Problem 1:
An electron in the hydrogen atom occupies the combined position and spin state
R_{21}(r)[√⅓ Y_{10}(θ,φ)χ_{+} + √⅔ Y_{11}(θ,φ)χ_{-}].
(a) If you measure L^{2}, what value(s) might you get, and with what
probability(ies)?
(b) If you measure L_{z}, what value(s) might you get, and with what
probability(ies)?
(c) If you measure S^{2}, what value(s) might you get, and with what
probability(ies)?
(d) If you measure S_{z}, what value(s) might you get, and with what
probability(ies)?
(e) If you measured the position of the electron, what is the probability
density for finding the electron at r, θ, φ in terms of the variables given
above.
(f) If you measured both S_{z} and the distance of the electron from
the proton, what is the probability per unit length for finding the particle
with spin up a distance r from the proton in terms of the variables given above?
Useful integral: ∫_{0}^{π}sinθ dθ∫_{0}^{2π}dφ |Y_{lm}(θ,φ))|^{2} = 1.
Solution:
Concepts: The eigenfunctions of the orbital and spin angular momentum operators, the spherical harmonics, tensor product states | |
Reasoning: The common eigenfunctions of L^{2} and L_{z} are the spherical harmonics. | |
Details of the calculation: (a) L^{2} = 2ħ^{2}, probability 1. (b) L_{z} = 0, probability ⅓, L_{z} = ħ, probability ⅔. (c) S^{2} = ¾ħ^{2}, probability 1. (d) S_{z} = ħ/2, probability ⅓, S_{z} = -ħ/, probability ⅔. (e) Let P(r,θ,φ) the probability per unit volume of for finding the electron at r, θ, φ. P(r,θ,φ) = <ψ|P_{r,θ,φ} |ψ>, where the projector P_{r,θ,φ} = ∑_{i}(|r,θ,φ> |χ_{i}>)(<r,θ,φ| <χ_{i}|), and i = +, -. P(r,θ,φ) = (√⅓<210| <χ_{+}| + √⅔<211| <χ_{-}|) (|r,θ,φ> |χ_{+}>)(<r,θ,φ| <χ_{+}|)(√⅓|210> |χ_{+}> + √⅔|211> |χ_{-}>) + (√⅓<210| <χ_{+}| + √⅔<211| <χ_{-}|) (|r,θ,φ> |χ_{-}>)(<r,θ,φ| <χ_{-}|)(√⅓|210> |χ_{+}> + √⅔|211> |χ_{-}>) = ⅓|<210| r,θ,φ>|^{2} + ⅔|<211| r,θ,φ>|^{2}, since <χ_{i}|χ_{j}|> = δ_{ij}. P(r,θ,φ) = |R_{2l}(r)|^{2}[⅓|Y_{1}^{0}(θ,φ) |^{2} + ⅔|Y_{1}^{1}(θ,φ) |^{2}]. (f) P_{+}(r,θ,φ) = ⅓|R_{2l}(r)|^{2}|Y_{1}^{0}(θ,φ) |^{2}, Probability per unit length P_{+}(r) = ⅓r^{2}|R_{2l}(r)|^{2}. |
Problem 2:
Is each of the following sets a valid combination of quantum numbers (n, ℓ, m_{ℓ},
m_{s}) for the energy eigenstates of hydrogen? If not, explain why not.
(2, 2, -1, ½)
(3, 1, +2, -½)
(3, 1, 0, ½)
(4, 1, 1, -3/2)
(2, -1, 1, +1/2)
Solution:
Concepts: Addition of angular momentum | |
Reasoning: We are supposed to add the orbital and spin angular momentum of the electron in the hydrogen atom and check if the given combinations are valid. | |
Details of the calculation: The combinations below are not valid. (2, 2,-1, ½) The quantum number ℓ is non-negative and smaller than the quantum number n (3, 1, +2, -½) The quantum number m_{ℓ} can take on all integer values between -ℓ and ℓ. (4, 1, 1, -3/2) The quantum number m_{s} for the spin ½ electron is restricted to the values ±½. (2, -1, 1, +1/2) The quantum number ℓ is non-negative. |
Problem 3:
Find the condition that must be satisfied by the spherically symmetric square
well potential
U(r) = -|U_{0}| for r < a, U(r) = 0 for r >
a,
if it is just barely deep
enough to contain one bound state.
Solution:
Concepts: Three-dimensional square potentials | |
Reasoning: We have to investigate the properties of the ground state in a three dimensional square potential. | |
Details of the calculation: If the potential only supports one bound state, this state must be an l = 0 state. The radial Schroedinger equation then is [(-ħ^{2}/(2μ))(∂^{2}/∂r^{2})+ U(r)]u_{10}(r) = E_{10}u_{10}(r). Let k^{2} = (2μ/ħ^{2})(E + U_{0}), ρ^{2} = (2μ/ħ^{2})(-E), and k_{0}^{2} = (2μ/ħ^{2})U_{0}. Let r < a define region 1 and r > a define region 2. The coordinate r is never negative. Therefore Φ_{1}(r) = A sin(kr), Φ_{1}(0) = 0, Φ_{2}(r) = Bexp(-ρr), Φ_{2}(∞) = 0. At r = a we need hat Φ(a) and (∂/∂r)Φ(r)|_{a} are continuous. A sin(ka) = Bexp(-ρa). kA cos(ka) = -ρBexp(-ρa). Therefore cot(ka) = -ρ/k. 1/sin^{2}(ka) = 1 + cot^{2}(ka) = (k^{2} + ρ^{2})/k^{2} = k_{0}^{2}/k^{2}. We can find a graphical solution by plotting |sin(ka)| and k/k_{0} versus k. The intersections of the two plots in regions where cot(ka) < 0 gives the values of k for which a solution exist. We need nπ/2 < ka < (n + 1)π/2, n = odd. To only support one bound state we need π/2 < k_{0}a < 3π/2. With E near zero we have k ~ k_{0 }and π/2 ~(2μU_{0}/ħ^{2})^{1/2}a. The smallest possible value for U_{0} is U_{0} = π^{2}ħ^{2}/(8μa^{2}). |
Problem 4:
An exotic atom consists of a Helium nucleus (Z = 2) and an electron and an
antiproton p(bar) both in n = 2 states. Take the mass of the p(bar) to be 2000
electron masses and that of the helium nucleus to be 8000 m_{e}. For an
electron in the n = 1 state of hydrogen E = -13.6 eV.
(a) How much energy is required to remove the electron from this atom?
(b) How much energy is required to remove the p(bar) from this atom?
(c) Assume both the p(bar) and the electron are in 2p states. Then each can
de-excite to their ground state. It is observed that radiation always
accompanies those transitions when the electron jumps first, but when the p(bar)
jumps first there is often no photon emitted. Explain!
Solution:
Concepts: Hydrogenic atoms | |
Reasoning: To find the eigenfunctions and eigenvalues of the Hamiltonian of a hydrogenic atom we replace in the eigenfunctions of the Hamiltonian of the hydrogen atom a_{0} by a_{0}' = ħ^{2}/(μ'Ze^{2}) = a_{0}(μ/μ')(1/Z), and in the eigenvalues of the Hamiltonian of the hydrogen atom we replace E_{I} by E_{I}' = μ'Z^{2}e^{4}/(2ħ^{2}) = E_{I}(μ'/μ)Z^{2}. Here μ is the reduced mass for the hydrogen atom. | |
Details of the calculation: |
Problem 5:
The Rydberg constant, R_{H} = 109737.568525/cm is
one of the most accurately known fundamental constants.
(a) Find the wave number of the Balmer alpha line (n = 3 to n' = 2) in atomic
hydrogen. Neglect fine structure.
(b) Is the Balmer alpha line in atomic deuterium shifted towards the blue or
towards the red compared to normal hydrogen?
(c) Calculate the shift in wave number between
deuterium and hydrogen.
Solution:
Concepts: The hydrogen atoms, reduced mass | |
Reasoning: The deuterium is a hydrogen atom with a slightly different reduced mass than ordinary hydrogen. | |
Details of the calculation: (a) For ordinary hydrogen: E_{n} = -μe^{4}/(2ħ^{2}n^{2}) with μ = m_{p}m_{e}/(m_{p} + m_{e}) E_{n} - E_{n'} = [μe^{4}/(2ħ^{2})](1/n'^{2} - 1/n^{2}) = hν = hc/λ = hcR_{H}(1/n'^{2} - 1/n^{2}). So R_{H} = [2πμe^{4}/(2ħ^{3}c)]. For the wave number we have 1/λ = R_{H}(1/3^{2} - ½^{2}) = 15241.328952 cm^{-1}. [Note: There are unfortunately two different definitions of the wave number, 1/λ or 2π/λ. Clearly state which definition you are using.] (d) For deuterium: E_{n} = -μ'e^{4}/(2ħ^{2}n^{2}) with μ' = (m_{p} + m_{n})m_{e}/(m_{p} + m_{n} + m_{e}) R_{D} = R_{H} μ'/μ > R_{H}. So 1/λ increases, λ decreases, the line is blue-shifted. (c) 1/λ' - 1/λ = (R_{D} – R_{H})(1/3^{2} - ½^{2}) = (μ'/μ – 1)R_{H}(1/3^{2} - ½^{2}) = -4.15 cm^{-1}. |