– Yes. It must be a multiple ofh ¯. To give this some perspective, consider a 1-kg rock sharing Earth’s orbit (R = 1.5 109 m, v = 3 104 m/s). L = mvr = (1 kg)(3 104 m/s)(1.5 109 m) = 4.5 1013 kg m2/s. To× get an ℓ value,× divide this byh ¯: × × ℓ×= L/h¯ = (4.5 1013/(1.05 10−34) = 4.3 1047. Since the radius of× the orbit scales× as ℓ2, the ratio× of orbital radii is (ℓ + 1)2/ℓ2 =1+2/ℓ + 1/ℓ2. Neglect the 1/ℓ2 to get the fractional change as 2/ℓ = 4.6 10−48. This means that the next allowed orbit is farther out by a distance of (4.6 10−48)(1.×5 109 m)=7 10−39 m. × × × Q4. Why is the analysis of the helium atom much more complex than that of the hydrogen atom, either in a Bohr type of model or using the Schr¨odinger equation? ————– In calculating allowed orbits of the “second” electron, the electrostatic influence of the first electron must be included. If the first electron is put into the lowest Schr¨odinger solution, that is only a start, because the second electron affects the first, so that the first electron orbit (state) is perturbed and must be re-claculated. Now, the wavefunction for the second must be recalculated with the perturbation. Which further perturbs the first .... Q7. In the Stern-Gerlach experiment, why is it essential for the magnetic field to be inhomoge- neous (that is, nonuniform)? ————– A homogeneous field only precesses the angular momentum and magnetic moment vectors. The field gradient (∂Bz/∂z) exerts a force on the dipole. Q11. Table 41.3 shows that for the ground state of the potassium atom, the outermost electron is in a 4s state. What does this tell you about the relative energies of the 3d and 4s levels for this atom? Explain. ————– Electrons fill in order of energy. Therefore, the 4s shell is lower than the 3d shell when the 1s, 2s, 2p, 3s, and 3p are filled. If all electrons were are stripped away and one given back, it would find the 3d to be lower in energy that the 4s. The presence of the lower electrons modifies this relationship. Q16. A small amount of magnetic-field splitting of spectral lines occurs even when the atoms are not in a magnetic field. What causes this? ————– The spin-orbit coupling. One way to look at it is that the orbit creates a magnetic field, and the electron spin interacts with this field. Q23. Can a hydrogen atom emit x rays? If so, how? If not, why not? ————– No Way. The highest energy photon it can emit is Eph = 13.6 eV λ = (1240 eV nm)/(13.6 eV) = 91 nm. This is still ultraviolet by anyone’s band division.⇒ · P8. An electron is in the hydrogen atom with n = 5 (a) Find the possible values of L and Lz for this electron, in units ofh ¯. (b) For each value of L, find all the possible angles between L~ and the z-axis. (c) What are the maximum and minimum values of the magnitude of the angle between and the z-axis? ————– For n = 5, ℓ may take on the values 0, 1, 2, 3, and 4. L (in units ofh ¯) is calculated by L = pℓ(ℓ + 1) Lz in units ofh ¯ is the same as ml which may take on the values ℓ, ℓ + 1, ... ℓ 1, ℓ. −1 − − − The angle is calculated as Cos (Lz/L). (a), (b), and (c) combined ℓL Lz angle max/min 0 0.0000 0 indeterminate N/A 1 1.4142 -1 135.0 max for ℓ = 1 1 1.4142 0 90.0 1 1.4142 1 45.0 min for ℓ = 1 2 2.4495 -2 144.7 max for ℓ = 2 2 2.4495 -1 114.1 2 2.4495 0 90.0 2 2.4495 1 65.9 2 2.4495 2 35.3 min for ℓ = 2 3 3.4641 -3 150.0 max for ℓ = 3 3 3.4641 -2 125.3 3 3.4641 -1 106.8 3 3.4641 0 90.0 3 3.4641 1 73.2 3 3.4641 2 54.7 3 3.4641 3 30.0 min for ℓ = 3 4 4.4721 -4 153.4 max for ℓ = 4 4 4.4721 -3 132.1 4 4.4721 -2 116.6 4 4.4721 -1 102.9 4 4.4721 0 90.0 4 4.4721 1 77.1 4 4.4721 2 63.4 4 4.4721 3 47.9 4 4.4721 4 26.6 min for ℓ = 4 P9. The orbital angular momentum of an electron has a magnitude of 4.716 10−34 kgm2/s. × What is the angular-momentum quantum number for this electron? ————– Calculate how many timesh ¯ this L is; suppress the common factor of 10−34. 4.716/1.054 = 4.474. This is close enough to √20 = 4.472 to say that ℓ = 4 (since L = p4(4+1) = √20). P19. A hydrogen atom in the 5g state is placed in a magnetic field of 0.600 T that is in the z-direction. (a) Into how many levels is this state split by the interaction of the atom’s orbital magnetic dipole moment with the magnetic field? (b) What is the energy separation between adjacent levels? (c) What is the energy separation between the level of lowest energy and the level of highest energy? ————– (a) Recall that the g sublevel corresponds to ℓ = 4. The magnetic quantum number mℓ takes on integer values from -4 to +4, so the splitting is into 9 levels. (b) To be definite, take the splitting between mℓ = 1 and mℓ = 0. Using Eq 41.36: −24 −24 −5 ∆U = (1 0)µBB = (9.27 10 J/T)(0.6 T) = 5.56 10 J= 3.47 10 eV. (c) Since there− are 9 levels, the× difference from highest to× lowest is 8 times× the value above, or 2.78 10−4 eV. You should× take a look at the units, including the fact that the tesla can be written as 1 T = kg/(C s). · P21. Classical Electron Spin. (a) If you treat an electron as a classical spherical object with a radius of 1.0 10−17 m, what angular speed is necessary to produce a spin angular momentum of × magnitude p3/4 ¯h? (b) Use v = rω and the result of part (a) to calculate the speed v of a point at the electron’s equator. What does your result suggest about the validity of this model? ————– Take the electron’s moment of inertia as I = (2/5)mr2. Then Lspin = Iω = p3/4h ¯ ⇒ ω = p3/4h/ ¯ [(2/5)mr2] and v = rω = p3/4h/ ¯ [(2/5)mr. (a) ω = (0.866)(1.05 10−34/[0.4 (9.11 10−31 kg)(10−17 m)2] = 2.5 1030 rad/s. (b) v = 2.5 1013 m/×s = 105 c. Warp-warp-warp.· × Einstein disapproves.× My suggestion× for the classical electron radius reduces the speed to a mere 9 1010 m/s = 300 c. Einstein’s frown didn’t change much. × The spin of the electron is simply some intrinsic property it has, which isn’t explainable with any reasonable classical parameters..
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