Solutions to Midterm Review Problems - Quantum Mechanics | PHYSICS 137A, Exams of Quantum Mechanics

Download Solutions to Midterm Review Problems - Quantum Mechanics | PHYSICS 137A and more Exams Quantum Mechanics in PDF only on Docsity! PHY 137A (D. Budker) Solutions to Midterm Review Problems TA: Uday Varadarajan 1. We will consider a hydrogenic ion with a Carbon-12 core, so Z = 6 and A = 12. (a) Describe the kind of radiation I would need to use to completely ionize this ion in its ground state. Calculate the maximum wavelength of radiation (in nm) which could be used for this purpose and its energy per photon (in eV). Solution: The first Bohr radius in hydrogenic carbon ion can be found by just noting that the only modification to the hydrogen problem from having a carbon core is that we need to modify all terms involving e2 −→ Ze2. In particular, this means that the ionization energy must be modified as −E1 = me 4 2
2 −→ Z2E1 = mZ 2e4 2
2 = 36 · 13.6 eV = 490 eV. (1.1) Thus, we would need to use radiation with wavelength λ = hc E = 1.24× 103 eV · nm 490 eV = 2.53 nm. (1.2) These would correspond to soft X-Rays. (b) Find the energy (in eV), frequency (in Hz), and wavelength (in nm) of the analogue of the Paschen-α line for this hydrogenic ion. Solution: The Paschen-α line is the transition from ni = 4 to nf = 3. Using the results from the previous section, we can easily find the energy by noting that it is: E = −E1 ( 1 32 − 1 42 ) = 490 eV · ( 7 144 ) = 23.8 eV. (1.3) The wavelength is then just dilated by a factor of 1447 , which means that λ = 2.53 nm · 144 7 = 52.0 nm. (1.4) Finally, the frequency can be calculated remembering that ν = E/h, and that 1/h = 2.42 × 1014 Hz · eV−1 so we have ν = 23.8 eV · 2.42× 1014 Hz · eV−1 = 5.76× 1015 Hz (1.5) (c) Calculate the approximate Bohr radius of the electron before and after this transition. Solution: The Bohr radius of the ground state in a hydrogenic ion must be modified as a =
2 me2 −→ a/Z =
2 Zme2 = a0/6 = 8.82× 10−10 cm, (1.6) Now, for an excited state, we can find the answer at least approximately by considering the fact that these orbitals have radial wavefunctions of the form: Rnl(r) = rρl−1e−ρv(ρ) (1.7) where ρ = rZna0 . From the form of this wavefunction, we can immediately see that the expo- nential term ensures that the bulk of the wavefunction is supported in the region with ρ < 1, or r < na. Thus, we expect that the Bohr radius of the n-th orbital, which is the expectation value of r in that state, will be an integral over a function whose support is going to be to largely contained within the region r ≤ na0/Z. This suggests that the Bohr radius should in fact be approximately equal to na0/Z, so we find that the Bohr radius before and after are just: ai = 4a0/6 = 3.53× 10−9 cm (1.8) af = 3a0/6 = 2.65× 10−9 cm. (1.9) 1 (d) What be the change in frequency (in MHz) of this line due to the isotope effect if we had a Carbon-14 ion (with A = 14 and the same Z) instead of a Carbon-12 ion? Solution: The change in frequency can be found simply by finding the difference in effective mass in the two situations, which is the only difference between the two cases. This is just: µ14 − µ12 ≈ me14mp me + 14mp − me12mp me + 12mp ≈ me 2 · me12 · 14mp = me 1 7 9.11× 10−31 g 1.67× 10−27 g = 7.79× 10 −5 · me (1.10) Thus, we get a difference in frequency which is just ∆ν = ∆E h = 1
( µ14Z 2e4 2
2 − µ12Z 2e4 2
2 )( 1 32 − 1 42 ) = µ14 − µ12 meh · ( meZ 2e4 2
2 )( 7 144 ) = 7.79× 10−5 · 23.8 eV h = 7.79× 10−5 · 5.76× 1015 Hz = 4.49× 105 MHz (1.11) (e) Suppose that I now apply various external fields and consider all relativistic (and other - excluding spin!) corrections such that I totally break all the degeneracies in the spectrum of the Carbon-12 Hydrogenic ion, but only slightly. That is, suppose that the correction to the energy of each state in the spectrum is different, but small. Then, if I use a very accurate spectrometer, I would find that the single Paschen-α line has actually split into many different lines. How many lines would I see? Solution: Thus, each of the energies Enlm are now different, though all the energies with the same n are still close to eachother. In this case, each line I would see would be coming from a transition from one the n2i = 4 2 = 16 states with ni = 4 to one of the n2f = 3 2 = 9 states with nf = 3. This would just give me 9 · 16 = 144 lines. (f) Find all transitions that are visible. Solution: In terms of energies of transitions, the visible range correspond to transitions with energies between 1.8 eV and 3.1 eV. To do this problem, we will start by finding two limiting cases. First, we need to find the first α line with a small enough energy to be in the visible range, and then second, the largest nf such that for ni very large, the energy of the transition is large enough to be in the visible spectrum. The smallest alpha line is found finding the smallest nf such that: 3.1 eV ≥ 490 eV ( 1 n2f − 1 (nf + 1)2 ) =⇒ nf ≥ 7 (1.12) (which I did by trial and error). The largest nf such that the transition has a large enough energy to be visible is found by considering the largest value of nf which satisfies (where we take ni → ∞): 1.8 eV ≤ 490 eV ( 1 n2 ) =⇒ nf ≤ 16 (1.13) Thus, we have found all of the allowed values of nf which might lead to light in the visible spectrum. We now simply need to compute, for each value of nf , the range of allowed values of ni which would lead to a transition in the visible spectrum. This can be done with the help of a calculator or spreadsheet. Essentially, given a value of nf , one is looking for all values of ni such that 1.8 490 = 3.673× 10−3 ≤ ( 1 n2f − 1 n2i ) ≤ 3.1 490 = 6.327× 10−3 (1.14) 2