ib chemistry on line emission spectrum, bohr model and electromagnetic spectrum
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IB Chemistry on Line Emission Spectrum, Bohr Model and Electromagnetic SpectrumTRANSCRIPT
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
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
Inverse relationship between- λ and f Wavelength, λ - long
Frequency, f - low
Wavelength, λ - short Frequency, f - high
Plank constant • proportionality constant bet energy and freq
Excellent video wave propagation Click here to view.
Electromagnetic Wave propagation.
Wave
Electromagnetic radiation
Electromagnetic radiation • 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
Wave – wavelength and frequency - travel at speed of light
Violet
λ = 410nm
Red
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
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
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Electromagnetic Wave propagation.
Wave
Electromagnetic radiation
Simulation on Electromagnetic Propagation
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Electromagnetic radiation • 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.
Click to view video -Wave-particle duality
Is it a particle or Wave?
Wave – wavelength and frequency - travel at speed of light
Electromagnetic Wave
Violet
λ = 410nm
Red
f = c/λ = 3 x 108/410 x 10-9
= 7.31 x 1014 Hz
λ = 700nm
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
Y
Red
X
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
Click here on excellent video red /violet wave
Light travel same speed Red flippers – long λ - less frequent Violet shoes – short λ - more frequent
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http://www.astrophys-assist.com/educate/orion/orion02.htm
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 Vs Atomic Absorption Spectroscopy
Ground state
Excited state Electrons from excited state
Emit radiation when drop to ground state
Radiation emitted
Emission Spectrum
Electrons from ground state
Absorb radiation to excited state
Electrons in excited state
Radiation absorbed
Continuous Spectrum Vs Line Spectrum
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)
N = 3-2, 656nm
N= 4-2 486nm
N= 5-2 434nm
N= 6-2 410nm
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
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Videos on line emission
Line Emission Spectroscopy
Ground state
Excited state
Hydrogen Emission Spectroscopy – Visible region (Balmer Series)
Line Emission Spectra for Hydrogen
Visible region Balmer Series n=∞ → n= 2
Hydrogen discharge tube Hydrogen Emission Spectroscopy
n = 3-2 n= 4-2 n= 5-2
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 4.57 x 1014
= 3.03 x 10-19 J
λ = 434nm λ = 486nm
f = c/λ = 3 x 108/434 x 10-9
= 6.90 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 Less energetic red line
2
1
3
4 5
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Bohr Model for Hydrogen Atom – Ionization Energy
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
Electronic Transition bet levels Energy level Bohr Model
1
2
3
4 5 ∞
Light emitted equal to difference bet energy levels, -ΔE = hf
Plank equation
Ionisation energy determined (Limit of convergence)
Line spectrum converge (get closer) with increase freq
Higher energy level n, smaller the difference in energy bet successive energy level.
Lines in spectrum converge- energy levels also converge
Visible region Balmer Series n=∞ → n= 2
Increase freq
UV region Lyman Series n=∞ → n= 1
ΔE = hf
Light energy - ΔE = hf Frequency = ΔE/h
Increase freq
Line spectrum converge (get closer) with increase freq
Ionization energy Transition electron from 1 ->∞
line converge line converge
Light given off
Energy Level/Ionization Energy Calculation
1
2
3
4
5 ∞
Formula - energy level, n (eV) Energy difference bet level 3 to 2
1
2
n = energy level
Lower energy level, n - more stable electron - more – ve (-13.6eV) - Less energetic
Higher energy level, n - more unstable electron - More + ve ( less negative) - More energetic
Energy level, n= 1 = -13.6/n2
= -13.6/1 = -13.6 eV
Energy level, n= 2 = -13.6/n2
= -13.6/22
= -3.4 eV
Energy level, n= 3 = -13.6/n2
= -13.6/32
= -1.51 eV
3
4
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
1eV – 1.6 x 10-19 J h = 6.626 x 10-34 Js
Light energy - ΔE = hf Frequency, f = ΔE/h
Frequency, f = ΔE/h f = 3.024 x 10-19 /6.626 x 10-34
= 4.56 x 1015 Hz
λ = c/f = 3 x 108/4.56 x 1015
= 657 x 10-9
= 657nm
Ionization energy Transition electron from 1 ->∞
constant
Light given off
Light given off
1
2
3
4
5
6
1
2
3
4
Ionization Energy for Hydrogen Atom
1
2
3
4
5 ∞
Ionization energy Min energy to remove 1 mole electron from 1 mole of element in gaseous state M(g) M+ (g) + e
Energy difference bet level 3 to 2
1
2
n = energy level
Energy level, n= 1 = -13.6/n2
= -13.6/1 = -13.6 eV
3
4
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
Light energy - ΔE = hf Frequency, f = ΔE/h
Frequency, f = ΔE/h f = 3.024 x 10-19 /6.626 x 10-34
= 4.56 x 1015 Hz
λ = c/f = 3 x 108/4.56 x 1015
= 657 x 10-9
= 657nm
Ionization energy Transition electron from 1 ->∞
Energy level, n= ∞ = -13.6/n2
= -13.6/∞ = o eV
∞
Energy Absorb
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
Energy Released
Light/photon ABSORB by electron
Light given off
Light given off
Light given off
electron
1
2
3
4
5
6
1
2
3
4
5
Energy/Wavelength – Plank/Rydberg Equation
ΔE = hf
1
2
3
4
5 ∞
Formula – Plank Equation Electron transition from 3 -> 2
1
2
n = energy level
R = Rydberg constant R = 1.097 x 107 m-1
3
4
5
Rydberg Equation to find wavelength
nf = 2, ni = 3 R = 1.097 x 107
λ = 657 x 10-9
= 657 nm
Nf = final n level Ni = initial n level
f = c/λ = 3 x 108/657 x 10-9
= 4.57 x 1014 Hz
Energy photon- high -> low level
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∞
Energy Level/Ionization Energy Calculation
Light given off
Light given off
Rydberg Eqn find wavelength emit
1
2
3
4
5
1
2
3
4
5 ∞
Electron transition from 3 -> 2
1
2
n = energy level
3
4
5
nf = 2, ni = 3 R = 1.097 x 107
λ = 657 x 10-9
= 657 nm
f = c/λ = 3 x 108/657 x 10-9
= 4.57 x 1014 Hz
Energy photon- high -> low level
∞
Ionization Energy for Hydrogen Atom
Ionization energy Min energy to remove 1 mole electron from 1 mole of element in gaseous state M(g) M+ (g) + e
Ionization energy Transition electron from 1 -> ∞
Energy Absorb
Rydberg Eqn find ionization energy
nf = ∞, ni = 1 R = 1.097 x 107
λ = 9.11 x 10-8
Energy absorb for 1 MOLE electron - 2.179 x 10-18 J - 1 electron - 2.179 x 10-18 x 6.02 x 1023 J - 1 mole - 1312kJ mol-1
Energy, E = hf = 6.626 x 10-34 x 3.29 x 1015
= 2.179 x 10-18 J for 1 electron
f = c/λ = 3 x 108/9.11 x 10-8
= 3.29 x 1015 Hz
Light/photon ABSORB by electron
electron
Light given off, electronic transition from high -> low level
Light given off
Light given off
Rydberg Eqn find wavelength emit
1
1
3
4
5 6 7
1
2
3
4
5
1
2
3
4
5
∞
1
2
n = energy level
3
4
5
Energy = hv (1 mole) = 6.626 x 10-34 x 32.26 x 1014 x 6.o2 x 1023
= 1312KJ mol-1
∞
Light/photon ABSORB by electron
electron
6 6 7 7
8 8
Calculating Ionization Energy from Lyman Series ( n = 1 to ∞ )
Excited energy level
Frequency, v / x 1014 s-1
ΔV / x 1014 s-1
2 24.66 4.57
3 29.23 1.60
4 30.83 0.74
5 31.57 0.40
6 31.97 0.24
7 32.21 0.16
8 32.37
Line spectrum converge (get closer) with increase freq
Ionization energy determined from (Limit of convergence)
Ionization energy Transition electron from 1 ->∞
Ionization energy Transition electron from 1 ->∞
Difference bet freq successive lines
Find freq v, at which it converge Δv = 0
Δv = 0
Plot graph v against Δv
Find Ionization Energy 1
2
3
Linear curve fit equation Δv = -0.5897v + 19.022
4
Freq, v when Δv = 0 -0.5897v + 19.022 = 0 v = 19.022 x 1014 s-1
0.5897 v = 32.26 x 1014s-1
5
Question adapted from Pearson
1
2
3
4
5
∞
1
2
n = energy level
3
4
5
Energy = hv (1 mole) = 6.626 x 10-34 x 32.86 x 1014 x 6.o2 x 1023
= 1312KJ mol-1
Energy for n level: ∞
6 6 7 7
8 8
Calculating Ionization Energy from Lyman Series ( n = 1 to ∞ )
Excited energy level, n
Excited energy level, 1/n2
Frequency V / x 1014 s-1
2 0.25 24.66
3 0.11 29.23
4 0.0625 30.83
5 0.04 31.57
6 0.027 31.97
7 0.02 32.21
8 0.015 32.37
Find freq v, when 1/n2 = 0
Plot graph v against 1/n2
Find Ionization Energy 1
2
3
Linear curve fit equation v = -32.84 (1/n2) + 32.86
4
Freq, v when 1/n2 = 0 v = -32.84 (1/n2) + 32.86 v = 32.86 x 1014 s-1
5
Question adapted from Pearson
2
6.13
nE
hvE
2
6.13
nhv
2
6.13
hnv
cnh
v
2
16.13
Plot graph v against 1/n2
Continuous Spectrum Light spectrum with all wavelength/frequency
Emission Line Spectrum • Spectrum with discrete wavelength/ frequency • Excited electrons drop from higher to lower energy level
Continuous Spectrum Vs Line Spectrum
Click here spectrum for diff elements Click here spectrum for diff element Click here on quantum mechanic, structure of atom
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Excellent simulation on emission spectrum
Emission line spectrum for different elements
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Video on quantum mechanics