Physics 1CModern Physics Lecture II

"Splitting the atom is like trying to shoot a gnat in the Albert Hall at night and using ten million

rounds of ammunition on the off chance of getting it. That should convince you that the atom will always be a sink of energy and never a reservoir of energy."

--Ernest Rutherford

Bohr AtomIn 1913, Neils Bohr explained atomic spectra by utilizing Rutherford’s Planetary model and quantization.

In Bohr’s postulate for the hydrogen atom, the electron moves in circular orbits around the proton.The Coulomb force provides the centripetal acceleration for continued motion.

Bohr AtomBut the electron could only have certain orbits that are stable.

In these orbits the atom does not emit energy in the form of electromagnetic radiation.Radiation is only emitted by the atom when the electron “jumps” between stable orbits.

Bohr AtomThe electron will move from a more energetic initial state to less energetic final state.

€

Ei − E f = hf

The frequency of the light emitted in the “jump” is related to the change in the atom’s energy.

If the electron is not “jumping” between allowed orbitals, then the energy of the atom remains constant.

Bohr AtomBohr then turned to conservation of energy of the atom in order to determine the allowed electron orbitals.The total energy of the atom will be:

€

Etot = KE + PEelec =12mev

2 − kee2

rBut the electron is undergoing centripetal acceleration:

€

Fcent = kee2

r2= me

v 2

r= macent

€

mev2 = ke

e2

r

Standing WavesRecall that the wavelength of an electron is given by the deBroglie formula:

€

λ =hp

=hmv

As the electron orbits the hydrogen atom and creates a stable orbit, a standing wave must form which restricts the circumference of the orbit to an integer number of deBroglie wavelengths.

Bohr AtomSolving the previous two equations for wavelength gives us:

we can define ħ to be h/2π.This gives us a velocity of:

€

v =nhmer

Substituting into the last equation from two slides before:

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menhmer⎛

⎝ ⎜

⎞

⎠ ⎟

2

= kee2

r

€

mev2 = ke

e2

r

Bohr AtomSolving for radius gives us:

The integer values of n give you the quantized Bohr orbits: Electrons can only exist in certain allowed orbits determined by the integer n.When n = 1, the orbit has the smallest radius, called the Bohr radius, ao.ao = 0.0529nm

€

rn =n2h2

mekee2

Bohr AtomRecall, from centripetal acceleration we found that:

Putting this back into energy we get:

€

mev2 = ke

e2

r

€

Etot =12mev

2 − kee2

r=12kee2

r⎛

⎝ ⎜

⎞

⎠ ⎟ − ke

e2

r

€

Etot = −12kee2

r⎛

⎝ ⎜

⎞

⎠ ⎟

But we can go back to the result for the radius (rn = n2ao) to get a numerical result.

Bohr Atom

This is energy of any orbit. Please note the negative sign in the equation.When n = 1, the total energy is –13.6eV.This is the lowest energy state and is called the ground state.The ionization energy is the energy needed to completely remove the electron from the atom.The ionization energy for hydrogen is 13.6eV.

€

Etot = −12ke

e2

n2ao

⎛

⎝ ⎜

⎞

⎠ ⎟ =

−13.6 eVn2

Bohr AtomSo, a general expression for the radius of any orbit in a hydrogen atom is: rn = n2ao

The energy of any orbit is:

If you would like to completely remove the electron from the atom it requires 13.6eV of energy.

Bohr AtomWhat are the first four energy levels for the hydrogen atom?When n = 1 => E1 = –13.6eV.When n = 2 => E2 = -13.6eV/22 = -3.40eV.When n = 3 => E3 = -13.6eV/32 = -1.51eV.When n = 4 => E4 = -13.6eV/42 = -0.850eV.Note that the energy levels get closer together as n increases (similar to how the wavelengths got closer in atomic spectra).When the atom releases a photon it will move from a final higher energy level (ni) to a final lower energy level (nf).

Bohr AtomThe energies can be compiled in an energy level diagram.As the atom is in a higher energy state and moves to a lower energy state it will release energy (in the form of a photon).The wavelength of this photon will be determined by the starting and ending energy levels.

Bohr AtomThe photon will have a wavelength λ and a frequency f:

To find the wavelengths between any energy jump (nf and ni) we can generalize the equation discovered by Balmer to:

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f =Ei − E f

h

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

λ 2i

2f

H n1

n1R1

Bohr AtomThe wavelength will be represented by a different series depending on your final energy level (nf).

For nf = 1 it is called the Lyman series (ni = 2,3,4...).

For nf = 2 it is called the Balmer series (ni = 3,4,5...).

For nf = 3 it is called the Paschen series (ni = 4,5,6...).

Atomic SpectraExampleWhat are the first four wavelengths for the Lyman, Balmer, and Paschen series for the hydrogen atom?

AnswerThe final energy level for either series will be nf = 1 (Lyman), nf = 2 (Balmer), and nf = 3 (Paschen).

Atomic SpectraAnswerTurn to the generalized Balmer equation:

For the Lyman series we have:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

λ 2i

2f

H n1

n1R1

€

1λ

= RH11−1ni2

⎛

⎝ ⎜

⎞

⎠ ⎟ = RH

ni2

ni2 −

1ni2

⎛

⎝ ⎜

⎞

⎠ ⎟

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1λ

= RHni2 −1ni2

⎛

⎝ ⎜

⎞

⎠ ⎟

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λn =1RH

ni2

ni2 −1

⎛

⎝ ⎜

⎞

⎠ ⎟ =

11.097 ×107 m-1

ni2

ni2 −1

⎛

⎝ ⎜

⎞

⎠ ⎟

Atomic SpectraAnswerFinally for the Lyman series:

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λ1 =1

1.097 ×107 m-122

22 −1⎛

⎝ ⎜

⎞

⎠ ⎟ =121 nm

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λ2 =1

1.097 ×107 m-132

32 −1⎛

⎝ ⎜

⎞

⎠ ⎟ =103 nm

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λ3 =1

1.097 ×107 m-142

42 −1⎛

⎝ ⎜

⎞

⎠ ⎟ = 97.2 nm

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λ4 =1

1.097 ×107 m-152

52 −1⎛

⎝ ⎜

⎞

⎠ ⎟ = 95.0 nm

For the Balmer series we have from before:λ1 = 656nm, λ2 = 486nm, λ3 = 434nm, λ4 = 410nm.

Atomic SpectraAnswerFor the Paschen series we have:

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1λ

= RH19−1ni2

⎛

⎝ ⎜

⎞

⎠ ⎟ = RH

ni2

9ni2 −

99ni

2

⎛

⎝ ⎜

⎞

⎠ ⎟

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λn =1

1.097 ×107 m-19ni

2

ni2 − 9

⎛

⎝ ⎜

⎞

⎠ ⎟

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λ1 =1

1.097 ×107 m-1

9 42( )42 − 9

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =1880 nm

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λ2 =1

1.097 ×107 m-1

9 52( )52 − 9

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =1280 nm

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λ3 =1

1.097 ×107 m-1

9 62( )62 − 9

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =1090 nm

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λ4 =1

1.097 ×107 m-1

9 72( )72 − 9

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =1010 nm

Atomic SpectraSo, the only series that lies in the visible range is the Balmer series.The Lyman series lies in the ultraviolet range and the Paschen series lies in the infrared range.We can extend the Bohr hydrogen atom to fully describe atoms that are “close” to hydrogen.These “hydrogen-like” atoms are those that only contain one electron.In those cases, when you have Z as the atomic number of the element (number of protons in the atom), you substitute Ze2 into where e2 was in the equations.

Quantum Numbersn is considered the principle quantum number. n can range from 1 to infinity in integer steps.But other quantum numbers were added later on to explain other subatomic effects (such as elliptical orbits).