Physics 249 Homework 8
Due Nov 9th
1) Five identical non-interacting particles are put in an infinite potential 1D square well
with L=1.0nm. Compare the ground state energy if they are muons (heavy version of the
electron) or pion (pions have symmetric wave functions).
Muon like electrons are fermions. They will have anti-symmetric wave functions and
obey the Pauli exclusion principle. Pions do not obey the Pauli exclusion principle and
can all be in the same ground state.
𝑘𝑛2 ℏ2
𝜋 2 ℏ2
2
𝐸=
=𝑛
2𝑚
2𝑚𝐿2
For muons. The ground state will be:
n=1 ms=1/2, n=1 ms=-1/2, n=2 ms=1/2, n=2 ms=-1/2, n=3 ms=1/2
𝐸 = (2 ∗ 12 + 2 ∗ 22 + 1 ∗ 32 )
= 19
𝜋 2 ℏ2
𝜋 2 ℏ2
𝜋 2 (ℏ𝑐)2
=
19
=
19
2𝑚𝜇 𝐿2
2𝑚𝜇 𝐿2
2𝑚𝜇 𝑐 2 𝐿2
𝜋 2 (197.3𝑒𝑉𝑛𝑚)2
= 0.0345𝑒𝑉
2(105.66𝑥106 𝑒𝑉)(1.0𝑛𝑚)2
For pions n=1 for 5.
𝐸 = (5 ∗ 12 )
𝜋 2 ℏ2
𝜋 2 ℏ2
𝜋 2 (197.3𝑒𝑉𝑛𝑚)2
=
5
=
5
= 0.00712𝑒𝑉
2𝑚𝜋 𝐿2
2𝑚𝜋 𝐿2
2(134.98𝑥106 𝑒𝑉)(1.0𝑛𝑚)2
2) Which of the following atoms would you expect to have its ground state split by the
spin orbit interaction. Li, B, Sc, Cu. For the atoms that have ground state splitting how
many split energies are there?
Not Li, or Cu. For both the valence electron is a s state orbital and further in the case of
Cu the d orbital is completely filled by taking an electron out of the s state.
B and Sc will have 2 split energies. The energy only depends on either the spin is aligned
or not aligned with the nuclear magnetic moment. In all cases where the electron has
total anglar momentum that is not zero the electron sees the nucleus rotating around it in
it’s rest from and has two splitting’s depending on whether it is aligned or not aligned. In
angular momentum terms: J: |l-1/2| … |1+1/2|. As long as l>0 there are two options.
3) For LiCl
a) Calculate the potential energy of attraction at the equilibrium separation
b) Find the disassociation energy neglecting the energy of repulsion.
c) What is the value of the energy of repulsion?
a)
𝑘𝑒 2 8.99𝑥109 (1.60𝑥10−19 )
𝑒𝑉
=
= −1.1393𝑥10−18 𝐽 ∗ 6.2416𝑥1018
−9
𝑟0
0.202𝑥10
𝐽
= −7.11𝑒𝑉
𝑈=−
b)
𝑈(𝑟0 ) = 𝐸𝐸𝑀 + 𝐸𝑒𝑥 + 𝐸𝑖𝑜𝑛 − 𝐸𝑎𝑓𝑓 = −7.1111 + 0 + 5.39 − 3.62 = −5.3411𝑒𝑉
5.34eV
c) the full dissociation energy is 4.86. Therefore:
𝐸𝑒𝑥 = 5.3411 − 4.86 = 0.481𝑒𝑉
4) Hydrogen can bond covalently with sulfur, tellurium, phosphorus and antimony. Use
the electron configuration tables in appendix C to write the chemical formulas for these
covalent molecules.
H2S, H2Te, H3P, H3Sb
5) The equilibrium separation in HF is 0.0917 nm and it’s electric dipole moment is
6.40x10-30 Cm. What percentage of the bonding is ionic
𝑝𝑖𝑜𝑛𝑖𝑐 = 𝑒𝑟0 = 1.60𝑥10−19 ∗0.0917x10−9 = 1.4672𝑥10−29
43%
6) The equilibrium potential energy between two atoms in a molecule can be well
described by the formula.
𝑎 12
𝑎 6
𝑈(𝑟) = 𝑈0 [( ) − 2 ( ) ]
𝑟
𝑟
Find r and U at the minimum in terms of a and 𝑈0 .
If r = 0.074nm and U= -32.8eV at the minimum graph the potential. On the same graph
include an electrostatic attraction potential energy assuming the H2 molecule acts as ionic
bond with an H+ attracting and H-. How do the two potentials compare at high radius?
What would you conclude about the ionic bonding disassociation energy from this
comparison and is ionic bonding a good model for this system? For the x axis use r/rmin
for the units.
𝑑𝑈
1 𝑎 12
1 𝑎 6
1 𝑎 6
𝑎 6
= 𝑈0 [−12 ( ) + 12 ( ) ] = 12𝑈0 ( ) [1 − ( ) ] = 0
𝑑𝑟
𝑟 𝑟
𝑟 𝑟
𝑟 𝑟
𝑟
r=a, then at r=a
𝑎 12
𝑎 6
𝑈(𝑎) = 𝑈0 [( ) − 2 ( ) ] = −𝑈0
𝑎
𝑎
r = 0.074nm and U= -32.8eV, then 𝑈0 = 32.8
0.074 12
0.074 6
𝑈(𝑟) = 32.8 [(
) − 2(
) ]
𝑟
𝑟
for the coulomb potential:
𝑘𝑒 2
8.99𝑥109 (1.60𝑥10−19 )2
𝑒𝑉
1.4364𝑥10−9 𝑒𝑉𝑚
=−
∗ 6.2416𝑥1018
=−
𝑟
𝑟
𝐽
𝑟
𝑚
1.4364 𝑒𝑉𝑛𝑚
=−
𝑟
𝑛𝑚
𝑈=−
1.4364 𝑒𝑉𝑛𝑚
Considering the coulomb potential it will be at = − 0.074 𝑛𝑚 = −19.4𝑒𝑉 at the
equilibrium separation. The potential well is 1.5 deeper. As we could infer from the fact
that covalent bond have shorter separation distances the covalent bond forms a deeper
potential well. An ionic model is not good. (note that I haven’t superimposed the
coulomb curve – I didn’t have a chance to graph that yet)