Physics 249 Midterm Exam 2
Nov 16 th 2012
1) List of the quantum numbers of all the electrons in ground state Sodium (Z=11) in the energy order they are generally filled. For each n and l combination state how many radial probability distribution minimum and maximum there are.
2
3
4
5
(15 points) electron #
1 n
1
1
2
2
2 l
0
0
0
0
1 l type s s s s p ml
0
0
0
0
-1 ms
½
-½
½
-½
½ min (n-l)+1
2
3
2 max n-l
1
2
1
6
6
6
6
6
6
2
2
2
2
2
3
1
1
1
1
1
0 p p p p p s
0
1
1
0
-1
0
½
½
-½
-½
-½
½ 4 3
For use in the next problem I’ve listed the atoms next to each of the 11 electrons in Sodium where that electron is the outermost “valence” electron for each corresponding atom.
Note the order of ms states in the s orbital is arbitrary.
Note the p orbital can start with ml = +1 or -1. A correct answer will have the spin projection ms oriented in the opposite direction and the next two spins with the same orientation to minimize magnetic potential energy from spin orbit coupling. Also note that the reverse ordering of ml in the second three also happens to minimize spin orbit coupling for the initial electrons that are filled.
Minimum’s = (n-l)+1 counting the minimums at r=0 and infinity. If you didn’t include the minimums at r=0 and infinity the answer was still counted as correct if the minimum numbering was consistent.
2) Which atoms through Z=11 (you can refer to them by Z) have the highest ionization energies, lowest ionization energies, and highest electron affinity. Explain why they have these properties in each of the three cases
(15 points)
Highest ionization energies 2 He(highest), 10 Ne(second highest – also a maximum in the series). For Ne: the p orbitals have high radial penetration such that those electrons see the proton charges less shielded and are harder to ionize. For He and Ne, when the final orbital in a shell is filled that electron has high ionization energy.
Lowest Ionization energy. 3 Li(very near Na) and 11 Na(lowest). The first orbital after filling a complete shell have low ionization energy because the completed shell of electrons is approximately spherically symmetric and shields these electron from the nucleus. Also the s orbitals have less radial penetration enhancing the screening effect lowering the Zeff.
Highest electron affinity: Cl The p orbitals have deep radial penetration exposing it to a higher Zeff for the nucleus. The effect is a negative potential and an attractive force. Filling the last open slot before filling up the full atomic shell which also has large ionization energies.

3) Many of the standard atomic energy levels are split into doublet energy levels. Which type of orbital does not split and why?
(15 points)
The s orbitals are not split into doublets. The s orbital has no orbital angular momentum as there nothing that the spin can align or not align with to lower or raise potential energy.
4) Using the quantum version of the argument that energy of a rotational orbit is
𝐸 =
𝐿 2
2𝐼
Compare the energy of the rotational transitions between the lowest rotational energy states in an H2 molecule and the H atom using this treatment. (Treat the H atom as a rotating system of proton and electron using the same equation for rotational energy above.) Express your answer in eV. Comment on why this treatment is valid for the H2 molecule and not the H atom.
Proton mass: 1.6726x10
-27 kg or 938.27 MeV/c 2
Electron mass: 9.1094x10
-31 kg or 0.51100 MeV/c 2
Bohr radius: 0.52918x10
H
-10 m
2
separation distance: 0.074nm ℏ = 1.0546x10
c = 3.00x10
8
-34 J*s, 6.5821x10
-16 eV*s
m/s ℎ𝑐 = 1240𝑒𝑉𝑛𝑚 ℏ c = 197.3eVnm
(25 points)
For H2.
𝐿 2 𝑙(𝑙 + 1)ℏ 2
𝐸 =
2𝐼
=
2𝐼 using the general formula
[(𝑙 + 1)(𝑙 + 2) − 𝑙(𝑙 + 1)]ℏ 2
Δ𝐸 𝑙,𝑙+1
=
2𝐼
For l=0
Δ𝐸
0,0+1
=
(0 + 1)ℏ 2
𝐼
= ℏ 2
𝐼
6 points
For H2
=
(𝑙 + 1)ℏ
𝐼
2
𝐼 = 𝑚 𝑝
(
1
2 𝑟
0
)
2
+ 𝑚 𝑝
(
1
2 𝑟
0
)
2
=
1
2 𝑚 𝑝 𝑟
0
2
Δ𝐸
0,0+1
=
2ℏ 2 𝑚 𝑝 𝑟
0
2
=
2(ℎ𝑐)
4𝜋 2 𝑚 𝑝 𝑐
2
2 𝑟
0
2
=
2𝜋 2
(1240𝑒𝑉𝑛𝑚)
938.27x10
6
2 𝑒𝑉(0.074𝑛𝑚) 2
= 0.015𝑒𝑉
7 points
For H
The reduced mass and electron mass are the same to 2 digits of precision.

𝐼 = 𝑚 𝑒
(𝑎 𝑜
) 2
Δ𝐸
0,0+1
= ℏ 2 𝑚 𝑒 𝑎 2
0
=
(ℎ𝑐)
4𝜋 2
2 𝑚 𝑒 𝑐 2 𝑎 2
0
=
4𝜋 2
(1240𝑒𝑉𝑛𝑚)
0.511x10
6
2 𝑒𝑉(0.052918𝑛𝑚) 2
= 27𝑒𝑉
7 points
In the case of H2 the vibrational and rotational energy states are separated by an order of magnitude.
Though not as large as some other case this means the rotational quantum state would have only a small effect on the radial wave equation. In the case of the H atom the rotational energy has a larger magnitude than the typical energies of the ground state. The angular momentum has a large effect on the radial wave equation and they have both be solved simultaneously to understand the true solution.
5 points
5) Calculate the rate of transmission for E>0 particles through a one dimensional delta function barrier potential at x=0: 𝑉(𝑥) = 𝜆𝛿(𝑥) where 𝜆 >0. The delta function is infinite at x=0 and 0 everywhere else.
This case is representative of the tunneling between a thin barrier between conductors since the potential is very high for an electron to be outside the wire.
Qualitatively you can think of the delta "function" as the limit of a process, which starts with a square barrier, and reduces the thickness of the barrier, but increases the height of the barrier so that the area under the barrier function remains the same and the barrier is finally infinitesimally thin in space and infinitely high in space. Since, product of width times height of the potential barrier is finite the wave function does not have to be zero at x=0.
(30 points, 5 points for each part) a) Write down the Schrödinger equation for this system.
− ℏ 2
2𝑚
𝜕 2 𝜓(𝑥)
𝜕𝑥 2
+ 𝜆𝛿(𝑥)𝜓(𝑥) = 𝐸𝜓(𝑥) b) Write down a proposed wave equation solution.
Hint: Since, product of width times height of the delta function potential barrier is finite, as in the finite potential barrier problem we solved, the wave function does not have to be zero at x=0. 𝜓
1
(𝑥) = 𝐴𝑒 𝑖𝑘𝑥 𝜓
2
(𝑥) = 𝐶𝑒 𝑖𝑘𝑥
+ 𝐵𝑒 −𝑖𝑘𝑥
√2𝑚𝐸 𝑘 = ℏ
(note: technically you could also solve the finite barrier problem with three regions and then take the limit, though this is harder.) c) Write down an equation relating the wave functions at the region boundary.
At x=0
𝐴 + 𝐵 = 𝐶 d) Write down an equation relating the first derivatives of the wave function at the region boundary.

Hint: You can evaluate the form of the first derivative accounting for the impact of the delta function the by integrating the Schrodinger equation in an infinitesimally small region, 𝑥 = −𝜀 to 𝑥 = −𝜀 around x=0 and considering the limit as that region shrinks to zero.
Hint: By definition of the delta function: ∫ 𝑓(𝑥) 𝛿(𝑥)𝑑𝑥 = 𝑓(0) over any range including x=0.
− ℏ 2
2𝑚
∫ 𝜀
−𝜀
𝜕
𝜕𝑥
(
𝜕𝜓(𝑥)
𝜕𝑥
) 𝜀 𝑑𝑥 + 𝜆 ∫ 𝛿(𝑥)𝜓(𝑥)𝑑𝑥
−𝜀 𝜀
= 𝐸 ∫ 𝜓(𝑥)
−𝜀 𝑑𝑥 as 𝜀 → 0 the integral over the continuous wave function must vanish. Therefore the right side is zero. taking the limit as 𝜀 → 0 ℏ 2
−
2𝑚 𝑖𝑘(𝐶𝑒 𝑖𝑘𝜀 − 𝐴𝑒 −𝑖𝑘𝜀 + 𝐵𝑒 +𝑖𝑘𝜀 ) + 𝜆(𝐴 + 𝐵) = 0 ℏ
2 𝑖𝑘(𝐶 − 𝐴 + 𝐵) = (𝐴 + 𝐵)
2𝑚𝜆 e) Solve the system of equations to understand the normalization constants of the transmitted wave compared to the incoming wave (the overall normalization constant does not need to be understood). ℏ 2 𝑖𝑘(𝐶 − 𝐴 + 𝐵) = (𝐴 + 𝐵)
2𝑚𝜆
𝐴 + 𝐵 = 𝐶 we want C in terms of A. Remove B since we are not interested in reflection. ℏ 2 𝑖𝑘(𝐶 − 𝐴 + 𝐶 − 𝐴) = (𝐴 + 𝐶 − 𝐴)
2𝑚𝜆 ℏ 2 𝑖𝑘(𝐶 − 𝐴 + 𝐶 − 𝐴) = (𝐴 + 𝐶 − 𝐴)
2𝑚𝜆 ℏ 2
( 𝑚𝜆 𝑖𝑘 − 1) 𝐶 = ℏ 2 𝑚𝜆 𝑖𝑘𝐴
𝐶 = ℏ 2
( 𝑚𝜆
1 𝑖𝑘 − 1) ℏ 2 𝑚𝜆 𝑖𝑘𝐴
𝐶 =
(1 −
1 𝑚𝜆 𝑖𝑘ℏ 2
)
𝐴 f) Write down the transmission rate.

𝑇 = 𝑅𝑒 ( 𝑘 𝑘
1
2
|𝐶| 2
|𝐴| 2
1
𝑇 =
(1 + ( 𝑚𝜆 𝑘ℏ 2
)
)
2
)