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Emission Series and Emitting Quantum States: Visible H Atom Emission Spectrum Experiment 6.
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Transcript of Emission Series and Emitting Quantum States: Visible H Atom Emission Spectrum Experiment 6.
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Emission Series and Emitting Quantum States:
Visible H Atom Emission Spectrum
Experiment 6
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#6 Emission Series and Emitting Quantum States: Visible H Atom Emission Spectrum
Goal: To determine information regarding the quantum
states of the H atom
Method: Calibrate a spectrometer using He emission lines Observe the visible emission lines of H atoms Determine the initial and final quantum states
responsible for the visible emission spectrum, as well as the Rydberg constant
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Electromagnetic Radiation
Oscillating electric and magnetic fields
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Light Energy
Wavelength distance peak-to-peak
Frequency oscillations per second
Energy E faster oscillation = more E
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Electromagnetic Spectrum
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Visible Emission
Wavelengths, , increaseEnergies decrease
Electronic transitions (“e- jumps”)
400 nm 500 nm 600 nm 700 nm
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Dual Nature of Light/Relationships
h Planck’s constant = 6.626×10-34 J.s
Units J = (J.s) (s-1)
c speed of light = 2.998×108 m.s-1
Units s-1 = (m.s-1)/(m)
hνE
λ
cν
1. Wave wavelength, frequency,
2. Particle photon = “packet”
E = h
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Using the Equations
(a) Calculate the frequency of 460nm blue light.
1-14
9
8
1052.6
nm101m1
nm) (460
)sm
1099.2(c
s
(b) Calculate the energy of 460 nm blue light.
J
ssJ19-
11434-
1032.4
)106.52)(10(6.626
hhc
E
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Spectroscopy
Spectroscopy: study of interaction of light with matter
h: photon
1. Absorption: matter + h → matter*
2. Emission: matter* → matter + h
Energy change in matter: Ematter = Eh
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Discrete Energy Levels
Observed energy level changes:
E = Eh= Efinal – Einitial
Ground state atom Absorption Emission
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“Discrete” Atomic Emission
Atomic absorption: electrons excited to higher energy levels
Atomic emission: excited electrons lose energy
Incandescent
Hot Gas
Cold Gas
Continuous
Discrete Emission
Discrete Absorption
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Quantized Energy Levels
Eh = ElevelsE = Ef – Ei
Absorption: Ef > Ei
Emission: Ef < Ei
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Hydrogen Emission Spectrum©The McGraw-Hill Companies. Permission required for reproduction or display
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H atom emission1) Electrical energy excites H H + energy
H* initial quantum state ni = 2, 3, 4, 5, 6,
…
2) H* loses energy H* H + h final quantum state nf = 1, 2, 3, …
nf < ni
You observe several Etransitions visible s ni’s levels > nf
nf end at same nf You determine ni’s and nf
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Hydrogen Atom and Emission
Lyman
Balmer
Paschen
Ground State:
n = 1
Excited States:
n = 2, 3, 4, …
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Rydberg Equation
A “series” is associated with two quantum numbers:
Lyman: ni = 2, 3, 4, … nf = 1
Balmer: ni = 3, 4, 5, … nf = 2
Paschen: ni = 4, 5, 6, … nf = 3
RH = 1.096776×107m-1 = 2.180×10-18 J = 2e4m/h3c
22
11
ifHhυ nn
RE
levelsifhν ΔEEEE General transition eq’n:
Hydrogen atomic
emission lines fit
(Rydberg eq’n):
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Incr
easing
(decr
easing E
,
smalle
r E)
n = principal E states (principal quantum #s)
Hydrogen Atomic Emission
22
11
ifHhυ nn
RE
En
erg
y
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Part 1 Correlate color with wavelength Use lucite rod 20 nm intervals, 400–700 nm , color Boundary s short, long
of max. intensity max
Observe Hg atomic emission (handheld specs)
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Part 2 Calibrate Spectrometer
Determine if measured wavelengths are “true”
Use He emissionRecord msr for lines
Plot true vs msr
7 or 8 lines
ColorAccepted (nm)
Measured (nm)
red 728.1 730red 706.5 710red 667.8 670
yellow 587.5 590green 501.5 500green 492.2 490
blue-green 471.3 470blue-violet 447.1 450
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Calibration Plot
msrd
true
xx
yyslope
12
12
H atom emission:•Multiply:msrd by slope•Converts: measured true
728.1730
706.5710
667.8670
587.5590
447.1450 471.3
470
492.2490
501.5500
y = 1.0021xR2 = 0.9996
440
480
520
560
600
640
680
720
440 480 520 560 600 640 680 720
Measured (nm)
Tru
e (
nm)
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Part 3 Record H emission s
Record color, msr (3 or 4 lines) color, msr
Determine true true
Calculate Eh from true Eh
Units: E in J
h in J.s
c in m/s
in m
)()(*
)(*
)(2)(2 22 linesggge
g hHHHH
)()(2*
)(2 bandsgg hHH
λ
hcEhν
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Questions/Data Analysis
1) Which set of lines?
Balmer or Paschen (nfinal = ?)
2) What is ninitial for each line?
3) What is your experimental RH?
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Hydrogen Lines / Analysis
Color (nm) E (J)
red 660 3.0×10-19
blue-green 490 4.1×10-19
blue-violet 430 4.6×10-19
violet 410 4.8×10-19
atomif
Hhυ ΔEnn
RE
22
11
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One way to think about the dataWhich series are we observing – Balmer or Paschen?Balmer: nf = 2 32, 42, 52
Paschen: nf = 3 43, 53, 63These would be the three lowest energy transitionsExample data:
Literature Observed (nm) Color (nm) E (J)410.1 violet 400 5.0E-19434.0 blue-violet 430 4.6E-19486.1 blue-green 500 4.0E-19656.2 red 650 3.1E-19
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Compare calculated ΔE to observed ΔE
EH atom 1/n2 = RH/n2 so calculate En=1, En=2, etc.Find ΔE between levels and compare to observed E’s
RH
2.18E-18 Experiment Balmer Paschen
n 1/n2En(J) measured n = 2 n = 3
7 0.020 4.45E-20 E (J) ninf
E (J) ninf
E (J)6 0.028 6.06E-20 72 5.0E-19 73 2.0E-19
5 0.040 8.72E-20 5.0E-19 6 4.8E-19 6 1.8E-194 0.063 1.36E-19 4.6E-19 52 4.6E-19 53 1.6E-193 0.111 2.42E-19 4.0E-19 4 4.1E-19 4 1.1E-192 0.250 5.45E-19 3.1E-19 32 3.0E-19 33 0.0E+00
Literature Observed (nm) Color (nm) E (J)410.1 violet 400 5.0E-19434.0 blue-violet 430 4.6E-19486.1 blue-green 500 4.0E-19656.2 red 650 3.1E-19
Experiment matches Balmer best
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How? Plot Eatom vs. 1/ni2
Rearranged Rydberg equation fits:
HRslope
22
1
f
H
iHatom n
R
nRΔE
bxmy
2intercepty
f
H
n
R
22
11 :
0 :interceptx
if nnso
ΔE
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Balmer Series
y = -2x10-18x + 6x10-19
R2 = 0.9878
0.0E+00
1.0E-19
2.0E-19
3.0E-19
4.0E-19
5.0E-19
6.0E-19
0.000 0.050 0.100 0.150 0.200 0.2501/ n2
Ef-E
i
Example Balmer Rydberg Plot
Slope (~RH):
2×10-18JClose to RH
2.18×10-18J
x-intercept:~0.24Close to 0.25~1/22
(nm) Color (nm) E (J)410.1 violet 400 5.0E-19434.0 blue-violet 430 4.6E-19486.1 blue-green 500 4.0E-19656.2 red 650 3.1E-19
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Balmer (nf = 2) – plot ΔE vs. 1/ni2
y = -2x10-18x + 5x10-19
R2 = 0.9988
0.0E+00
1.0E-19
2.0E-19
3.0E-19
4.0E-19
5.0E-19
0.000 0.050 0.100 0.150 0.200 0.250
1/ni2
E
Good:Slope –RH
x-intercept: 1 22
so nf = 2
~ 0.25 =
This plot verifies our data – we observed the Balmer series!
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An alternative way to analyze the data1) Data is for:
Balmer (nf = 2) or Paschen (nf = 3)
Be sure to correct wavelengths (measured true)
2) Transitions are 3 lowest energy:
Balmer (ni = 5, 4, 3) or Paschen (ni = 6, 5, 4)
nm Ef-Ei (J) ni nf 1/ni2 ni nf 1/ni
2
0 0 2 2 x-intercept 3 3 x-intercept660 3.0E-19 3 2 0.111 4 3 0.063490 4.1E-19 4 2 0.063 5 3 0.040430 4.6E-19 5 2 0.040 6 3 0.028
Balmer Paschen
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Graphs
Prepare two graphs (Balmer and Paschen) x-axis should extend to x-intercept (y = 0) y-axis should be appropriate
Draw best-fit straight line Find slope (one should be close to –RH)
Find relative error in experimental RH
Match and color to ni and nf
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Paschen (nf = 3)
Not too good
Slope ≠ RH
x-int. ≠ 1/32
y = -4x10-18
x + 6x10-19
R2 = 0.9969
0.0E+00
1.0E-19
2.0E-19
3.0E-19
4.0E-19
5.0E-19
0.000 0.020 0.040 0.060 0.080 0.100 0.120 0.140
1/ni2
E
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Balmer (nf = 2)
y = -2x10-18x + 5x10-19
R2 = 0.9988
0.0E+00
1.0E-19
2.0E-19
3.0E-19
4.0E-19
5.0E-19
0.000 0.050 0.100 0.150 0.200 0.250
1/ni2
E
Good:
Slope –RH
x-intercept: 1
22
so
nf = 2
~ 0.25 =
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Balmer Series
y = -2x10-18x + 6x10-19
R2 = 0.9878
0.0E+00
1.0E-19
2.0E-19
3.0E-19
4.0E-19
5.0E-19
6.0E-19
0.000 0.050 0.100 0.150 0.200 0.2501/ n2
Ef-E
i
Example Balmer Rydberg Plot
Slope (~RH):
2×10-18JClose to RH
2.18×10-18J
x-intercept:~0.24Close to 0.25~1/22
(nm) Color (nm) E (J)410.1 violet 400 5.0E-19434.0 blue-violet 430 4.6E-19486.1 blue-green 500 4.0E-19656.2 red 650 3.1E-19
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Data
color nm Ef-Ei (J) ni nf
--- 0 0 2 2red 660 3.0E-19 3 2
blue-green 490 4.1E-19 4 2blue-violet 430 4.6E-19 5 2
Balmer
Experimental RH: 2 ×10-18 J
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1/ vs. 1/ni
2
Atomic Hydrogen Emission Lines
0
20000
40000
60000
80000
100000
120000
0.000 0.200 0.400 0.600 0.800 1.000
1/n2
cm
Lyman
Balmer
Paschen
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ReportAbstractResults
2a: Calibration data and plot 2b: Table Series plot (or Balmer and Paschen plots)
depending on your analysis choice RH and error from literature Predicted wavelengths and error
Sample calculations of: photon energy and Rydberg slope
Discussion/review questions