CHEM 113 UND Physical Chemistry Problem Set

2023/01/12Chem 113A Problem Set 01: Lectures 01–03
Due 2023/01/19 @ 10 PM
We encourage you to work with others in the class on the problem set, but the end product must
represent your effort. Formulas and constants can be found on the last page.
Part 1 (To be done in Discuss Section. If you do not finish it in section, please finish at home)
1) Blackbody Radiation
a. Two of the many classes of stars are called blue dwarfs and red dwarfs because of their apparent
colors. Here’s the visible spectrum and its corresponding wavelengths:
Which class of stars have higher temperature? Explain your reasoning. Which class of stars is
b. In spectroscopy, we use terms like “redshift” and “blueshift” to describe how the spectrum
changes. Redshift means the peak of the spectrum is shifting to longer wavelength (becoming
redder); and blueshift means the peak of the spectrum is shifting to shorter wavelength
(becoming bluer). For a blackbody, as temperature increases, does its spectrum blueshift or
pg. 1
Problem Set 01
Due 01/09
c. The figure below shows the solar spectrum at sea level (i.e., the intensity of sun light as a function
of wavelength measured at sea level). The dotted curve shows the extraterrestrial spectrum (i.e.,
measured outside the atmosphere). Estimate the temperature of the sun. (Hint: You’ll have to
use one of the two laws we learned in the lecture, Wien’s Displacement Law or Stefan-Boltzmann
pg. 2
Problem Set 01
Due 01/09
2) Photoelectric Effect
a. 300-nm light is shone on a piece of calcium. The ejected electron has a de Broglie wavelength of
1.10 nm. What is the work function of calcium?
b. What is the maximum wavelength (i.e. minimum frequency) of light needed to eject an electron
from calcium?
pg. 3
Problem Set 01
Due 01/09
3) de Broglie wavelength The Tolbert group in the Chemistry Department uses X-ray diffraction to
examine the structures of their polymer films. The X-ray beam they use in their experiments has an
energy of 12.75 keV (1 eV = 1.602×10-19 J).
a. If we want to use a beam of electrons instead of X-ray, what should be the velocity of the
electrons so that would produce the same diffraction pattern?
b. What is the kinetic energy of the electron?
c. Why do you think that they used electrons instead of photons?
pg. 4
Problem Set 01
Due 01/09
Part 2 (Take-home)
1) Bohr Model
a. In class, we identified the 𝑛 = 2 → 𝑛 = 1 transition …
Calculate the frequencies for the following transitions and label them on the figure:
𝑛 = 4 → 𝑛 = 1, 𝑛 = 3 → 𝑛 = 2, and 𝑛 = 4 → 𝑛 = 2
Do you observe a trend? Label all the transitions without doing additional calculations.
pg. 5
Problem Set 01
Due 01/09
b. For a hydrogen atom with its electron in the 𝑛 = 1 energy level, what wavelength of light would
remove (eject) the electron? (i.e., what wavelength of light would ionize ground state hydrogen?)
(Hint: removing electron means putting the electrons infinitely far away from the nucleus. What
value of 𝑛 would cause an electron to be placed infinitely far away from the nucleus?)
pg. 6
Problem Set 01
Due 01/09
2) Uncertainty Principle
a. What is the uncertainty of the momentum of an electron if we know its position is somewhere in
a 52.9 pm interval (the radius of the 𝑛 = 1 orbit for the Bohr model)?
b. How does the value compare to the momentum of an electron in the first Bohr orbit?
c. Based on your answers in parts a & b, how much confidence can we have in the position and the
momentum of an electron in the first Bohr orbit?
pg. 7
Problem Set 01
Due 01/09
3) Origin of Q.M. Complete the following chart
Bohr’s explanation for
blackbody radiation
Einstein’s explanation
for photoelectric effect
Bohr’s atomic model
Two-slit experiment
with electrons
pg. 8
Problem Set 01
Due 01/09
Formulas and constants:
𝑐 = 𝜈𝜆
𝐸 = ℎ𝜈
𝜆max 𝑇 = 2.90 × 10−3 m ⋅ K
𝑅 = (5.6697 × 10−8
) 𝑇4
m2 K 4 s

𝐾𝐸 = ℎ𝜈 − Φ
𝐾𝐸 = 𝑚v 2
4𝜋𝜀0 ℏ 2
𝑚𝑒 𝑒 2
𝑚𝑒 𝑟
𝑚𝑒 𝑒 4 1
𝐸=− 2 2 2
8𝜀0 ℎ 𝑛
𝑐 = 2.9989 × 108 m/s
ℎ = 6.626 × 10−34 J ⋅ s
ℏ = 1.055 × 10−34 J ⋅ s
Charge of an electron, 𝑒 = 1.60 × 10−19 C
Mass of an electron, 𝑚𝑒 = 9.11 × 10−31 kg
𝜀0 = 8.854 × 10−12 F/m
pg. 9

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