lecture 16: bohr model of the atom reading: zumdahl 12.3, 12.4 outline –emission spectrum of...
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Lecture 16: Bohr Model of the Atom
• Reading: Zumdahl 12.3, 12.4
• Outline– Emission spectrum of atomic hydrogen.
– The Bohr model.– Extension to higher atomic number.
Photon Emission
System drops from a higher energy level to a lower one by spontaneously emitting a photon.
E = hc/
• If = 440 nm,= 4.5 x 10-19 J
Emission
“Continuous” spectrum “Quantized” spectrum
Any E ispossible
Only certain E are ‘allowed’transitions
E E
Emission spectrum of atomic H
Light Bulb:Continuous spectrum
Hydrogen Lamp:Discrete lines only
Quantized, not continuous
Balmer Model• Joseph Balmer (1885) first noticed that the frequency of visible lines (transitions) in the H atom spectrum could be reproduced by a formula where frequency (v) varies according to:
€
ν ∝1
22−
1
n2n = 3, 4, 5, …..
• The above equation predicts that as n increases, the frequencies become more closely spaced.
Rydberg Model
• Johann Rydberg extends the Balmer model by finding more emission lines outside the visible region of the spectrum (uv, ir):
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ν =Ry1
n12
−1
n22
⎛
⎝ ⎜
⎞
⎠ ⎟n1 = 1, 2, 3, …..
• This suggested that the energy levels of the H atom are proportional to 1/n2
n2 = n1+1, n1+2, …
Ry = 3.29 x 1015 1/s
The Bohr Model
• Niels Bohr uses the emission spectrum of hydrogen to develop a quantum model for H.
• Central idea: electron circles the “nucleus” in only certain allowed circular orbitals.
• Bohr postulates that there is Coulombic attraction between e- and nucleus (+). However, classical physics is unable to explain why an H atom doesn’t simply collapse, with the electron spiraling into the nucleus.
The Bohr model for the H atom is capable of reproducing the energy levels given by the empirical formulas of Balmer and Rydberg.
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E = −2.178x10−18JZ 2
n2
⎛
⎝ ⎜
⎞
⎠ ⎟ Z = atomic no. (1 for H)n = integer (1, 2, ….)
Note: Ry x h = -2.178 x 10-18 J
€
E = −2.178x10−18JZ 2
n2
⎛
⎝ ⎜
⎞
⎠ ⎟
Energy levels get closer together as n increases
for n = infinity, E = 0, so reference state is electron completely removed from the H atom.
We can use the Bohr model to predict what E is for any two energy levels:
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E = E final − E initial
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E = −2.178x10−18J1
n final2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟− (−2.178x10−18J)
1
ninitial2
⎛
⎝ ⎜
⎞
⎠ ⎟
€
E = −2.178x10−18J1
n final2
−1
ninitial2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
• Example: At what wavelength will emission from n = 4 to n = 1 for the H atom be observed?
€
E = −2.178x10−18J1
n final2
−1
ninitial2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
1 4
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E = −2.178x10−18J 1−1
16
⎛
⎝ ⎜
⎞
⎠ ⎟= −2.04x10−18J
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E = 2.04x10−18J =hc
λ
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=9.74x10−8m = 97.4nm
• Example: What is the longest wavelength of light that will result in removal of the e- from H?
€
E = −2.178x10−18J1
n final2
−1
ninitial2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
1
€
E = −2.178x10−18J 0 −1( ) = 2.178x10−18J
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E = 2.178x10−18J =hc
λ
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=9.13x10−8m = 91.3nm
Extension to Higher Z• The Bohr model can be extended to any single electron system….must keep track of Z.
Examples: He+ (Z = 2), Li+2 (Z = 3), etc.€
E = −2.178x10−18JZ 2
n2
⎛
⎝ ⎜
⎞
⎠ ⎟Z = atomic number
n = integer (1, 2, ….)
• Example: At what wavelength will emission from n = 4 to n = 1 for the He+ atom be observed?
€
E = −2.178x10−18J Z 2( )
1
n final2
−1
ninitial2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2 1 4
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E = −2.178x10−18J 4( ) 1−1
16
⎛
⎝ ⎜
⎞
⎠ ⎟= −8.16x10−18J
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E = 8.16x10−18J =hc
λ
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=2.43x10−8m = 24.3nm
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H > λHe+Note
: