IB Chemistry on Line Emission Spectrum, Bohr Model and Electromagnetic Spectrum

Tutorial on Electromagnetic
Radiation, Emission Line spectrum and
Bohr Model.

Prepared by
Lawrence Kok
http://lawrencekok.blogspot.com

Electromagnetic Spectrum
Electromagnetic spectrum ranges from Radiowaves to Gamma waves.
- Form of energy
- Shorter wavelength -> Higher frequency -> Higher energy
- Longer wavelength -> Lower frequency -> Lower energy

- Form of energy

Wavelength, λ - long 
Frequency, f

- low 

Wavelength, λ - short 
Inverse relationship between- λ and f

Frequency, f

- high 

- Form of energy

Wavelength, λ - long 
Frequency, f

- low 

Wavelength, λ - short 
Inverse relationship between- λ and f

Frequency, f

- high 

Electromagnetic radiation
•
Travel at speed of light, c = fλ -> 3.0 x 108 m/s
•
Light Particle – photon have energy given by -> E = hf
•
Energy photon - proportional to frequency

Plank constant
• proportionality constant bet energy and freq

Excellent video wave propagation
Click here to view.

Electromagnetic Wave propagation.
•
•
•

Moving charges/particles through space
Oscillating wave like property of electric and magnetic field
Electric and magnetic field oscillate perpendicular to each other and perpendicular to
direction of wave propagation.

Electromagnetic wave propagation

Click here to view video

•
•
•




Wave

Wave – wavelength and frequency
- travel at speed of light

•
•
•




Violet

λ = 410nm
f = c/λ
= 3 x 108/410 x 10-9
= 7.31 x 1014 Hz
E = hf
= 6.626 x 10-34 x 7.31 x 1014
= 4.84 x 10-19 J

Wave


Red

λ = 700nm
f = c/λ
= 3 x 108/700 x 10-9
= 4.28 x 1014 Hz
E = hf
= 6.626 x 10-34 x 4.28 x 1014
= 2.83 x 10-19 J

•
•
•
Is it a particle or Wave?

Click to view video -Wave-particle duality

Wave


•
•
•
Is it a particle or Wave?

Click to view video -Wave-particle duality

Wave


Simulation on Electromagnetic Propagation

Click here to view simulation



Electromagnetic Wave
Violet

Red

λ = 410nm

λ = 700nm

f = c/λ
= 3 x 108/410 x 10-9
= 7.31 x 1014 Hz

f = c/λ
= 3 x 108/700 x 10-9
= 4.28 x 1014 Hz

Wavelength – Distance bet two point with same phase, bet crest/troughs – unit nm
Frequency – Number of cycle/repeat per unit time (cycles in 1 second) – unit Hz

Violet

Red

λ = 410nm

λ = 700nm

f = c/λ
= 3 x 108/410 x 10-9
= 7.31 x 1014 Hz

f = c/λ
= 3 x 108/700 x 10-9
= 4.28 x 1014 Hz


Which wave have higher frequency, if both have same speed reaching Y same time?
Violet

X

Y

Red

Violet

Red

λ = 410nm

λ = 700nm

f = c/λ
= 3 x 108/410 x 10-9
= 7.31 x 1014 Hz

f = c/λ
= 3 x 108/700 x 10-9
= 4.28 x 1014 Hz


Which wave have higher frequency, if both have same speed reaching Y same time?
Violet

X

Click here on excellent video red /violet wave

Click here to view video energy photon

Y
Light travel same speed
Red flippers – long λ - less frequent
Violet shoes – short λ - more frequent

Red

Continuous Spectrum Vs Line Spectrum

Continuous Spectrum :
Light spectrum with all wavelength/frequency
Emission Line Spectrum :
• Spectrum with discrete wavelength/ frequency
• Emitted when excited electrons drop from higher to lower energy level
Absorption Line Spectrum :
• Spectrum with discrete wavelength/frequency
• Absorbed when ground state electrons are excited


Atomic Emission
Electrons from excited state

Excited state

Emit radiation
when drop to ground state

Radiation emitted

Emission Spectrum

Ground state

http://www.astrophys-assist.com/educate/orion/orion02.htm


Atomic Emission Vs Atomic Absorption Spectroscopy
Electrons from excited state

Excited state

Electrons in excited state

Emit radiation
when drop to ground state

Radiation absorbed
Radiation emitted
Absorb radiation
to excited state

Emission Spectrum

Ground state

http://www.astrophys-assist.com/educate/orion/orion02.htm

Electrons from ground state

Line Emission Spectroscopy
Line Emission Spectra for Hydrogen
Energy supplied to atoms
• Electrons excited - ground to excited states
• Electrons exist fixed energy level (quantum)
• Electrons transition from higher to lower,
emit energy of particular wavelength/frequency - photon
• Higher the energy level, smaller the difference in energy
bet successive energy level.
• Spectrum converge (get closer) with increase freq.
• Lines spectrum converge- energy levels also converge
• Ionisation energy determined (Limit of convergence)

UV region
Lyman Series
n=∞ → n= 1

Visible region
Balmer Series
n=∞ → n= 2

IR region
Paschen Series
n=∞ → n= 3


UV region
Lyman Series
n=∞ → n= 1

Visible region
Balmer Series
n=∞ → n= 2

IR region
Paschen Series
n=∞ → n= 3

Line Emission Spectra
• Energy supplied
• Electrons surround nucleus in allowed energy states (quantum)
• Excited electron return to lower energy level, photon with
discrete energy/wavelength (colour) given out.
• Light pass through spectroscope (prism/diffraction grating) to separate
out diff colours

N= 6-2
410nm

N= 5-2
434nm

N= 4-2
486nm

N = 3-2,
656nm

Visible region- Balmer Series


UV region
Lyman Series
n=∞ → n= 1

Visible region
Balmer Series
n=∞ → n= 2

IR region
Paschen Series
n=∞ → n= 3

Line Emission Spectra
• Energy supplied
• Electrons surround nucleus in allowed energy states (quantum)
• Excited electron return to lower energy level, photon with
discrete energy/wavelength (colour) given out.
• Light pass through spectroscope (prism/diffraction grating) to separate
out diff colours
Videos on line emission

N= 6-2
410nm



N= 5-2
434nm

N= 4-2
486nm

N = 3-2,
656nm

Visible region- Balmer Series

Hydrogen Emission Spectroscopy – Visible region (Balmer Series)

Excited state

5
4
3

2
Visible region
Balmer Series
n=∞ → n= 2
Ground state

Click here for detail notes

1

Click here video line emission spectrum

Hydrogen discharge tube

Excited state

5
4
3

2
Visible region
Balmer Series
n=∞ → n= 2
Ground state


1


Hydrogen Emission Spectroscopy

Hydrogen discharge tube

Excited state

Hydrogen Emission Spectroscopy

5
4
3
n= 5-2

n = 3-2

n= 4-2

2
λ = 434nm

Visible region
Balmer Series
n=∞ → n= 2
Ground state

1

f = c/λ
= 3 x 108/434 x 10-9
= 6.90 x 1014 Hz

λ = 486nm

λ = 656nm

f = c/λ
= 3 x 108/656 x 10-9
= 4.57 x 1014 Hz

E = hf
= 6.62 x 10-34 x 6.90 x 1014
= 4.56 x 10-19 J

More energetic violet line



E = hf
= 6.62 x 10-34 x 4.57 x 1014
= 3.03 x 10-19 J

Less energetic red line

Bohr Model for Hydrogen Atom – Ionization Energy
Bohr Model

Energy level

Electronic Transition bet levels

Niels Bohr Model (1913)
•
•
•

Electrons orbit nucleus.
Orbits with discrete energy levels – Quantized.
Transition electron bet diff levels by absorb/emit radiation
with frequency, f determined by energy diff bet levels -ΔE = hf
• Energy light emit/absorb equal to diff bet energy levels

Bohr Model

Energy level


•
•
•


Light emitted equal to difference
bet energy levels, -ΔE = hf

Ionization energy
Transition electron from 1 ->∞

∞

Plank equation

Higher energy level n, smaller the difference
in energy bet successive energy level.

5
4
3

ΔE = hf

Light given off

2

Light energy - ΔE = hf
Frequency = ΔE/h

1

Energy level

Bohr Model


•
•
•


Light emitted equal to difference
bet energy levels, -ΔE = hf

Ionization energy

∞

Plank equation

Higher energy level n, smaller the difference
in energy bet successive energy level.

5
4
3

ΔE = hf

2

Light given off

Frequency = ΔE/h

1
line converge

UV region
Lyman Series
n=∞ → n= 1

Increase freq 
Line spectrum converge (get closer) with increase freq
Ionisation energy determined (Limit of convergence)

line converge

Visible region
Balmer Series
n=∞ → n= 2

Increase freq 
Line spectrum converge (get closer) with increase freq

Lines in spectrum converge- energy levels also converge

Energy Level/Ionization Energy Calculation

∞

Formula - energy level, n (eV)

n = energy level

5

5

4

4

3

3

1

2

2

Energy level, n= 3
= -13.6/n2
= -13.6/32
= -1.51 eV

3

Energy level, n= 2
= -13.6/n2
= -13.6/22
= -3.4 eV

4

Energy level, n= 1
= -13.6/n2
= -13.6/1
= -13.6 eV

2

constant
10-19 J

1eV – 1.6 x
h = 6.626 x 10-34 Js

1

1


∞


n = energy level

5

5

4

4

3

3

1

2

2

Energy level, n= 3
= -13.6/n2
= -13.6/32
= -1.51 eV

3

Energy level, n= 2
= -13.6/n2
= -13.6/22
= -3.4 eV

4

Energy level, n= 1
= -13.6/n2
= -13.6/1
= -13.6 eV

2

constant
10-19 J

1eV – 1.6 x
h = 6.626 x 10-34 Js

1

1

Higher energy level, n
- more unstable electron
- More + ve ( less negative)
- More energetic

5

6

Ionization energy

Lower energy level, n
- more stable electron
- more – ve (-13.6eV)
- Less energetic

Energy difference bet level 3 to 2

∞


n = energy level

5
1

4

4

3

Energy difference, n= 3-2
= -1.51 – (-3.4) eV
= 1.89 eV
= 1.89 x 1.6 x 10-19 J
= 3.024 x 10-19 J

5

3

1

2

Energy level, n= 3
= -13.6/n2
= -13.6/32
= -1.51 eV

3

Energy level, n= 2
= -13.6/n2
= -13.6/22
= -3.4 eV

4

Energy level, n= 1
= -13.6/n2
= -13.6/1
= -13.6 eV

Light given off
2

Frequency, f = ΔE/h

2

2

constant
3

10-19 J

f = 3.024 x 10-19 /6.626 x 10-34
= 4.56 x 1015 Hz

4

λ = c/f
= 3 x 108/4.56 x 1015
= 657 x 10-9
= 657nm

1eV – 1.6 x
h = 6.626 x 10-34 Js

1

1

Higher energy level, n
- more unstable electron
- More + ve ( less negative)
- More energetic

5

Light given off

6

Ionization energy

Lower energy level, n
- more stable electron
- more – ve (-13.6eV)
- Less energetic

Ionization Energy for Hydrogen Atom

∞

1

n = energy level

5

∞
5

4

4

3

Ionization energy
Min energy to remove 1 mole electron from
1 mole of element in gaseous state
M(g)  M+ (g) + e

3

2

Ionization energy

Energy Absorb

2

2
3

Energy level, n= ∞
= -13.6/n2
= -13.6/∞
= o eV

4

Energy level, n= 1
= -13.6/n2
= -13.6/1
= -13.6 eV

electron
Light/photon ABSORB by electron

1

1


∞

1

n = energy level

5

∞
5

4

4

3

Ionization energy
M(g)  M+ (g) + e

3

2

Ionization energy

Energy Absorb

2

2
3

= -13.6/n2
= -13.6/∞
= o eV

4

Energy level, n= 1
= -13.6/n2
= -13.6/1
= -13.6 eV

electron

1

1

5

6

Energy difference, n= 1-> ∞
= 0 – (-13.6) eV
= 13.6 eV
= 13.6 x 1.6 x 10-19 J
= 2.176 x 10-18 J for 1 electron

Energy absorb for 1 MOLE electron
- 2.176 x 10-18 J - 1 electron
- 2.176 x 10-18 x 6.02 x 1023 J - 1 mole
- 1309kJ mol-1

Light given off, electronic transition from high -> low level
Light given off

Energy Released

∞

1

n = energy level

5

Energy difference, n= 3-2
= -1.51 – (-3.4) eV
= 1.89 eV
= 1.89 x 1.6 x 10-19 J
= 3.024 x 10-19 J

2

4

4

3

Energy difference bet level 3 to 2

1

∞
5

Ionization energy
M(g)  M+ (g) + e

3

2

Ionization energy

Light given off

Energy Absorb

2

2
3

3

f = 3.024 x 10-19 /6.626 x 10-34
= 4.56 x 1015 Hz

4

Energy level, n= 1
= -13.6/n2
= -13.6/1
= -13.6 eV


4

= -13.6/n2
= -13.6/∞
= o eV

electron

1
5

1

λ = c/f
= 3 x 108/4.56 x 1015
= 657 x 10-9
= 657nm

Light given off

5

6

Energy difference, n= 1-> ∞
= 0 – (-13.6) eV
= 13.6 eV
= 13.6 x 1.6 x 10-19 J

- 2.176 x 10-18 J - 1 electron
- 2.176 x 10-18 x 6.02 x 1023 J - 1 mole
- 1309kJ mol-1


n = energy level

Energy/Wavelength – Plank/Rydberg Equation

∞

Formula – Plank Equation

5

5

4

4

ΔE = hf

3

3

∞

Rydberg Equation to find wavelength

2

2

R = Rydberg constant
R = 1.097 x 107 m-1
1

1

Nf = final n level
Ni = initial n level

Energy photon- high -> low level
1

Electron transition from 3 -> 2


n = energy level


Light given off


5
Rydberg Eqn find wavelength emit

∞
5

4

4

ΔE = hf

3

3

∞


2

2

2

nf = 2, ni = 3
R = 1.097 x 107
3

R = 1.097 x 107 m-1
4

5

λ = 657 x 10-9
= 657 nm
f = c/λ
= 3 x 108/657 x 10-9
= 4.57 x 1014 Hz

Light given off

1

1

Nf = final n level

Energy photon- high -> low level
1



n = energy level


Light given off


5

∞
5

4

4

ΔE = hf

3

3

∞


2

2

2

nf = 2, ni = 3
R = 1.097 x 107
3

R = 1.097 x 107 m-1
4

5

λ = 657 x 10-9
= 657 nm
f = c/λ
= 3 x 108/657 x 10-9
= 4.57 x 1014 Hz

1

1

Click here on energy calculation

Light given off



Nf = final n level

Light given off, high -> low level
Energy photon-electronic transition from high -> low level
1


∞

n = energy level

5

Light given off

4

4

3

3

2


∞
5

2

2

nf = 2, ni = 3
R = 1.097 x 107
3

4

5

λ = 657 x 10-9
= 657 nm
f = c/λ
= 3 x 108/657 x 10-9
= 4.57 x 1014 Hz

Light given off

1

1

1


∞

1

n = energy level

5

Light given off

∞
5

4

4

3

Ionization energy
M(g)  M+ (g) + e

3

Ionization energy
Transition electron from 1 -> ∞

1

2

Energy Absorb

2

2

nf = 2, ni = 3
R = 1.097 x 107

Rydberg Eqn find ionization energy

3

3

electron
4

λ = 657 x 10-9
= 657 nm

1

nf = ∞, ni = 1
R = 1.097 x 107

1
4

5

f = c/λ
= 3 x 108/657 x 10-9
= 4.57 x 1014 Hz

Light given off

1


∞

1

n = energy level

5

Light given off

∞
5

4

4

3

Ionization energy
M(g)  M+ (g) + e

3

Ionization energy
Transition electron from 1 -> ∞

1

2

Energy Absorb

2

2

nf = 2, ni = 3
R = 1.097 x 107

Rydberg Eqn find ionization energy

3

3

electron
4

1

λ = 657 x 10-9
= 657 nm

nf = ∞, ni = 1
R = 1.097 x 107

1
4

5

f = c/λ
= 3 x 108/657 x 10-9
= 4.57 x 1014 Hz

λ = 9.11 x 10-8
7

Light given off

- 2.179 x 10-18 J - 1 electron
- 2.179 x 10-18 x 6.02 x 1023 J - 1 mole
- 1312kJ mol-1

6

Energy, E = hf
= 6.626 x 10-34 x 3.29 x 1015

5

f = c/λ
= 3 x 108/9.11 x 10-8
= 3.29 x 1015 Hz


Emission Line Spectrum
• Excited electrons drop from higher to lower energy level

Continuous Spectrum

Excellent simulation on emission spectrum

Click here to view excellent simulation


Emission line spectrum for different elements

Click here spectrum for diff elements

Click here spectrum for diff element

Video on quantum mechanics

Click here on quantum mechanic, structure of atom

Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/

Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com

IB Chemistry on Line Emission Spectrum, Bohr Model and Electromagnetic Spectrum

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