topic 7.1 extended c – the bohr theory of the hydrogen atom

When a gas in a tube is subjected to a voltage, the gas ionizes, and emits light.

We can analyze that light by looking at it through a spectroscope.

Topic 7.1 ExtendedC –The Bohr theory of the hydrogen atom

A spectroscope acts similar to a prism, in that it separates the incident light into its constituent wavelengths.For example, barium gas in a gas discharge tube will produce an emission spectrum that looks like this:

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The emission spectrum is really an elemental fingerprint - it uniquely identifies the element producing it.

FYI: The wavelengths are given in angstroms Å. 1 Å = 10-10 m.

Calcium gas produces this spectrum:


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Not only do glowing gases emit spectral lines, but cool gases absorb light in the same wavelengths and produce what is called an absorption spectrum.For example, the Sun produces a continuous spectrum that looks like this...

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...with characteristic absorption spectral lines, revealing what non-glowing elements are in the Sun's atmosphere.

FYI: Note the possible fingerprint for calcium as an absorbing constituent of the Sun's outer atmosphere.

In the late 1800s a Swedish physicist by the name of J.J. Balmer observed the spectrum of hydrogen - the simplest of all the elements:


The general spectral signature is divided up into natural groups of spectral lines called series, falling roughly in the UV (ultraviolet), Visible, and IR (infrared) ranges of wavelengths.

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Emission Spectra of Hydrogen

Lyman Series

UV

Balmer Series

Paschen Series

IR

FYI: Each series is characterized by spectral line "bunching" at the smaller wavelengths, and "spreading" at the larger wavelengths.

FYI: Balmer studied the middle series, which has parts in the visible spectrum. Obviously, the series is in his name.

In fact, Balmer found an empirical formula that predicted the allowed spectral wavelengths for the Balmer series of the hydrogen atom:


1

Balmer Series

= R -122

1n2

for n = 3,4,5,... Balmer Series

where R = 1.09710-2 nm-1 is called the Rydberg constant.

FYI: The visible spectrum for hydrogen was found to fit this formula, but it was NOT understood why.

FYI: Similar formulas were found to fit the other two series.

An explanation was finally given in 1913 by the Danish physicist Niels Bohr.


Bohr postulated that the single electron was held in a circular orbit about the single proton in the hydrogen nucleus by the Coulomb force:

+

r

Fc = FE

=mv2

rke2

r2

The total mechanical E energy of the hydrogen atom is given by E = K + U

E = mv2 - 12

ke2

r

so that

FYI: Note that we are using the relation U = -kqQ / r for two point charges. Recall that this energy is negative since q and Q are oppositely charged.

mv2 = 12

ke2

2r

E = - ke2

rke2

2r2 2

E = - ke2

2rEnergy inH Atom

FYI: Up to this point in Bohr's derivation, classical mechanics has been used.

Recall that the angular momentum l of a point mass moving in a circle of radius r is given by


l = mvr

+

r

Bohr then postulated the radical idea that the angular momentum of the electron was quantized, just like light. He stated that the angular momentum of the electron can only carry the discrete values given by

mvr = n h2

for n = 1,2,3,... Principal Quantum Number - H Atom

mv2 = 12

ke2

2rFrom and the previous equation we caneliminate v, solving for r:

r = n2h2

42ke2mfor n = 1,2,3,... Allowed Radii

- H Atomn

FYI: We subscript the r to indicate that the electron can only orbit the nucleus at certain quantized radii, determined by the principal quantum number.

We can then take our energy equation for the hydrogen atom and substitute our allowed values for r:


E = - ke2

2r

En = -

22k2e4mh2

1n2

for n = 1,2,3,...

Everything in the parentheses is a constant whose value we know. We can then rewrite both rn and En like this:

Energy Quantization in the Bohr Hydrogen Atom

En = eV-13.6n2

rn = 0.0529n2 nm

The Bohr Hydrogen Atom

What are the orbital radius and energy of an electron in a hydrogen atom characterized by principal quantum number 3?


rn = 0.0529n2 nm

r3 = 0.052932 nm

r3 = 0.4761 nm

En = eV-13.6n2

E3 = eV-13.632

E3 = -1.51 eV

What is the change in energy if the electron "drops" to the energy characterized by principal quantum number 2?

E2 = eV-13.622

E2 = -3.4 eV

E = (-3.4 - -1.51) eVE = -1.89 eV

FYI: In general, if an electron "drops" from a higher quantum state to a lower one, the hydrogen atom experiences a net loss of energy.

FYI: We call the lowest energy level (n = 1) the GROUND STATE. We call the next highest energy level (n = 2) the 1ST EXCITED STATE.We call the next highest energy level (n = 3) the 2ND EXCITED STATE.Et cetera.

What is the orbital velocity of an electron in the second excited state (n = 3)?


From the previous slide r3 = 0.4761 nm.

mvr = n h2

Then

v =nh2mr

v =3(6.6310-34)

2(9.1110-31)(0.476110-9)

v = 7.30105 m/sWhat then is the centripetal acceleration of the electron?

ac =v2

r=

(7.30105)2

0.476110-9 = 1.12 1021 m s-2

FYI: Classical theory predicts that electromagnetic radiation is created by accelerating charges. Since the hydrogen atom only radiates when its electron "drops" from one excited state to a less energetic state, Bohr postulated that "the hydrogen electron does NOT radiate energy when it is in one of its bound states (allowed by n). It only does so when "dropping" from a higher state to a lower state."

Consider a plot of energies for n = 1 to :


En = eV-13.6n2

n = 1 -13.6 eV

n = 2 -3.40 eV

n = 3 -1.51 eV

n = 4 -0.850 eVn = 5 -0.544 eV

n = 0.00 eV

Excited States

FYI: Bohr's theory only allows electrons in the hydrogen atom to absorb or emit photons having energies equal to the difference between any two of the allowed states shown here.

EXAMPLE: If an electron at r3 suddenly "drops" to the ground state, the hydrogen atom LOSES energy having a value of

E = (-13.6 - -1.51) = -12.09 eV. To conserve energy, a PHOTON having an energy of

E = +12.09 eV is released.

Ground State

EXAMPLE: If a photon having an energy of E = +12.09 eV is absorbed by a hydrogen atom in its ground state, the electron will "jump" up to the second excited state (n = 3) since E = (-1.51 - -13.6) = +12.09 eV.

So how do the three series of hydrogen spectra relate to the Bohr model?


n = 1 -13.6 eV

n = 2 -3.40 eV

n = 3 -1.51 eV

n = 4 -0.850 eVn = 5 -0.544 eV

n = 0.00 eV

First Excited State

Ground State

Second Excited State

Consider the following transi-tions of hydrogen from higher to lower states:

LymanSeries

UV

BalmerSeriesVisible

PaschenSeries

IR

Each transition gives off a photon of a different wavelength.

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Lyman Series

UV

Balmer Series

Paschen Series

IR

FYI: The Lyman Series has as its final state the GROUND STATE.The Balmer Series has as its final state the FIRST EXCITED STATE.The Paschen Series has as its final state the SECOND EXCITED STATE.

topic 7.1 extended c – the bohr theory of the hydrogen atom

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