The Rydberg Equation an Atomic Emi
Emissio
ssio
ssion
nSSpe
pe
pec
ctra Activit
Activityy
Part A, Determination o
off the Rydberg C
Constant
onstant
1. From the Beyond Labz portal, start Virtual Chem Lab, select Atomic Theory, and then select
‘The Rydberg Equation’ from the list of assignments. The link will take you to the Quantum
laboratory. The Spectrometer will be on the right of the lab table. The center of the table
has a gas discharge tube containing hydrogen gas. The hydrogen emission spectra is
displayed in the window in the upper right corner (shown as a graph of intensity vs.
wavelength). The colors of the corresponding peaks are shown in the window above the
spectra.
2. Click on the Visible/Full switch to magnify only the visible spectrum. You will see four peaks in
the spectrum, but may only be able to see the corresponding colors of three of them if your
eyes are like the authors. If you drag your cursor over a peak, it will identify the wavelength
(in nm) in the x-coordinate field in the bottom right corner of the detector window. You can
also use the cursor to select a region of the spectra to zoom-in on. You can reset the zoom
feature by toggling the visible/full switch.
3. Record the colors and wavelengths of the four peaks in the visible hydrogen spectrum in the
Table 1. (Round wavelength values to whole numbers). Even if you are unable to visibly see
the fourth peak, the spectrometer can detect it, you can just describe the peak by its
wavelength.
4. Convert the wavelength to units of meters and enter in Table 1.
ℎ끫뢠
5. The formula for the determination of the energy of light is E = where h is Planck’s
끫브
constant and c is the speed of light. Calculate the energy of each photon (J) and input that
value in Table 1.
What is the relationship between wavelength and energy?
Wavelength, frequency, and energy are all related to light. The higher the
frequency, the shorter the wavelengths, which corresponds to greater energy. If
there are longer wavelengths, with lower frequency, there will be lower energy.
The equation for energy is: E = h*c, with E = energy, v = frequency, and h = the
plank constant.
, Table 1
color Wavelength Wavelength Energy
(nm) (m) (J)
Purple 410. nm 410x10-9 m 4.85x10-19 J
Line #1 (left most)
Blue 434. nm 434x10-9 m 4.58x10-19 J
Line # 2
Cyan 486. nm 486x10-9 m 4.09x10-19 J
Line #3
Red 656. nm 656x10-9 m 3.03x10-19 J
Line #4 (right most)
6. Compare the energy values from Table 1. The greater the energy of the photon, the greater
the value of ni associated with the transition. Given that in the Balmer series nf is always 2,
assign an energy and wavelength (m) to each of the corresponding transitions in Table 2.
Table 2
Transitionni to nf Energy (J
(J)) Wavelength (m
(m))
6 – 2 (highest energy) 4.85x10-19 J 410x10-9 m
5–2 4.58x10-19 J 434x10-9 m
4–2 4.09x10-19 J 486x10-9 m
3 – 2 (lowest energy) 3.03x10-19 J 656x10-9 m
11
�−
7. Calculate the value of끫뢶 � for each of the transitions and input those values in Table 3.
2
끫뢦끫뢶
2
끫뢬
For the wavelength associated with each transition, calculate the inverse wavelength, 1/λ (m-
1) and input that date into Table 3.
Table 3
1 1
�2 − 끫뢶
�
Transition ni to nf 끫뢶
끫뢦 2끫뢬 1/λ (m-1)
1 1 (1)/(410x10-9 m-1) =
6–2 (highest energy) �− 62 � = 2/9 = 0.2222
22
2439024.39
1 1 -9 -1
5–2 �
22
− 52 � = 21/100 = 0.210(1)/(434x10 m ) =
2304147.47
1 1 -9 -1
4–2 �
22
− 42 � = 3/16 = 0.1875 (1)/(486x10 m ) =
2057613.17
1 1 -9 -1
3 – 2 (lowest energy) 22
�− 32 � = 5/36 = 0.1389 (1)/(656x10 m ) =
1524390.24
Emissio
ssio
ssion
nSSpe
pe
pec
ctra Activit
Activityy
Part A, Determination o
off the Rydberg C
Constant
onstant
1. From the Beyond Labz portal, start Virtual Chem Lab, select Atomic Theory, and then select
‘The Rydberg Equation’ from the list of assignments. The link will take you to the Quantum
laboratory. The Spectrometer will be on the right of the lab table. The center of the table
has a gas discharge tube containing hydrogen gas. The hydrogen emission spectra is
displayed in the window in the upper right corner (shown as a graph of intensity vs.
wavelength). The colors of the corresponding peaks are shown in the window above the
spectra.
2. Click on the Visible/Full switch to magnify only the visible spectrum. You will see four peaks in
the spectrum, but may only be able to see the corresponding colors of three of them if your
eyes are like the authors. If you drag your cursor over a peak, it will identify the wavelength
(in nm) in the x-coordinate field in the bottom right corner of the detector window. You can
also use the cursor to select a region of the spectra to zoom-in on. You can reset the zoom
feature by toggling the visible/full switch.
3. Record the colors and wavelengths of the four peaks in the visible hydrogen spectrum in the
Table 1. (Round wavelength values to whole numbers). Even if you are unable to visibly see
the fourth peak, the spectrometer can detect it, you can just describe the peak by its
wavelength.
4. Convert the wavelength to units of meters and enter in Table 1.
ℎ끫뢠
5. The formula for the determination of the energy of light is E = where h is Planck’s
끫브
constant and c is the speed of light. Calculate the energy of each photon (J) and input that
value in Table 1.
What is the relationship between wavelength and energy?
Wavelength, frequency, and energy are all related to light. The higher the
frequency, the shorter the wavelengths, which corresponds to greater energy. If
there are longer wavelengths, with lower frequency, there will be lower energy.
The equation for energy is: E = h*c, with E = energy, v = frequency, and h = the
plank constant.
, Table 1
color Wavelength Wavelength Energy
(nm) (m) (J)
Purple 410. nm 410x10-9 m 4.85x10-19 J
Line #1 (left most)
Blue 434. nm 434x10-9 m 4.58x10-19 J
Line # 2
Cyan 486. nm 486x10-9 m 4.09x10-19 J
Line #3
Red 656. nm 656x10-9 m 3.03x10-19 J
Line #4 (right most)
6. Compare the energy values from Table 1. The greater the energy of the photon, the greater
the value of ni associated with the transition. Given that in the Balmer series nf is always 2,
assign an energy and wavelength (m) to each of the corresponding transitions in Table 2.
Table 2
Transitionni to nf Energy (J
(J)) Wavelength (m
(m))
6 – 2 (highest energy) 4.85x10-19 J 410x10-9 m
5–2 4.58x10-19 J 434x10-9 m
4–2 4.09x10-19 J 486x10-9 m
3 – 2 (lowest energy) 3.03x10-19 J 656x10-9 m
11
�−
7. Calculate the value of끫뢶 � for each of the transitions and input those values in Table 3.
2
끫뢦끫뢶
2
끫뢬
For the wavelength associated with each transition, calculate the inverse wavelength, 1/λ (m-
1) and input that date into Table 3.
Table 3
1 1
�2 − 끫뢶
�
Transition ni to nf 끫뢶
끫뢦 2끫뢬 1/λ (m-1)
1 1 (1)/(410x10-9 m-1) =
6–2 (highest energy) �− 62 � = 2/9 = 0.2222
22
2439024.39
1 1 -9 -1
5–2 �
22
− 52 � = 21/100 = 0.210(1)/(434x10 m ) =
2304147.47
1 1 -9 -1
4–2 �
22
− 42 � = 3/16 = 0.1875 (1)/(486x10 m ) =
2057613.17
1 1 -9 -1
3 – 2 (lowest energy) 22
�− 32 � = 5/36 = 0.1389 (1)/(656x10 m ) =
1524390.24
