Tienne Myers Homework 5 - Page 1
MISE - Physical Basis of Chemistry
Fifth Set of Problems - Due February 12, 2006
Tienne Moriniere-Myers
October 1, 2007
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) = λ•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• 2
- 2  ; [For H, Z = 1].
ninitial nfinal
Energy balance for energy change of electron in H-like atom emitting/absorbing light:
∆Eelectron = Ephtoton where Ephtoton = h•f = h•c/λ
total energy from light source = Nphtotons•Ephtoton
Tienne Myers Homework 5 - Page 2
1 mole = 6.022 x 1023 “things” (atoms, molecules, photons)
Tienne Moriniere-Myers
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.
E total = power " time
600 j = 10.00 j / s " 60sec
hc
#
(6.63 "10$10 ) " 3.00 "1017 nm
E=
520nm
$16
1.989 "10
E=
= 3.825 "10$19 j /s
520nm
E total = N photon " E photon
E=
600 j = N photon " 3.825 "10$19
600 j
3.825 "10$19
N = 1.57 "10 21
(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).
N=
!
qwater= mc∆T
100g "1.00cal /gC°(Tf # 25°C )
0 = ((100cal /°CTf ) # 2500cal) + (#143.03cal)
= 100cal /°CTf # 2643.40cal
2643.403cal
=
100cal /°C
1cal = 4.1845 j
$ 1cal '
600 j&
) = 143.40cal
% 4.1845 (
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
! have to submit the diagram - only the reasoning and calculations used
do not
Tienne Myers Homework 5 - Page 3
A. to answer any questions.]
# 1
1 &
Ry%%" 2 + 2 (( = Ephoton
$ n final n initial '
#1 1&
2.18 )10"18 % 2 + 2 (
$1 2 '
#18
2.18 "10 (.75) = 1.635 "10#18 = 1.635 "10#18 j = hf
#18
6.63 "10#34
j
j / s = 1.635 "10
!
1.635 "10#18
6.63 "10#34
j /s
15
= 2.466 "10 /s
!
(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.]
see above
(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.]
%1( % 1 (
2.18 "10#18 ' * + ' 2 *
& $) & 2 )
%
1(
2.18 "10#18 '0 + * = 5.45 "10#19
&
4)
5.45 "10#19
= 8.22 "1014 /s
#34
6.63 "10
!
(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
Tienne Myers Homework 5 - Page 4
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.
In the emission spectrum energy is being zapped and given off by electrons falling from a higher
to a lower energy level.
In the absorption spectrum energy is being gained and electrons are moving from a lower to a
higher energy level.
The lines that are in the the same place on the spectrum show the light being emitted and
absorbed at a balanced energy rate.
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?
In order for Be to be like H, Be must lose 3 electrons because H only has 1 electron. The net
charge must be +3.
(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. ]
"1% "1%
Z 2 Ry$ 2 ' + $ 2 '
#n & #n &
" 1 1%
4 2 ( 2.18 (10)18 $ + '
# 4 9&
16 ( 2.18 (10)18 (.1388) = )4.84 (10)18
)4.84 (10)18
j /s
6.63 (10)34
)7.30 (1015 /s
C=f "#
3.0 "10 8 = #7.30 "1015s =
!
!
!
#4.1057 "10#8
m
= 41.1nm
10#9
Tienne Myers Homework 5 - Page 5
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
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
3.00 "10 8 = 9,192,631,770 s
100 " .0326348327 m = #
3.26cm = #
Ephoton=hf
6.63 "10#34 j • s
energy
#24
! Ephoton = 6.09 "10 j
(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,
Tienne Myers Homework 5 - Page 6
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.
"Ry
E= 2
n
2.18 #10"18
E=
= 6.06 #10"20
2
6
2
R = n ( A0 )
6 2 (0.0529nm)
= 1.9044nm
1
(c) The webpage discussion (in blue) reveals that the gaseous Cs atoms are kept
R=
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.
!
pv
=n
rt
n=
10"11 atm(2.00L)
2 #10"11
=
.0821L • atm • k • mole"1 (973k ) 0.0821# 973
2 #10"11
=
= 2.50 #10"13 moles
79.8833
6.02 #10 23
atoms
2.50 #10"13 #
= 1.50719 #1011
atoms
1mole
!
(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.
Tienne Myers Homework 5 - Page 7
Zeffx Nx
My
=
•
Zeffy Ny
Mx
x
My
=
y
Mx
20 H 1
=
80 Cs 4
1 133g /mole
4 4.0g /mole
1
33.25
4
1.44:1 ratio For every 1.44 H -1Cs passes through.
!
Helium-59%
Cesium-41%
Tienne Moriniere-Myers