General Chemistry I - CHEM 181 Critical Thinking Exercise #3 Due Thursday, Oct. 24 Name_________________________ Section _______ Why do excited hydrogen atoms emit only specific wavelengths of light? Introduction In the 1800’s scientists discovered that a glass cylinder containing a low pressure of a gas will emit light when high voltage electricity is applied to metal electrodes inserted at opposite ends of the cylinder. This principal is used in fluorescent lighting today. When the light emitted by a particular gas, for instance hydrogen, was analyzed by passing it through a prism, scientists were surprised to find that only a few specific wavelengths of visible light were emitted, rather than a continuous range of wavelengths. This mystery took over 70 years to fully solve. wavelength, (nm) Infrared Ultraviolet (uv) Over time, scientists discovered that the wavelengths of radiation emitted by hydrogen extended from the ultraviolet through the visible into the infrared, microwave, and radio wave portions of the electromagnetic (EM) spectrum. There appeared to be a pattern in the wavelengths that consisted of a series of series of lines. The first 4 series starting from the shortest wavelength line are named for the people that discovered them. In 1885 J.J. Balmer discovered a mathematical pattern in the wavelengths emitted by hydrogen in the visible and near ultraviolet regions of the electromagnetic spectrum: Each allowed value for n plugged into this equation produces one of the wavelengths of light emitted by hydrogen in the region of the spectrum that Balmer studied. Five years later another scientist, Johannes Rydberg, extended this equation so that it could predict all of the wavelengths emitted by hydrogen in all regions of the electromagnetic spectrum: (eq. 1) This equation contains two integers that must be greater than zero and the value of n must be greater than the value of m. The way it works is that the value of m specifies the series, e.g. m = 3 is necessary to generate the wavelengths observed in the Paschen series and n = 4,5,6… will generate the wavelengths emitted within that series. The constant R is known as the Rydberg constant. Though the equations from Balmer and Rydberg demonstrated mathematical skill, they did not answer the big question on the minds of many scientists: Why do the atoms of hydrogen in the gas phase emit only certain wavelengths of radiation and why do the wavelengths have the pattern described by these clever equations. As it turned out, the solution to this riddle completely altered civilization on this planet. Part I – Putting together pieces of the puzzle Part of the reason that it took so long to solve hydrogen line spectrum puzzle was that scientists at the time had an incomplete understanding of light. Specifically, they were not aware of any relationship between the energy of light and the wavelength. It was assumed that the ene.

1. General Chemistry I - CHEM 181
Critical Thinking Exercise #3
Due Thursday, Oct. 24
Name_________________________
Section _______
Why do excited hydrogen atoms emit only specific wavelengths
of light?
Introduction
In the 1800’s scientists discovered that a glass cylinder
containing a low pressure of a gas will emit light when high
voltage electricity is applied to metal electrodes inserted at
opposite ends of the cylinder. This principal is used in
fluorescent lighting today. When the light emitted by a
particular gas, for instance hydrogen, was analyzed by passing
it through a prism, scientists were surprised to find that only a
few specific wavelengths of visible light were emitted, rather
than a continuous range of wavelengths. This mystery took
over 70 years to fully solve.
wavelength, (nm)

2. Infrared
Ultraviolet (uv)
Over time, scientists discovered that the wavelengths of
radiation emitted by hydrogen extended from the ultraviolet
through the visible into the infrared, microwave, and radio wave
portions of the electromagnetic (EM) spectrum. There appeared
to be a pattern in the wavelengths that consisted of a series of
series of lines. The first 4 series starting from the shortest
wavelength line are named for the people that discovered them.
In 1885 J.J. Balmer discovered a mathematical pattern in the
wavelengths emitted by hydrogen in the visible and near
ultraviolet regions of the electromagnetic spectrum:

3. Each allowed value for n plugged into this equation produces
one of the wavelengths of light emitted by hydrogen in the
region of the spectrum that Balmer studied.
Five years later another scientist, Johannes Rydberg, extended
this equation so that it could predict all of the wavelengths
emitted by hydrogen in all regions of the electromagnetic
spectrum:
(eq. 1)
This equation contains two integers that must be greater than
zero and the value of n must be greater than the value of m.
The way it works is that the value of m specifies the series, e.g.
m = 3 is necessary to generate the wavelengths observed in the
Paschen series and n = 4,5,6… will generate the wavelengths
emitted within that series. The constant R is known as the
Rydberg constant.
Though the equations from Balmer and Rydberg demonstrated
mathematical skill, they did not answer the big question on the
minds of many scientists: Why do the atoms of hydrogen in the
gas phase emit only certain wavelengths of radiation and why
do the wavelengths have the pattern described by these clever
equations. As it turned out, the solution to this riddle
completely altered civilization on this planet.
Part I – Putting together pieces of the puzzle

4. Part of the reason that it took so long to solve hydrogen line
spectrum puzzle was that scientists at the time had an
incomplete understanding of light. Specifically, they were not
aware of any relationship between the energy of light and the
wavelength. It was assumed that the energy of light depended
entirely on the intensity, which was the square of the amplitude
of the light waves according to accepted theory.
This understanding was changed drastically by 1905 due to
experiments that revealed that light had a particle nature in
addition to its wave nature and that particles of light, which we
now call photons, have an energy that is directly proportional to
frequency and inversely proportional to the wavelength of the
light.
Ephoton = where h is a constant called Plank’s constant
Since , we can also write Eph = (we’ll abbreviate
photon as ph)
Step 1: Notice that the left hand side of the Rydberg equation
(eq. 1) can easily be converted from into photon energy, Eph,
just by multiplying by hc. Naturally, the right hand side of the
equation will also need to be multiplied by hc to keeps things
equal. Rewrite the Rydberg equation (eq. 1) below in terms of
photon energy. The left side of the equation should look like;
Eph =
From the equation Ephoton = hc/λ, re-arrange it to get the
expression 1/λ.

5. Step 2: The right side of your revised Rydberg equation above
is really a difference between two terms (in parentheses). To
make this more apparent, multiply both terms by to remove the
parentheses and rewrite the equation below.
Just distribute the term hcR∞ in the two terms in the
parenthesis. The equation will be:
Step 3: What is the unit for each of these two terms on the right
side of the equation. The values (with units) for h and c are
6.626 X 10-34 Js and 2.998 X 108 m/s, respectively. The unit
for the Rydberg constant is given with eq. 1 on page 2. Use
unit cancellation to obtain the final unit for each term and write
that below.
The values of m and n are unitless, so it will not be included.
The units will be like:
=J So that unit of Eph is in Joules.
It should be clear from the unit you obtained above, that the
right side of the equation is simply the difference between two
energies. So right now the equation is telling us that the energy
of the photon of light emitted by a hydrogen atom for some
allowed combination of the integers m and n is equal to the
difference between the two energies given by the terms on the
right hand side. What do the energies on the right refer to? The
Law of Conservation of Energy can help us here. “During any

6. process, energy can be neither created nor destroyed.” Energy
may be converted from one form into another, however. So if a
hydrogen atom emits a specific amount of energy in the form of
a photon of light, Eph, then what must happen to the energy of
the atom? It must decrease by an amount that is exactly equal
the energy of the photon. We can represent this relationship
with a simple equation:
Eph = ∆Eatom = Einitial - Efinal (eq. 2)
If you compare this equation with your equation obtained in
step 2, you might be tempted to conclude that
and
However we run into a little problem here that becomes evident
if we ask ourselves what kind of energy would an atom of
hydrogen have?
Step 4: Work through the following questions to determine the
form(s) of energy existing within an atom of hydrogen.
a) What does an atom of the common isotope of hydrogen
consist of, i.e. what subatomic particles are present? For
hydrogen (H1), it consists of 1 electron and 1 proton. For
deuterium (H2), 1 electron, 1 proton and 1 neutron. For tritium
(H3), 2 neutrons, 1 electron and 1 proton.
b) What type of force, if any, would exist between these
particles? There’s this attractive electrostatic force of
interaction that exist between electrons and protons. Between

7. protons and neutrons, there’s what they call strong nuclear
force.
c) Should an atom of hydrogen have potential energy? (Note:
Whenever a force is acting on an object, the object must contain
potential energy.) Yes.
d) Are there other forms of energy that would likely exist within
a hydrogen atom? Yes. There should be kinetic energy.
Taking potential energy to be the primary form of energy
existing in an atom of hydrogen due to the attractive force
acting between the electron and the proton, we can consult a
physics source to find the equation for this form of energy,
which is known as Coulomb’s Law:
This tells us that the potential energy is proportional to the
product of the two charges and inversely proportional to the
distance between the charged objects. Note that k is simply a
proportionality constant.
The thing we need to notice here is that when the charges are
opposite and force is attractive, the value of the potential
energy must be a negative number. The equations that we
inferred earlier (restated below) from the comparison of the
Conservation of Energy equation with your revised Rydberg
equation from step 2 give positive values for the energy of the
atom, because all the constants are positive as well as the
integers m and n.
and
(eqs. 3)

8. Energies that are negative, rather than positive, can be obtained
by first switching the order of the two terms in the revised
Rydberg equation (step 2), as follows:
Then replacing the “+” with “-(-)” as follows:
(eq. 4)
Now each of the energy terms on the right must always have a
negative value, as it should to represent the energy of a
hydrogen atom. Note that since the integer n must always be
larger than the integer m, as defined in the original Rydberg
equation, the first term must have a smaller negative value than
the second term and thus represents a larger energy value. (If
this is confusing, just consider that the number -2 is greater
than -4.)
Step 5: Now with the exchange of the terms on the right,
compare eq. 4 with the Conservation of Energy equation, eq. 2,
and write two new equations for Einitial and Efinal similar to
eqs. 3 above, such that the initial and final energies generated
by the equations will necessarily have negative values.

9. Equation 4 is . Equation 2 is Eph = ∆Eatom = Einitial – Efinal.
Therefore, Einitial is -hcR∞/n2 and Efinal is -hcR∞/m2.
Compare the two equations just written. How are they
different? The numerators on the right hand side are exactly the
same, and the denominators must each be some positive integer
squared. The only difference between the two equations is that
the value of the integer for the initial energy of the atom must
be larger than the value of the integer for the final energy.
There would be nothing wrong with using the same integer
variable n for both equations if we subscript the variable to
distinguish initial and final values of n as ni and nf.
We can consider these two equations as special cases of a more
general equation that gives us all the possible energies that a
hydrogen atom can contain
where n =1,2,3,… (eq. 5)
Step 6: At this point you have derived an equation for the
energy of a hydrogen atom that must be satisfied in order to
account for the observed emission wavelengths in hydrogen’s
atomic emission spectrum. How does this equation limit the
energy that a hydrogen atom can contain due to the interaction
between the electron and the proton? Is any energy possible for
the atom over a broad range of values, as is the case for golf
balls, bicycles, and objects orbiting the planet (everyday life in
general)? Describe two ways in which this equation limits the
energy of a hydrogen atom. {Hint: The answer to this question
may be clearer after you complete Part II of this exercise.}

10. Limitation 1: The value of En approaches a certain number as
the value of n approaches infinity. That means that is the
minimum energy the H atom can have.
Limitation 2: The value of n should not be equal to zero. So the
lowest energy level that is allowed to H is n=1.
Part II – Testing the equation derived in part I for the allowed
energy levels for an atom of hydrogen.
Please show your work on all calculations.
1) Use the final equation from step 7 along with values for h
and c given in step 3, and the value of R given in eq. 1 to
calculate values for the first 10 allowed energies, n =
1,2,3,…10. List these values below (or include a copy of the
relevant cells if you used Excel to perform the calculations).
Be sure to include the appropriate unit for your values.
Using equation 5, the following data were obtained using Excel.
h
6.63E-34
J-s
n
En, Joules
c
3.00E+08
m/s
1
-2.17916E-18
R
1.10E+07
1/m
2

11. -5.44791E-19
3
-2.42129E-19
4
-1.36198E-19
5
-8.71665E-20
6
-6.05323E-20
7
-4.44727E-20
8
-3.40494E-20
9
-2.69032E-20

12. 10
-2.17916E-20
infinity
1E+40
-2.17916E-98
2) Next, use your energy values to create an energy level
diagram akin to Figure 3.22 in the textbook (but without the
vertical arrows) on a full size sheet of paper that you can attach
to this exercise. Try to make the diagram accurate to scale
using either graph paper or Excel. Be sure to label and scale
your energy axis appropriately.
Plotting the values of n versus wavelength, the graph is shown
below.
3) Recall that the Balmer series in the hydrogen spectrum
extends from the visible region into the uv portion of the
electromagnetic spectrum.
a. What do the electron transitions giving rise to the Balmer
series wavelengths have in common? {Hint: A transition is
specified by giving the integer values of the initial and final
energy levels, e.g. ni = 4 nf = 3. You may want to look up the
Balmer series in your textbook.}
The electron transitions that give rise to the Balmer series are
from the energy that is released from the atom instead of having
supplied to it. This is indicated by the negative Eph values.
b. Refer to the figure on the 1st page of this exercise to notice
that there are only 4 emission wavelengths from the Balmer

13. series that are in the visible portion of the EM spectrum. Use
the allowed energies that you calculated above in (1) to predict
the wavelengths (in nm) for the two longest wavelength
emission lines in the series. (Be sure to show your
calculations.) {Hint: The longest wavelength corresponds to
the smallest photon energy, which in turn corresponds to the
smallest energy difference between the initial level and final
level of the electron jump.}
The longest wavelength emission lines are 656 and 486 nm.
The wavelengths that are calculated for n=1 to 10 is
summarized below. 121 is not part of the UV-Vis spectrum.
En, J
delE, J
lambda, m
lambda, nm
-2.17916E-18
-5.44791E-19
1.63437E-18
1.21544E-07
121.5436038
-2.42129E-19
3.02662E-19
6.56335E-07
656.3354603
-1.36198E-19
4.08593E-19
4.86174E-07
486.1744151
-8.71665E-20
4.57624E-19
4.34084E-07
434.0842992

14. -6.05323E-20
4.84258E-19
4.1021E-07
410.2096627
-4.44727E-20
5.00318E-19
3.97042E-07
397.042439
-3.40494E-20
5.10741E-19
3.8894E-07
388.9395321
-2.69032E-20
5.17887E-19
3.83573E-07
383.5726716
-2.17916E-20
5.22999E-19
3.79824E-07
379.8237618
-2.17916E-98
5.44791E-19
3.64631E-07
364.6308113
The solution below is how to calculate wavelengths.
En is calculated using equation 5. Example of calculation:
N= 3 En = [(6.626x10-34 J-s)(3.0x108m/s)(1.10x107/m)]/3
= -2.42129E-19 Joules
∆E(32) = -2.42129E-19 – (-5.44791E-19) = 3.02662E-19
Joules
λ = hc/∆En = [(6.626x10-34 J-s)(3.0x108m/s)/ 3.02662E-
19 Joules] = 6.56335E-07m
in nm 6.56335E-07m (1nm/1.0x10-9m) = 656.3354603nm.

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