Line Spectra and Energy Levels - CCSF...500 Wavelength (nm) 350 400 450 550 600 650 700 750 When we...

Line Spectra and Energy LevelsA Chem 101A Tutorial

A normalincandescent lightbulb contains a hotpiece of metal wire,

which produceswhite light. A hydrogen

discharge tubecontains hot

hydrogen gas, whichproduces lilac-colored light.

Let’s explore howthese differ!

500

Wavelength (nm)

750700650600550450400350

When we pass the white light from the normal light bulb through aprism, we see the entire visible spectrum (the “colors of the rainbow”).

The colors we see are our brain’s way of representing wavelengths.

Note that our eyes can only detect wavelengths from roughly 400 nm to700 nm. Light bulbs also produce infrared and ultraviolet light, but we

cannot see these types of radiation.

ultravioletregion

infraredregion

500

Wavelength (nm)

750700650600550450400350

When we pass the lilac-colored light from the hydrogendischarge tube through a prism, we only observe four specific

colors, which appear as colored vertical lines.

This pattern is called the line spectrum of hydrogen.

500

Wavelength (nm)

750700650600550450400350

656nm

486nm

434nm

410nm

Here is the line spectrum of hydrogen superimposed on theentire visible spectrum.

500

Wavelength (nm)

750700650600550450400350

Other elements also produce line spectra when we pass anelectrical discharge through them.

For example, helium produces bright blue-white light.

When we pass this light through a prism, we separate it intothe line spectrum of helium.

500

Wavelength (nm)750700650600550450400350

500

Wavelength (nm)750700650600550450400350

The line spectrum of every element is unique.

For example, both mercury and krypton produce light bluelight in a discharge tube, but their line spectra are obviously

different.

As a result, we can use line spectra to identify the elements inan unknown substance.

Appearance of thelight produced byeither mercury or

kryptonLine spectrum

of krypton

Line spectrumof mercury

500

Wavelength (nm)750700650600550450400350

500

Wavelength (nm)750700650600550450400350

Some line spectra are very complex, while others are simple.

Neon produces many visible lines, while sodium producesonly one.

A sodiumdischarge tube

Line spectrumof sodium

Line spectrumof neon

A neondischarge tube

Before we explore line spectra in more detail, we need to change the way we observethem.

Instead of looking at wavelengths, we will now focus our attention on the photonenergy of the light.

Photon energy is inversely proportional to wavelength:

The photon energy of visible light ranges from 170 kJ/mol to 300 kJ/mol. Here is thevisible spectrum plotted using energy as the scale. Wavelength is shown below the

spectrum for comparison. Note that higher energies correspond to shorterwavelengths.

Ephoton = hcλ

Energy (kJ/mol)100 250 350200150 3000 50

34240050060070080090010001500 120020005000 300010000

Wavelength (nm)∞

infraredregion

ultravioletregion

Energy (kJ/mol)100 250 350200150 3000 50

34240050060070080090010001500 120020005000 300010000

Wavelength (nm)∞

Here is the hydrogen line spectrum plotted on our energy scale.

Problem: our eyes can only detect radiation whose wavelength is inthe range 400 – 700 nm.

Does the hydrogen discharge tube produce other wavelengths that wecannot observe with our eyes?

???? ????range of visible spectrum

Energy (kJ/mol)100 250 350200150 3000 50

34240050060070080090010001500 120020005000 300010000

Wavelength (nm)∞

Yes, it does!!If we use film that is sensitive to wavelengths outside

the visible range, we observe a vast number ofadditional lines that we cannot see with our eyes.

Some of these lines are widely spaced, while othersare bunched tightly together.

Energy (kJ/mol)

250 500 750 1000 12500

85.410015020030040050010002000∞Wavelength (nm)

1400

If we look farther into theultraviolet region, we find still

more spectral lines.

This diagram shows the completeline spectrum of hydrogen.

Area covered by theprevious figure

Energy (kJ/mol)

250 500 750 1000 12500

85.410015020030040050010002000∞Wavelength (nm)

1400

When we look at the line spectrum of hydrogen, wenotice several groups of lines.

In each group, the lines start out widely spaced, then getcloser and closer until they blur together.

Each of these groups of lines is called a series.

Energy (kJ/mol)

250 500 750 1000 12500

85.410015020030040050010002000∞Wavelength (nm)

1400

The first group of lines, lying at theshortest wavelengths and highest

energies, is called the Lymanseries.

Energy (kJ/mol)

250 500 750 1000 12500

85.410015020030040050010002000∞Wavelength (nm)

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The second group,which includes the

lines we can see withour eyes, is called the

Balmer series.

Lyman series

Energy (kJ/mol)

250 500 750 1000 12500

85.410015020030040050010002000∞Wavelength (nm)

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The third group,which lies in the

infrared region andoverlaps the fourthgroup, is called thePaschen series.

Lyman seriesBalmerseries

Why does the line spectrum of hydrogen look like this?

To understand, we must start with the fact that the electron in a hydrogenatom can only have certain energies.

These allowed energies can be calculated from the following formula:

€

E = -1313 kJ/moln2

The symbol n here stands for a counting number (1, 2, 3…).

Note that all of the possible energies for the hydrogen electron arenegative numbers.

There are infinite possible energies for the hydrogen atom, becausethere are infinite counting numbers.

However, the electron cannot have any energy we choose. Forinstance, the electron cannot have E = -1000 kJ/mol.

€

-1000 kJ/mol = -1313 kJ/moln2

n2 = -1313 kJ/mol-1000 kJ/mol

= 1.313

n = 1.146 (not a counting number)

Here are the first few energies that the electron can have in a hydrogen atom:

€

E1 = -1313 kJ/mol12 = -1313 kJ/mol

€

E2 = -1313 kJ/mol22 = -328.2 kJ/mol

€

E3 = -1313 kJ/mol32 = -145.9 kJ/mol

€

E4 = -1313 kJ/mol42 = -82.1 kJ/mol

€

E5 = -1313 kJ/mol52 = -52.5 kJ/mol

€

E6 = -1313 kJ/mol62 = -36.5 kJ/mol

If we let n become infinitely large, E becomes infinitesimally small (it gets closerand closer to zero):

€

limn→∞ En = 0 kJ/mol

E2 = -328.2 kJ/mol (n = 2)

E1 = -1313 kJ/mol (n = 1)

E3 = -145.9 kJ/mol (n = 3)

E6 = -36.5 kJ/mol (n = 6)E5 = -52.5 kJ/mol (n = 5)E4 = -82.1 kJ/mol (n = 4)

0

-100

-300

-200

-400

-500

-600

-700

-800

-900

-1000

-1100

-1200

-1300

We can represent theseallowed energies by

plotting them on a verticalgraph.

This graph is called theenergy level diagram for

the hydrogen atom.

Remember that everyallowed energy for a

hydrogen atom fits theformula

all other levels (n = 7 to infinity)

E = -1313 kJ/moln2

ElectronEnergy(kJ/mol)

E2 = -328.2 kJ/mol

E3 = -145.9 kJ/mol

E6 = -36.5 kJ/molE5 = -52.5 kJ/molE4 = -82.1 kJ/mol

0

-100

-300

-200

-400

-500

-600

-700

-800

-900

-1000

-1100

-1200

-1300


E1 = -1313 kJ/mol (ground state)

If we have a collection ofcold hydrogen atoms, each

atom’s electron will havethe lowest possible energy

(-1313 kJ/mol),because that is the most

stable state for theelectron.

This lowest energy level iscalled the ground state.

e– e– e– e– e– e– e– e–

all other levels

E2 = -328.2 kJ/mol

E3 = -145.9 kJ/mol


0

-100

-300

-200

-400

-500

-600

-700

-800

-900

-1000

-1100

-1200

-1300


E1 = -1313 kJ/mol

If we add a lot ofenergy, though,

many of thehydrogen atomswill absorb this

energy and theirelectrons will riseto higher energy

levels, calledexcited states.We will end up

with electrons in avariety of levels.

e– e– e–

e– e–

e– e– e– e–e–

excitedstates

all other levels

E2 = -328.2 kJ/mol

E3 = -145.9 kJ/mol


0

-100

-300

-200

-400

-500

-600

-700

-800

-900

-1000

-1100

-1200

-1300


e–

Any electron that isabove the ground state

will eventually drop backto the ground state. It

can do so in one step orin several.

For instance, an electronthat is in level 5 mightdrop to level 4, then tolevel 2, and finally to

level 1.

E1 = -1313 kJ/mol

0

-100

-300

-200

-400

-500

-600

-700

-800

-900

-1000

-1100

-1200

-1300


e–

As the electrondrops from one

level to another, itloses energy.This energy

comes out in theform of a photon

of electromagneticradiation (light).

The energy of thephoton exactly

equals the size ofthe energy

change for theelectron.

Low energy photon(small energy change)

Medium energy photon(moderate energy change)

High energy photon(large energy change)

0

-100

-300

-200

-400

-500

-600

-700

-800

-900

-1000

-1100

-1200

-1300


e–

We can calculatethe energy of

each photon bycalculating the

differencebetween the

starting and finalenergy levels for

each jump.

ΔE = -82.1 – -52.5 = -29.6 kJ/mol

ΔE = -328.2 – -82.1 = -246.1 kJ/mol

ΔE = -1313 – -328.2 = -985 kJ/mol

-52.5-82.1

-328.2

-145.9

-1313

0

-100

-300

-200

-400

-500

-600

-700

-800

-900

-1000

-1100

-1200

-1300


e–ΔE = -29.6 kJ/mol

ΔE = -246.1 kJ/mol

ΔE = -985 kJ/mol

The photonenergy equals theenergy change forthe electron, but it

is a positivenumber.

Ephoton = 29.6 kJ/mol


Ephoton = 985 kJ/mol

Energy (kJ/mol)

250 500 750 1000 12500

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Here is what we will observe. Wewill see three specific

wavelengths of light beingemitted from the atom,

corresponding to the threephoton energies we calculated.

E = 29.6kJ/mol

(4040 nm)

E = 246.1kJ/mol

(486 nm)

E = 985kJ/mol

(121 nm)

Energy (kJ/mol)250 500 750 1000 12500

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250 500 750 1000 12500

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If we compare these wavelengthswith the actual line spectrum of

hydrogen, we find an exactmatch!

This line wasproducedwhen theelectronjumped fromlevel 2 tolevel 1



Energy (kJ/mol)250 500 750 1000 12500

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Wavelength (nm)

In fact, every line in the hydrogen line spectrum corresponds to an electron jump.

The energy of the photon equals the amount of energy the electron loses (thechange in the electron energy).

Ephoton = |ΔEelectron |

Energy (kJ/mol)250 500 750 1000 12500

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Wavelength (nm)

Let’s see how the lines in the spectrum match up with electron transitions (jumpsbetween energy levels).

We’ll begin with the Lyman series.

Energy (kJ/mol)1000 1050 1100 1150 1250950

88.61100110120 115 105125.9

1200

Wavelength (nm)

Here is an expanded view of the Lyman series, withthe wavelengths and energies of some lines.

We can match each of these lines with an electrontransition.

1300 1350

90

121.5 nm984.7 kJ/mol

102.5 nm1167 kJ/mol

97.18 nm1231 kJ/mol

94.90 nm1260 kJ/mol

93.71 nm1277 kJ/mol

93.01 nm1286 kJ/mol

series limit91.11 nm

1313 kJ/mol

95


Line 1

121.5 nm

The first linein the Lymanseries…

-328.2 kJ/mol

0 kJ/mol

ΔEelectron = -984.8 kJ/mol

-1313 kJ/mol1

2

∞

…is produced byelectrons movingfrom level 2 tolevel 1.


Line 2

102.5 nm

The second linein the Lymanseries…

-145.9 kJ/mol

0 kJ/mol

ΔEelectron = -1167 kJ/mol

-1313 kJ/mol1

3

∞



Line 3

97.18 nm

The third line inthe Lymanseries…

-82.1 kJ/mol0 kJ/mol


-1313 kJ/mol1

4∞



Line 4

94.90 nm

The fourth linein the Lymanseries…

-52.5 kJ/mol0 kJ/mol


-1313 kJ/mol1

5∞



series limit

91.11 nm

The series limitfor the Lymanseries…

0 kJ/mol


-1313 kJ/mol1

∞

…is produced byelectrons movingfrom the “infinitieth”energy level tolevel 1.

Energy (kJ/mol)

1000 1050 1100 1150 1250950

88.61100110120 115 105125.9

1200

Any line in the line spectrum corresponds to an electrontransition (a jump from one level to another).

For the Lyman series, the electron always ends up in level 1.

This diagram shows the electron transitions that correspondto the first six lines of the Lyman series, plus the series limit.

1300 1350

9095

level 2↓

level 1

5↓1

6↓1

∞↓1

7↓1

level 3↓

level 1

level 4↓

level 1

wavelength (nm)

Energy (kJ/mol)

250 500 750 1000 12500

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Now that we’ve explored the Lyman series,let’s apply the same ideas to the other serieswe observe in the line spectrum of hydrogen.

If will help us see the individual series if weexpand that section of the line spectrum….

All other series(Balmer, Paschen,

etc.)

Lyman series

Energy (kJ/mol)100 250 350200150 3000 50

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Wavelength (nm)∞

Here is the region of the spectrumfrom 0 to 350 kJ/mol.

The four lines we can observe withour eyes are shown in their actual

colors.

Energy (kJ/mol)100 250 350200150 3000 50

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Wavelength (nm)∞

Let’s start with the Balmer series.This series contains the second mostenergetic photons (after the Lyman

series).

Balmer series

Energy (kJ/mol)100 250 350200150 3000 50

34240050060070080090010001500 120020005000 300010000

Wavelength (nm)∞ 656 nm182.3 kJ/mol

486 nm246.2 kJ/mol

434 nm275.7 kJ/mol

410 nm291.8 kJ/mol

397 nm301.5 kJ/mol

series limit364 nm

328.2 kJ/mol

Here is the Balmer series with the wavelengthsand energies of a few of its lines.

Let’s match these photon energies with the ΔEvalues for the electron.


Line 1

656 nm

The first linein the Balmerseries…

0 kJ/mol

ΔEelectron = -182.3 kJ/mol-328.2 kJ/mol2

∞


-145.9 kJ/mol3


Line 2

486 nm

The second linein the Balmerseries…

0 kJ/mol

ΔEelectron = -246.1 kJ/mol-328.2 kJ/mol2

∞


-82.1 kJ/mol4


Line 3

434 nm

The third linein the Balmerseries…

0 kJ/mol


-328.2 kJ/mol2

∞


-52.5 kJ/mol5

Energy (kJ/mol)100 250 350200150 3000 50

34240050060070080090010001500 120020005000 300010000

Wavelength (nm)∞

level 3↓

level 2

6↓2

7↓2

∞↓2

level 4↓

level 2

5↓2

The Balmer series corresponds to a new set of electron transitions.

For the Balmer series, the electron always ends up in level 2.

This diagram shows the electron transitions that correspond to thefirst five lines of the Balmer series, plus the series limit.

Energy (kJ/mol)250 500 750 1000 12500

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Wavelength (nm)

We now know how the Lyman and Balmer series lines are formed.

The Lyman series is produced by electrons dropping from higherlevels into level 1.

The Balmer series is produced by electrons dropping from higherlevels into level 2.

2↓1

3↓1

4↓1

5↓1

3↓2

4↓2

5↓2

Balmer series Lyman series

250 500 750 1000 12500Energy (kJ/mol)

Balmer series182 – 328 kJ/mol

Lyman series985 – 1313 kJ/mol

2

∞

SmallΔE values

1

∞

Very largeΔE values

The Lyman series falls in a muchhigher energy range than the

Balmer series, because electronslose a very large amount of

energy when they drop into level1. All of the possible drops intolevel 2 release far less energy.

Energy (kJ/mol)250 500 750 1000 12500

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Wavelength (nm)

We can analyze the remaining series in the same fashion.However, all of the remaining series overlap with one another tosome extent, so it becomes difficult to tell which lines belong to

which series.

Let’s expand the remaining series to make them a little easier to see.

All otherseries

(Paschen,Brackett,Pfund,etc.)

Lyman series(All transitions thatend up in the 1st

level.)

Balmer series(All transitions thatend up in the 2nd

level.)

Paschen series

Brackett series

This diagram highlights the Paschen series (series#3) and the Brackett series (series #4).

Notice that these two series overlap one another, andthe Brackett series overlaps with some of the lower-

energy series.

Energy (kJ/mol)

Wavelength (nm)

∞

100 1500 50

90010001500 120020005000 300010000

Here is the Paschen series with all of the other seriesremoved.

Energy (kJ/mol)

Wavelength (nm)∞

100 1500 50

90010001500 120020005000 300010000

Energy (kJ/mol)

Wavelength (nm)∞

The Paschen series contains all of the photons produced when theelectron drops into level 3.

For example, the first line in this series (at 63.8 kJ/mol = 1874 nm) isproduced when the electron drops from level 4 to level 3.

100 1500 50

80090010001500 120020005000 300010000

level 4↓

level 3

level 5↓

level 3

level 6↓

level 3

7↓3

8↓3

4

3 -145.9 kJ/mol

-82.1 kJ/mol

ΔE = -63.8 kJ/mol

Here is the Brackett series with all other series removed.

Energy (kJ/mol)

Wavelength (nm)∞

100 1500 50

90010001500 120020005000 300010000

The Brackett series contains all of the photons produced when theelectron drops into level 4.

For example, the second line in this series (at 45.6 kJ/mol = 2624 nm)is produced when the electron drops from level 6 to level 4.

6

4 -82.1 kJ/mol

-36.5 kJ/mol5 ΔE = -45.6 kJ/mol

Energy (kJ/mol)

Wavelength (nm)∞

100 1500 50

90010001500 120020005000 300010000

level 5↓

level 4

level 6↓

level 4

7↓4

8↓4

9↓4

Lyman? → 1

Balmer? → 2

Paschen? → 3

Brackett? → 4

Pfund? → 5

Each series in the line spectrum of hydrogen correlates with thefinal energy level for an electron transition.

Lyman series: transitions that end up in level 1Balmer series: transitions that end up in level 2

Paschen series: transitions that end up in level 3Brackett series: transitions that end up in level 4

Pfund series: transitions that end up in level 5

Let’s practice!Can we answer these questions…

Where is the fifth line of the Paschen series?What is the corresponding electron transition?

What is the energy of this line?

First, we need to find the Paschen series. It is the third series,counting from the right (i.e. it has the lines with the third-highest

energies).

You should learn the sequence: Lyman, Balmer, Paschen (fromhighest energy to lowest).

Paschenseries(in red)

Lymanseries

Balmerseries

Next, we need to find the fifth line in the series. The lines in eachseries are numbered from left to right (from lowest energy to highest

energy).

Now we need to identify the electron transition that produces this line.

1 2 3 4 5 (in yellow)

The light we observe is the result of the electron moving between twolevels. The Paschen series is the third series, which means that forall lines in the Paschen series, the electron ends up in level 3. But

where does it start?

??

3 -145.9 kJ/mol

We can work this out systematically.For the first line, the electron makes the smallest possible jump: level 4 to level 3

For the second line, the electron jumps from level 5 to level 3Third line: level 6 to level 3

Fourth line: level 7 to level 3Fifth line: level 8 to level 3

8

3

4

5

67

12

3 4 5

To get the energy of this light, we subtract the energies of the two levels.ΔEelectron = (-145.9 kJ/mol) – (-20.5 kJ/mol) = -125.4 kJ/mol

The photon energy is positive: Ephoton = 125.4 kJ/mol

8

3

-20.5 kJ/mol

-145.9 kJ/mol


All other elements have line spectra, as do all ions that contain atleast one electron.

What can we tell from their line spectra?

???????

Ions that have only one electron show line spectra that are similar tothat of hydrogen.

Examples are He+, Li2+, and Be3+.

However, the energies are much larger.

The energies of all lines are increased by a factor of Z2, where Z isthe atomic number of the element.

1000 2000 3000 4000 50000

∞

1000 2000 3000 4000 50000

255075100200500∞

Here is a comparison of the line spectra of H and He+

For He, the atomic number (Z) is 2.Therefore, the energies of all helium lines are 4 times as large as they

are for hydrogen (22 = 4)

Wavelength (nm)

Energy (kJ/mol)

H

He+

3040

255075100200 3040

1000 2000 3000 4000 50000

∞

1000 2000 3000 4000 50000

255075100200500∞

For example, the energies of the Lyman series for H rangefrom 984.8 kJ/mol to 1313 kJ/mol.

The energies of the corresponding series in the He+ spectrum rangefrom 3939 kJ/mol to 5252 kJ/mol.

984.8 kJ/mol x 22 = 3939 kJ/mol 1313 kJ/mol x 22 = 5252 kJ/mol

Wavelength (nm)

Energy (kJ/mol)

H

He+

3040

255075100200 3040

x 22

Energy (kJ/mol)100 250 350200150 3000 50

34240050060070080090010001500 120020005000 300010000

Wavelength (nm)∞

However, line spectra for uncharged elements other thanhydrogen do not resemble the line spectrum for hydrogen,

and they do not fit any simple pattern.

Here is the line spectrum of uncharged helium as it appearsto our eyes. Remember that we can only see wavelengths

from 400 nm to 700 nm.

Energy (kJ/mol)100 250 350200150 3000 50

34240050060070080090010001500 120020005000 300010000

Wavelength (nm)∞

The line spectrum of sodium (shown here) is unusually simple.

There are many other lines, but they are so faint and theprominent yellow line at 590 nm is so bright that the other lines are

usually not visible.

What can we tell from this spectrum?

Energy (kJ/mol)100 250 350200150 3000 50

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Wavelength (nm)∞

The photon energy of 590 nm light is 203 kJ/mol.

This tells us that sodium atoms can lose 203 kJ/mol of energy.

Somewhere among the allowed energy levels of Na, there mustbe two levels that are 203 kJ/mol apart.

But we cannot determine the energies of these two levels, orwhether there are other levels between them.

ΔEelectron =-203 kJ/mol

Einitial = ????

Efinal = ????


Energy (kJ/mol)100 250 350200150 3000 50

34240050060070080090010001500 120020005000 300010000

Wavelength (nm)∞

However, other measurements have shown that the lowerenergy level for this electron transition is -495 kJ/mol.

Therefore, we can calculate that the upper energy level must be-292 kJ/mol.

Line spectra tell us how far apart energy levels are, but theydo not give us the actual energies of the levels.

ΔE = -203 kJ/mol

Einitial = -292 kJ/mol

Efinal = -495 kJ/mol

Line Spectra and Energy Levels - CCSF...500 Wavelength (nm) 350 400 450 550 600 650 700 750 When we...

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