Use the measured ionization energy of Helium (E_{ionization} = 24.6
eV) to calculate the energy of interaction of the two electrons in the ground
state of the atom.

Solution:

- Concepts:

Hydrogenic atoms - Reasoning:

If the electrons were not interacting with each other but only with the Z = 2 nucleus, the ionization energy of He would be that of a Hydrogenic atom with a Z = 2 nucleus. The difference this ionization energy and the measured ionization energy is equal to the interaction energy of the two electrons. - Details of the calculation:

For Hydrogenic atoms:

E_{I}' = Z^{2}E_{I}(μ'/μ) . If (μ'/μ) ~ 1, then E_{I}' = Z^{2}13.6 eV.

For non-interacting electrons and Z = 2 we have E_{I}' = 54.4 eV.

(The binding energy is the negative of the ionization energy.)

The interaction energy of the two electrons therefore is +29.8 eV.

Consider an "atom" of positronium made up of an electron and a
positron.

Calculate the energies of the lowest three states of the system and briefly
enumerate correction terms you have neglected.

Solution:

- Concepts:

The hydrogenic atom - Reasoning:

Positronium, made up of an electron and a positron, is a hydrogenic atom. -
Details of the calculation:

H_{0}= p^{2}/(2μ) - e^{2}/r is the Hamiltonian for the relative motion of the two particles.**r**=**r**_{1}-**r**_{2}is the vector pointing from particle 2 to particle 1.

μ = m_{e-}m_{e+}/(m_{e-}+ m_{e+}) = m_{e}/2 is the reduced mass of the system. H_{0}is symmetric under the interchange of the two particles. H_{0}neglects relativistic effects.

Eigenvalues of H_{0}: E_{n}= -me^{4}/(2ħ^{2}n^{2}) = -13.6/(2n^{2}) eV.

E_{n}depends only on n.

Lowest energy level: n = 1, l = 0.

First excited state: n = 2, l = 0, or l = 1.

Relativistic effects, and spin-orbit and spin-spin couplings have been neglected.

The ground state wave function for a hydrogen like atom is

Φ_{100}(r) = (1/√π)(Z/a_{0})^{3/2
}exp(-Zr/a_{0}),

where a_{0} = ħ^{2}/(μe^{2})
and μ is the reduced mass, μ ~
m_{e} = mass of the electron.

(a) What is the ground state wave function of tritium?

(b) What is the ground state wave function of
^{3}He^{+}?

(c) An electron is in the ground state of tritium. A nuclear reaction
instantaneously changes the nucleus to ^{3}He^{+}. Assume the beta particle and the
neutrino are immediately removed from the
system. Calculate the probability that the electron remains in the ground
state of ^{3}He^{+}.

The emission spectrum of hydrogenic Lithium ions is measured and the wavelength of a series of emission lines are recorded. The longest line in that series has a wavelength of 450.25 nm. What is the wavelength of the shortest line in that series?

Solution:

- Concepts:

The spectrum of hydrogenic atoms - Reasoning:

Longest wavelength in an emission series of hydrogenic ions:

hc/λ_{long}= Z^{2}*13.6 eV(1/n^{2}– 1/(n+1)^{2}). - Details of the calculation:

Z = 3 and hc = 1240 eV nm.

λ_{long}= 450.25 nm = (10.13 nm) /(1/n^{2}– 1/(n+1)^{2}. Solve for n, n = 4.

For the shortest wavelength we have hc/λ_{short}= (Z^{2}*13.6 eV)/n^{2}.

λ_{short}= (10.13 nm)*16 = 162.09 nm.

A certain species of ionized atoms produces an emission line spectrum according to the Bohr model, but the number of protons in the nucleus is unknown. A group of lines in the spectrum forms a series in which the shortest wavelength is 22.79 nm and the longest wavelength is 41.02 nm. Find the next-to-longest wavelength in the series of lines.

Solution:

- Concepts:

Line spectrum of a hydrogenic atom - Reasoning:

E_{n'}- E_{n}= Z^{2}*13.6 eV(1/n^{2}- 1/n'^{2}) = (hc/λ)(1/n^{2}- 1/n'^{2}). -
Details of the calculation:

1/λ = (Z^{2}*13.6/hc)(1/n^{2}- 1/n'^{2}) = C(1/n^{2}- 1/n'^{2}).

1/λ_{shortest}= C/n^{2}, 1/λ_{longest}= C(1/n^{2}- 1/(n + 1)^{2}), λ_{shortest}/λ_{longest}= 1 – n^{2}/(n + 1)^{2}= 22.79/41.02.

n^{2}/(n + 1)^{2}= 0.444, n = (n + 1)*0.666, n = 2.

λ_{shortest}/λ_{nl = }1 – n^{2}/(n + 2)^{2}= ¾. λ_{nl}= (4/3) 22.79 nm = 65.69 nm.

Certain nuclei can occasionally de-excite by internal conversion, which is a process whereby the excitation energy is transferred directly to one of the atomic electrons, causing it to be ejected from the atom. (This process competes with de-excitation by photon emission.) It is a reasonable assumption that the probability of this occurrence on a particular electron is directly proportional to the probability of that electron being at the nucleus. Of the n = 2 electrons, which are the most likely to undergo internal conversion and why? Estimate the ratio of conversion of 1s electrons to that of 2s electrons. Assume that the nuclear excitation energy is much greater than the ionization energy of the 1s electron.

Hydrogenic atoms wave functions:

R_{10}(r) = 2 (Z/a)^{3/2 }exp(-Zr/a),

R_{20}(r) = (Z/(2a))^{3/2 }(2 - Zr/a) exp(-Zr/(2a)),

R_{21}(r) = 3^{-½}(Z/(2a))^{3/2}(Zr/a) exp(-Zr/(2a)),

Y_{00} = (4π)^{-½}, Y_{1±1} =
∓(3/8π)^{½}sinθ
exp(±iφ), Y_{10} = (¾π)^{½}cosθ.

Solution:

- Concepts:

Hydrogenic atoms, fundamental assumptions of Quantum Mechanics - Reasoning:

The probability that the electron is found at the nucleus is proportional to |ψ_{nlm}(**r**=0)|^{2}. -
Details of the calculation:

ψ_{2p}(0) = 0, ψ_{2s}(0) ≠ 0. The 2s electron is more likely to undergo internal conversion than the 2p electron.

Ratio: (1s internal conversion)/(2s internal conversion) = |ψ_{1s}(0)|^{2}/|ψ_{2s}(0)|^{2}= 8.

For hydrogen-like atoms, such as
the alkali atoms, the screening effect of the "closed-shell" electrons can be
accounted for by considering the electron to move in the potential
V(r) = -(e^{2}/r)(1 + α/r),
where α is a
constant. Find the energy eigenvalues for this potential.

Solution:

- Concepts:

Hydrogen-like atoms, the energy eigenfunctions and eigenvalues of the hydrogen atom - Reasoning:

The same equations have the same solutions. -
Details of the calculation:

The radial equation for u(r) = rR(r) is

[(∂^{2}/∂r^{2}) - l(l + 1)/r^{2}+ (2μ/ħ^{2})(e^{2}/r)(1 + α/r) - k^{2}_{kl}] u_{kl}(r) = 0,

with k^{2}_{kl}= -2μE_{kl}/ħ^{2}.

We can write

[(∂^{2}/∂r^{2}) - C_{l}/r^{2}+ (2μ/ħ^{2})(e^{2}/r) - k^{2}_{kl}] u_{kl}(r) = 0,

with C_{l}= l(l + 1) - (2μe^{2}α/ħ^{2}).

Changing to the variable ρ = r/a with a = ħ^{2}/(μe^{2}) we have

[(∂^{2}/∂ρ^{2}) - C_{l}/ρ^{2}+ (2/ρ) - λ_{kl}^{2})] u_{kl}(ρ) = 0,

with λ_{kl}^{2}= -2E_{kl}a/e^{2}.

If we write C_{l}= l'(l' + 1), then this is the same equation we obtain for the hydrogen atom with l replaced by l'.For the hydrogen atom we have acceptable solutions if λ

_{kl}= 1/(k + l), k = 1, 2, 3, ... .

For the hydrogen atom l = 0, 1, 2, ... , so 1/(k + l) = 1/n, n = 1, 2, 3, ... .

For our problem we have acceptable solutions if λ_{kl}= 1/(k + l').l'(l' + 1) = l(l + 1) - (2μe

^{2}α/ħ^{2}) = l(l + 1) - 2α/a.

l' = -½ + (¼ + l(l + 1) - 2α/a)^{½}= -½ + ((l + ½)^{2}- 2α/a)^{½}.

Let n = k + l, k = n - l. Then λ_{kl}= 1/(n - l + l').

λ_{kl}= 1/[n + ((l + ½)^{2}- 2α/a)^{½}- (l + ½)].

Let D(l) = ((l + ½)^{2}- 2α/a)^{½}- (l + ½). ThenE

_{kl }= [e^{2}/(2a)] [1/(n + D(l))^{2}].

The accidental degeneracy of the energy levels of hydrogen with respect to l is removed.

Consider a quasi-stable muonic atom composed of a muon in a 1s
state and 9 electrons around a Ne nucleus (Z = 10).

(a) Calculate the approximate energy and radius of the lowest energy bound
state of the muon. Comment on the effect of the outer electrons and of the
finite size of the nucleus on the wave functions of this state.

(b) Calculate the approximate energy difference between the lowest energy **
electron states** in the muonic atom and in ordinary neon.

Solution:

- Concepts:

The hydrogenic atom - 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}.

m_{muon }= 207 m_{e}. Therefore the average radius of the 1s state of the muon in Ne is approximately (½00) times the average radius of the 1s state of the electron in Ne. For the muon the nuclear charge is hardly screened and we can use the hydrogenic atom formalism to solve for its energy.

The muon, however screens the nuclear charge, and the outer electron of Ne move in the potential of a nucleus with charge Z-1 = 9. -
Details of the calculation:

(a) The reduced mass of the electron-Ne system is ~m_{e}and the reduced mass of the muon-Ne system is ~m_{muon }= 207 m_{e}. For the 1s state of the muon in Ne we have

a_{0}' = a_{0}(μ/μ')(1/Z) = a_{0}(½070) = 25 fm,

and E_{I}' = E_{I}(μ'/μ)Z^{2}= E_{I}(207*100) = 0.28 MeV.

The energy of the lowest energy bound state of the muon is -E_{I}.

Since the size of the nucleus is on the order of 10^{-15}m, the radius of the region in which the potential differs from a pure Coulomb potential is not completely negligible, and we may want to use perturbation theory to find correction terms for the energy and the wave function.

(b) In ordinary Ne the energy of the lowest energy bound state of an electron is -13.6*100 eV.

In Ne with a muon in a 1s state it is -13.6*81 eV.

The absolute value of the difference is 258 eV.