Physics 7A FNTs due DL 10
1) (application) Consider a substance made up of two different types of atoms, A and B, with
three different Lennard-Jones potential interactions, A-A, B-B, and A-B. The potential well depth
for the pair-wise interactions are given as: εAA = 20 × 10−21 J, εBB = 5 × 10−21 J, and εAB =
3 × 10−21 J. Atoms A have a diameter of σA = 1 × 10−10 m and atoms B have a diameter of σB =
2 × 10−10 m. Sketch all three potential plots on the same graph so you can compare both the
and the r0 values.
2) (definition)
a) Explain briefly, in your own words (and not more than one or two sentences), the meaning of
the construct “bond energy” (from the Energy-Interaction Model) as we used it at the beginning
of the course.
b) Explain briefly, in your own words (and not more than one or two sentences), what bond
energy is in our Particle Model of Bond Energy.
3) (application) In this FNT you will be comparing the approximation found in Activity 3.5, 𝜟𝑯𝒗𝒂𝒑 ≈
𝟔𝑵𝑨 𝜺, to some real substances. Note: 𝑁𝐴 = 6.02 × 1023 𝑚𝑜𝑙−1 is Avogadro’s number.
a) Construct a graph for the substances in the table below by plotting 𝜟𝑯𝒗𝒂𝒑 vs. 𝑵𝑨 𝜺 using
the values in the table for each of the five substances. Then, find the slope of the best-fit
line. (Careful: don’t neglect the units of 𝜺 and don’t forget to multiply 𝜺 by 𝑵𝑨 for the x-axis.)
b) What do you expect the value of the slope to be based on the equation 𝜟𝑯𝒗𝒂𝒑 ≈ 𝟔𝑵𝑨 𝜺? If
your expected slope does not agree with your measured slope, think about why that is?
These are molecule-molecule potentials, not atom-atom potentials!
Substance Well-depth (x 10-21 J) () ∆HVAPORIZATION TBP (Kelvin)
(Joules)
N2 1.28 5560 77
CH4 2.05 8180 112
O2 1.62 6820 90
C2 H6 3.35 14,720 185
Ar 1.66 6447 87
4) (application) Say you have a monatomic substance simulated with 100 atoms on the computer
simulation program. You see that it is in its solid phase and when all KE is removed and the
temperature reads 0 K, the total PE of these 100 atoms is -400x10-21 J. You therefore know
that the bond energy of these 100 atoms is -400x10-21 J.
-
a) The well depth, 𝜺, for these atoms is 0.8 x 10 -21 J. Determine the average number of
-
nearest-neighbors for each atom (refer to Activity 3.5 or the Particle Model of Bond
-
Energy foldout).
b) Why do you think that the average number of nearest-neighbors per atom turns out to be
less than 12 for this sample of 100 atoms?
N
400x18-2 Ex
-
=
FOX X0 .
50XM =
Continue to Next Page
Unit 3: Applying Particle Models to Matter DL 9
, Physics 7A FNTs due DL 10
5) (challenge) Consider the two structures drawn below. Because of the Lennard-Jones potential,
the bond energy for a configuration of particles will always be negative. The amount of energy
it takes to break apart the configuration will always be the amount required to raise the total
bond energy to zero.
a) Using the Pythagorean theorem, determine the distance between next-nearest-neighbor
bonds in the square configuration. Use this distance and a graph of the Lennard-Jones
potential to estimate the amount of energy stored in each of these next-nearest neighbor
bonds. Recall that r0 = 1.12σ.
b) Considering both nearest and next-nearest
neighbors, estimate the bond energy for these two
structures. Is the distance between next-nearest
neighbors the same for both configurations? Report
your answer in terms of the well depth, 𝜀. How much
energy would you have to add to break these
structures apart?
c) Based on what you found in b), which structure would you expect to be more stable if both
configurations were made of the same kind of particle?
d) If the square atomic structure was made up of atoms B from FNT 1), with 𝜀𝐵𝐵 = 5 × 10−21 𝐽,
and the linear structure was made up of atoms A from FNT 1), with 𝜀𝐴𝐴 = 20 × 10−21 𝐽,
which structure would be more stable?
e) Calculate the bond energy of the square structure shown to the right using
atoms A and B, considering all neighbors as in a). Assume the atoms are the
same size, but with potential well-depths as in FNT 1):
𝜀𝐴𝐴 = 20 × 10−21 𝐽, 𝜀𝐵𝐵 = 5 × 10−21 𝐽, and 𝜀𝐴𝐵 = 3 × 10−21 𝐽.
Unit 3: Applying Particle Models to Matter DL 9
1) (application) Consider a substance made up of two different types of atoms, A and B, with
three different Lennard-Jones potential interactions, A-A, B-B, and A-B. The potential well depth
for the pair-wise interactions are given as: εAA = 20 × 10−21 J, εBB = 5 × 10−21 J, and εAB =
3 × 10−21 J. Atoms A have a diameter of σA = 1 × 10−10 m and atoms B have a diameter of σB =
2 × 10−10 m. Sketch all three potential plots on the same graph so you can compare both the
and the r0 values.
2) (definition)
a) Explain briefly, in your own words (and not more than one or two sentences), the meaning of
the construct “bond energy” (from the Energy-Interaction Model) as we used it at the beginning
of the course.
b) Explain briefly, in your own words (and not more than one or two sentences), what bond
energy is in our Particle Model of Bond Energy.
3) (application) In this FNT you will be comparing the approximation found in Activity 3.5, 𝜟𝑯𝒗𝒂𝒑 ≈
𝟔𝑵𝑨 𝜺, to some real substances. Note: 𝑁𝐴 = 6.02 × 1023 𝑚𝑜𝑙−1 is Avogadro’s number.
a) Construct a graph for the substances in the table below by plotting 𝜟𝑯𝒗𝒂𝒑 vs. 𝑵𝑨 𝜺 using
the values in the table for each of the five substances. Then, find the slope of the best-fit
line. (Careful: don’t neglect the units of 𝜺 and don’t forget to multiply 𝜺 by 𝑵𝑨 for the x-axis.)
b) What do you expect the value of the slope to be based on the equation 𝜟𝑯𝒗𝒂𝒑 ≈ 𝟔𝑵𝑨 𝜺? If
your expected slope does not agree with your measured slope, think about why that is?
These are molecule-molecule potentials, not atom-atom potentials!
Substance Well-depth (x 10-21 J) () ∆HVAPORIZATION TBP (Kelvin)
(Joules)
N2 1.28 5560 77
CH4 2.05 8180 112
O2 1.62 6820 90
C2 H6 3.35 14,720 185
Ar 1.66 6447 87
4) (application) Say you have a monatomic substance simulated with 100 atoms on the computer
simulation program. You see that it is in its solid phase and when all KE is removed and the
temperature reads 0 K, the total PE of these 100 atoms is -400x10-21 J. You therefore know
that the bond energy of these 100 atoms is -400x10-21 J.
-
a) The well depth, 𝜺, for these atoms is 0.8 x 10 -21 J. Determine the average number of
-
nearest-neighbors for each atom (refer to Activity 3.5 or the Particle Model of Bond
-
Energy foldout).
b) Why do you think that the average number of nearest-neighbors per atom turns out to be
less than 12 for this sample of 100 atoms?
N
400x18-2 Ex
-
=
FOX X0 .
50XM =
Continue to Next Page
Unit 3: Applying Particle Models to Matter DL 9
, Physics 7A FNTs due DL 10
5) (challenge) Consider the two structures drawn below. Because of the Lennard-Jones potential,
the bond energy for a configuration of particles will always be negative. The amount of energy
it takes to break apart the configuration will always be the amount required to raise the total
bond energy to zero.
a) Using the Pythagorean theorem, determine the distance between next-nearest-neighbor
bonds in the square configuration. Use this distance and a graph of the Lennard-Jones
potential to estimate the amount of energy stored in each of these next-nearest neighbor
bonds. Recall that r0 = 1.12σ.
b) Considering both nearest and next-nearest
neighbors, estimate the bond energy for these two
structures. Is the distance between next-nearest
neighbors the same for both configurations? Report
your answer in terms of the well depth, 𝜀. How much
energy would you have to add to break these
structures apart?
c) Based on what you found in b), which structure would you expect to be more stable if both
configurations were made of the same kind of particle?
d) If the square atomic structure was made up of atoms B from FNT 1), with 𝜀𝐵𝐵 = 5 × 10−21 𝐽,
and the linear structure was made up of atoms A from FNT 1), with 𝜀𝐴𝐴 = 20 × 10−21 𝐽,
which structure would be more stable?
e) Calculate the bond energy of the square structure shown to the right using
atoms A and B, considering all neighbors as in a). Assume the atoms are the
same size, but with potential well-depths as in FNT 1):
𝜀𝐴𝐴 = 20 × 10−21 𝐽, 𝜀𝐵𝐵 = 5 × 10−21 𝐽, and 𝜀𝐴𝐵 = 3 × 10−21 𝐽.
Unit 3: Applying Particle Models to Matter DL 9
