Niels Bohr and the quantum atom Contents: Problems in nucleus land Spectral lines and Rydberg’s formula Photon wavelengths from transition energies Electron.

Niels Bohr and the quantum atom

Contents:•Problems in nucleus land•Spectral lines and Rydberg’s formula•Photon wavelengths from transition energies•Electron in a box•Schrödinger•Limitations of Bohr’s model

Niels Bohr1881 - 1962

Problems with the Rutherford Atom

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• Acceleration/Radiation• Spectral Lines

Spectral lines•Energy from excited atoms•demo

H

He

Sun

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Rydberg’s Formula: (FYI)

1/ = R(1/22 - 1/n2), n = 3, 4, ...(Balmer) (Visible)1/ = R(1/12 - 1/n2), n = 2, 3, ...(Lyman) (UV)1/ = R(1/32 - 1/n2), n = 4, 5, ...(Paschen) (IR)(R = 1.097 x 10-7 m-1)

Bohr’s Quantum Atom• Only certain orbits are allowed “stationary states”• Electron transitions create photons

1/ = R(1/22 - 1/n2), n = 3, 4, ...(Balmer) (Visible)1/ = R(1/12 - 1/n2), n = 2, 3, ...(Lyman) (UV)1/ = R(1/32 - 1/n2), n = 4, 5, ...(Paschen) (IR)

Example: What is the wavelength of the first Lyman line?

The first Lyman line is a transition from -3.4 eV to -13.6 eV, so it releases 10.2 eV of energy. A photon with this energy has this wavelength:E = (10.2)(1.602E-19) = 1.63404E-18 JE = hc/λ, λ = hc/E = (6.626E-34)(3.00E8)/(1.63404E-18) = 1.21649E-07 m = 122 nm

Whiteboards: Bohr Photons

1 | 2 | 3 | 4

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What possible photon energies can you get from these energy levels? (there are 6 different ones)

-14.0 eV

-9.0 eV

-6.0 eV-5.0 eV

1, 4, 9, 3, 8, and 5 eV

1 4 9 3 8 5

97 nm W

E = hf = hc/ E = -.85 - -13.6 = 12.75 eV E = (12.75 eV)(1.602E-19J/eV) = 2.04E-18J = hc/E = 97.3 = 97 nm

What is the wavelength of the photon released from the third Lyman spectral line (from -.85 to -13.6 eV)?

490 nm W

E = hf = hc/ E = -0.85 -3.4 = 2.55 eV E = (2.55 eV)(1.602E-19J/eV) = 4.09E-19J = hc/E = 487 = 490 nm

What is the wavelength of the photon released from the second Balmer spectral line (from –0.85 to -3.4 eV)?

12.1 eV W

E = hf = hc/ (6.626E-34)(3.00E8)/(86.4E-9) = 2.30069E-18 J(2.30069E-18 J)/(1.602E-19) = 12.1 eV

This could be the second Lyman line

An 102.5 nm photon is emitted. What is the energy of this photon in eV, and what transition occurred?

“Electron in a box”

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• Why are only certain orbits allowed? (Demo)

nL2

h

p mp

Ek 2

2

2

22

8 Lm

hnE

ek

6 x 10-18 J W

Ek = 12(6.626E-34)2/(8(9.11E-31)(.1E-9)2) = 6.02413E-18 J (≈ 38 eV)

Try this: What would be the kinetic energy of an electron in the ground state of a “box” that is about 0.1 nm in length?me = 9.11x10-31 kg

2

22

8 Lm

hnE

ek

Schrödinger and the quantum atom

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Schrödinger solves for hydrogen atomThe electron is represented by a waveCan only be solved for H, singly ionised He

Limitations of Bohr’s model

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• Works well for H, but doesn’t even work for He• Did not explain

• Spectral fine structure• Brightness of lines• Molecular bonds

• Theory was not complete.• But otherwise it generally kicked tuckus