Redding Homework 5 - Page 1
MISE - Physical Basis of Chemistry
Fifth Set of Problems - Due February 12, 2006
Submit electronically (digital drop box) by Sunday, February 12 – by 6 pm.
Note: When submitting to digital Drop Box label your files with your name first and
then the label - HmwkFive. Please put you name in the ‘Header’ along with the
already-inserted page #.
The following information may be useful when solving the problems:
Ideal gas law: PV = nRT ; where R = 0.0821 L•atm•mole-1•K-1.
1 nm = 10-9 m ; 1 Ångstrom (Å) = 10-10 m ; 1 calorie (cal) = 4.184 joules (J)
density of (liquid) water (H2O) = 1.00 g/mL ; 1000 mL = 1 L
specific heat of (liquid) water (H2O) = 1.00 cal/gºC.
q = C∆T = mc∆T = nCmolar∆T
watt (W) = joule/s ; total energy delivered = watts • time (in s).
Speed of light = 3.00 x 108 m/s = (wavelength)•(frequency) = •f
Planck’s constant = 6.63 x 10-34 J•s ; Unit of frequency (f) = hertz (Hz) = 1/s
Bohr radius constant = ao = 0.0529 nm = 5.29 x 10-11 meters.
A H-like atom has one (1) electron & Z protons , Z is the atomic #.
[For H, Z =1 and is electrically neutral ; other H-like atoms are ions with + charge.]
Radius of electron orbit for H-like atom = n2•ao/Z ; n = 1, 2, 3, … [For H, Z = 1].
Rydberg energy constant = Ry = 2.18 x 10-18 joules (J).
Energy of electron in H-like atom (E) = - Z2•Ry/n2 ; n = 1, 2, 3, … [For H, Z = 1].
Change in energy of electron in H-like atom (∆Eelectron):
1 
 1
∆Eelectron = Efinal - Einitial = Z2•Ry•
; [For H, Z = 1].
2
2 
n

n
 initial final
Energy balance for energy change of electron in H-like atom emitting/absorbing light:
∆Eelectron = Ephtoton where Ephtoton = h•f = h•c/
total energy from light source = Nphtotons•Ephtoton
1 mole = 6.022 x 1023 “things” (atoms, molecules, photons)
Redding Homework 5 - Page 2
1. The power output of a laser is measured by its wattage (W), the number of
joules it radiates per second ( 1 W = 1 J/s). A 10.00 W laser produces a beam
of green light with a wavelength of 520.0 nm.
(a) Calculate the number of photons emitted in one minute by the laser.
(520nm)(10-9m/1nm)= 5.20x10-7m
Watts * seconds= (10.0 W) (60s) = Nphotons * (6.63x10-34)(3.0x108)
5.20x10-7m
-19
600J= Nphotons * 3.825 x 10
Nphotons= 1.57x1021
(b) If the laser output is converted completely to heat and is used to heat 100.0 grams of
water, initially at 25ºC, for a period of one minute, calculate the temperature change
∆T (in ºC) and the final temperature (in ºC) of the water. The idea is that the photons
provide all of the energy to heat the sample. In other words, presume that heat
transfer occurs in a insulated system - such that the sum of the total (heat) energy
released by the photons plus the heat absorbed by the 100.0 grams of water equals
zero (0).
(600J)(1cal/4.184J) = 143 cal
143 cal = (100g)(1cal/g0C)(Tf-25oC)
143cal=100Tf -2500goC
2643.4=100Tf
Tf= 26.4oC
ΔT=1.4oC
2. Consider the hydrogen atom (Z = 1) when answering the following. [For this
problem it may be useful to construct an energy level diagram like we did in
class to “follow the electron” as it “jumps” between energy levels. You
do not have to submit the diagram - only the reasoning and calculations used
to answer any questions.]
(a) Calculate the minimum (i.e., lowest) frequency (in Hz) that can be observed in
the emission spectrum for the Lyman series. [Recall, the Lyman emission series
results from transitions in which the electron in initially in any energy level
in which n > 1 and the final energy level is n = 1.]
ΔEelectron=(2.18x10-18J) (1/4 – 1/1)
6.63x10-34J*s f=(2.18x10-18J) (1/4 – 1/1)
6.63x10-34J*s f=(2.18x10-18J) (.25 – 1)
6.63x10-34J*s f=(2.18x10-18J) (-.75)
6.63x10-34J*s f=(-1.635x10-18J)
f=-2.47x1015htz
Redding Homework 5 - Page 3
(b) Calculate the maximum (i.e., longest) wavelength (in nm) that can be observed in
the absorption spectrum for the Balmer series. [An absorption spectrum is one
in which the electron absorbs a photon from an external white light source and
undergoes a transition from a lower energy level to a higher energy level. Thus,
the Balmer absorption series results from transitions in which the electron in initially
in an energy level in which n = 2 and the final energy level could be any energy level
in which n ≥ 3.]
(6.63x10-34J*s)(3.0x108m/s)=(2.18x10-18J) (1/4 – 1/9)

(6.63x10-34J*s)(3.0x108m/s)=(2.18x10-18J) (.13)

(1.989x10-25J*m)=(2.834x10-19J)

-25
(1.989x10 J*m)=(2.834x10-19J) 
(1.989x10-25J*m)= 
(2.834x10-19J)
7.02x10-7m=
(7.02x10-7m)(1nm/10-9m)=701.8nm
=701.8nm
(c) Please refer to the pictures of the emission and absorption spectrum on page 7
of handout #4 - “The Emission of Light by Matter...the Revolution Begins...”
Note how the Balmer absorption spectrum “looks” compared to the Balmer
emission spectrum. The “lines” in the absorption spectrum appear as absences of
color in an otherwise continuous background of colors - the exact “opposite” of the
emission spectrum. Absorption line spectra are thus sometimes referred to as
“dark line” spectra. The dark lines are exactly at the same locations as the colored
lines in the emission spectrum. Given the description in (b) of how an absorption
spectrum is set up, please explan why the hydrogen Balmer absorption
spectrum “looks” the way it does - compared to the hydrogen Balmer emission
spectrum. Base your explanation on the “Bohr model” of quantized energy
levels, electron transitions, and their relationship to photons.
The hydrogen Balmer absorption spectrum “looks” the way it does due to
the electrons transitioning from level to level. Firstly, according to Bohr,
there are restrictions placed on microscopic behavior of the electrons. The
radii, or energy levels (n), may only be whole numbers. He hypothesized that
the levels looked similar to planets rotating around the sun, thus it is
sometimes referred to as the planetary model. The levels must be
energetically stable; otherwise the electrons could spiral into the nucleus. The
radius may be represented using the formula r=n2 (0.0529nm). The levels are
represented by whole numbers such as 1, 2, 3, etc.
Redding Homework 5 - Page 4
It must also be understood absorption and emission of light only occur when
the electrons are transitioning. The energy flow is positive or negative
depending upon the “direction” in which the electron is transitioning. For
example, if the electron is in a higher level, due to electricity, it could then
makes its way back to a lower energy level by transitioning from level four
and ending at level two. That energy loss is negative and seen as light. Bohr
represented this mathematically as follows: ΔE= Efinal – Einitial or more
specifically ΔEelectron = (2.18x10-18J) (1/ni2 - 1/nf2). The energy changes are
known as a photons and are equal to the energy change of the electron.
As stated in part b, in the absorption spectrum, the “electron absorbs a
photon from external white light source.” The electron is transitioning from a
lower energy level to a higher energy level. Instead of emitting light, the light
is now absorbed and is seen as dark lines.
3. For this question, imagine the hydrogen-like atom (ion) that can be created from
elemental beryllium (Be) for which Z = 4.
[For this problem it may be useful for some of the questions below to construct
an energy level diagram like we did in class to “follow the electron” as it “jumps”
between energy levels. You do not have to submit the diagram - only the reasoning
and calculations used to answer any related questions.]
(a) What must be the net charge (+1, +2, …) on the Be atom for it to become
hydrogen-like? How do you know?
The Be atom must lose 3 electrons to become hydrogen like because hydrogen
only has 1 electron. If Be has 4 protons, it must have 4 electrons (assuming it
is not an ion). In order for it to become like hydrogen, it would have to lose 3
electrons, giving it an overall charge of +3 because it would have 3 more
protons than electrons. It would no longer be neutral.
(b) Determine the wavelength (in nm) and the frequency (in Hz) of the first spectral
line in the Balmer emission series.
[The first spectral line, is the one of longest possible wavelength (i.e., least photon
energy). Recall, that all Balmer photon emissions involve the electron undergoing an
energy level transition in which the final n = 2 and the initial energy level can be any
n ≥ 3. ]
Redding Homework 5 - Page 5
1 
 1
h•c/=∆Eelectron = Efinal - Einitial = Z2•Ry•
2
2 
n

n
 initial final
(6.63x10-34J*s)(3.0x108m/s)=16 (2.18x10-18J) (1/9 – 1/4)

(6.63x10-34J*s)(3.0x108m/s)=16 (2.18x10-18J) (-.139)

(1.989x10-25J*m)=(-4.848x10-8J)

(1.989x10-25J*m)=(-4.848x10-8J) 
(1.989x10-25J*m)= 
(-4.848x10-8J)
=-4.103x10-8m
=-41.02nm
This does not seem correct… why is it negative? Why is it so small?
f=c/
f=3.0x108m/s
-4.102x10-8m
f=-7.31x1015htz
This does not seem correct… why is it negative?
4. Quantum Theory at work…telling the time…Cesium (Cs) clocks…
“…The most accurate realization of a unit that mankind has achieved…”
A bit of Déjà Vu…
The excerpts below are from the following URL: http://tycho.usno.navy.mil/cesium.html
“A “cesium(-beam) atomic clock” (or “cesium-beam frequency standard”) is a device that uses
as a reference the exact frequency of the microwave spectral line emitted by atoms of the
metallic element cesium, in particular its isotope of atomic weight 133 (“Cs-133”). The integral
of frequency is time, so this frequency, 9,192,631,770 hertz (Hz = cycles/second), provides the
fundamental unit of time, which may thus be measured by cesium clocks.
Today, cesium clocks measure frequency with an accuracy of from 2 to 3 parts in 10 to the
14th, i.e. 0.00000000000002 Hz; this corresponds to a time measurement accuracy of 2
nanoseconds per day or one second in 1,400,000 years. It is the most accurate realization of a
unit that mankind has yet achieved. A cesium clock operates by exposing cesium atoms to
microwaves until they vibrate at one of their resonant frequencies and then counting the
corresponding cycles as a measure of time. The frequency involved is that of the energy
absorbed from the incident photons when they excite the outermost electron in a cesium
Redding Homework 5 - Page 6
atom to jump (“transition”) from a lower to a higher orbit.
According to quantum theory, atoms can only exist in certain discrete (“quantized”)
energy states depending on what orbits about their nuclei are occupied by their electrons. …
Cesium is the best choice of atom for such a measurement because all of its 55 electrons but
the outermost are confined to orbits in stable shells of electromagnetic force. Thus, the
outermost electron is not disturbed much by the others. The cesium atoms are kept in a very
good vacuum of about 10 trillionths of an atmosphere so that the cesium atoms are little
affected by other particles. All this means that they radiate in a narrow spectral line whose
wavelength or frequency can be accurately determined.”
Further down the same webpage:
“In a cesium clock like these, liquid cesium is heated to a gaseous state in an oven. A
hole in the oven allows the atoms to escape at high speed….”
(a) Determine the wavelength (in cm) of the spectral line corresponding to this listed
frequency (9,192,631,770 Hz). Also, determine the energy (in J) of the photon
that has this frequency.
f=c/
9,192,631,770 Hz=3.0x108m/s

=.0326m=3.263cm
Energy in Joules:
H*f=6.63x10-34J*s)( 9,192,631,770 Hz)
Ephoton=6.095x10-24J
(b) This outermost electron in Cs that is mentioned above, is in an energy level that the
Bohr model would designate as n = 6. However, in this situation we are dealing
with an electrically neutral Cs atom, there are as many protons as electrons. Thus,
this outermost electron does not “feel” the full nuclear charge attraction of 55
protons (Z = 55) - because of the repulsion of the other 54 electrons. To a decent
level of approximation, this outermost electron “feels” a nuclear charge of only
one proton, i.e., “Z = 1” - just like hydrogen itself! Determine the energy (in J)
and the orbital radius (in nm) of this outmost electron - assuming Z = 1 according to the Bohr model.
Radius of electron orbit for H-like atom = n2•ao/Z
(36)(.0529nm)=1.9044nm
1
Energy:
Eelectron= -Z2*Ry/n2
Redding Homework 5 - Page 7
Eelectron= -12*2.18x10-18J/36
Eelectron= 6.056x10-20J
(c) The webpage discussion (in blue) reveals that the gaseous Cs atoms are kept
at a pressure of 10 trillionths of an atmosphere (i.e., 10-11 atmospheres) .
Assuming that these Cs atoms are in a 2.00 L chamber maintained at 700ºC,
determine the number of Cs atoms present if they exhibit ideal gas behavior.
n=PV/RT
n=(1x10-11)(2.0L)/(0.0821L*atm*mol-1*K)(973K)
n=2.5x10-13
This seems like a really small amount!?!?!
(d) The webpage discussion (in violet) reveals that the Cs atoms are heated in an oven to
become a gas and then allowed to escape through a hole in the oven…sounds like
effusion! :). [Recall Homework Set #3 - Problem #2 and the kinetic theory lecture
slides (especially #15).] Assume a particular oven of fixed volume - maintained at
constant temperature - contains some gaseous “Cs-133 atoms”
(atomic weight = 133 g/mole) that are contaminated with some gaseous He atoms
(atomic weight = 4.0 g/mole). The contaminated mixture contains 80 mole % of
Cs-133 and 20 mole % He.
• Determine the ratio of moles He : moles Cs-133 a few moments after the gaseous
mixture begins to effuse through the hole in the oven into an evacuated chamber.
• Now, determine the mole percent of each gas in this effused mixture.
√My = √133g.mol = 5.77moles Cs: 1 He
Mx
4g/mol
Mole % Cs=5.77moles Cs = 85.23%
5.77 + 1.0
Mole % He=1.0moles He = 14.77%
5.77 + 1.0
I am not completely confident with this answer either…I feel as if there should be
less Cs because it is a larger atom…