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LECTURE # 32

HYDROGEN ATOMPARTICLE DOUBLE-SLIT

PROBABILITY

PHYS 270-SPRING 2010

Dennis Papadopoulos

MAY 3 2010

Bohr’s Model of Atomic Quantization1. An atom consists of negative electrons orbiting a very

small positive nucleus.2. Atoms can exist only in certain stationary states. Each

stationary state corresponds to a particular set of electron orbits around the nucleus. These states can be numbered 2, 3, 4, . . . , where n is the quantum number.

3. Each stationary state has an energy En. The stationary states of an atom are numbered in order of increasing energy: E1 < E2 < E3 < …

4. The lowest energy state of the atom E1 is stable and can persist indefinitely. It is called the ground state of the atom. Other stationary states with energies E2, E3, E4,. . . are called excited states of the atom.

Bohr’s Model of Atomic Quantization

5. An atom can “jump” from one stationary state to another by emitting or absorbing a photon of frequency

where h is Planck’s constant and and ΔEatom = |Ef – Ei|.

Ef and Ei are the energies of the initial and final states. Such a jump is called a transition or, sometimes, a quantum jump.

r=n2aB ,

aB= (h/2)2/(ke2m)=.0529 nm

En=-E1/n2

E1=13.6 eV

Bohr Atom1 2

2

22

2

2

2

ˆ

0

cb

q qF k r

r

vek mr r

ker

mv

2 2 2

/ 2

, 1,2,3,4...

/

( / )

h

mv r n n

v n mr

r n kme

rmr

vmF e

elcentripeta

20

20

€

L = mevr = nh n =1,2,3...

h = h /2π

an integer number of wavelengths fits into the circular orbit

rn 2

where

p

h

is the de Broglie wavelength

Photonsp=h/c=

h/

What is the quantum number of this particle confined in a box?

A. n = 8B. n = 6C. n = 5D. n = 4E. n = 3

A. n = 8B. n = 6C. n = 5D. n = 4E. n = 3

What is the quantum number of this particle confined in a box?

What is the quantum number of this hydrogen atom?

A. n = 5B. n = 4C. n = 3D. n = 2E. n = 1

A. n = 5B. n = 4C. n = 3D. n = 2E. n = 1

What is the quantum number of this hydrogen atom?

The Bohr Hydrogen AtomThe radius of the electron’s orbit in Bohr’s hydrogen atom is

where aB is the Bohr radius, defined as

The possible electron speeds and energies are

The Hydrogen SpectrumAccording to the fifth assumption of Bohr’s model of atomic quantization, the frequency of the photon emitted in an n → m transition is

The corresponding wavelengths in the hydrogen spectrum are then

A photon with a wavelength of 414 nm has energy Ephoton = 3.0 eV. Do you expect to see a spectral line with = 414 nm in the emission spectrum of the atom represented by this energy-level diagram? If so, what transition or transitions will emit it?

A photon with a wavelength of 414 nm has energy Ephoton = 3.0 eV. Do you expect to see a spectral line with λ = 414 nm in the absorption spectrum of the atom represented by this energy-level diagram? If so, what transition or transitions will absorb it?

A photon with a wavelength of 414 nm has energy Ephoton = 3.0 eV. Do you expect to see a spectral line with = 414 nm in the emission spectrum of the atom represented by this energy-level diagram? If so, what transition or transitions will emit it?

We said earlier that Bohr was mostly right…so where did he go wrong?

• Failed to account for why some spectral lines are stronger than others. (To determine transition probabilities, you need QUANTUM MECHANICS!) Auugh!

• Treats an electron like a miniature planet…but is an electron a particle…or a wave?

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PA = lim(NA /NT ),NT → ∞

€

PAorB = PA + PB

€

NA expected = NPA

€

P(in δx at x) = limN(in δx at x)

N tot

, N tot → ∞

N(in δx at x) = N × P(x,δx)

WAVE PICTURE

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E(x,δx) = I(x)A = I(x)Hδx

PHOTON PICTURE

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N(δx,x) =E(x,δx)

hf=H

hfI(x)δx

but

P(δx,x) =N(δx,x)

N tot

= (H /hfN tot )I(x)δx

Energy E intercepted by an area A=Hx

Connecting the Wave and Photon Connecting the Wave and Photon ViewsViews

The intensity of the light wave is correlated with the probability of detecting photons. That is, photons are more likely to be detected at those points where the wave intensity is high and less likely to be detected at those points where the wave intensity is low.The probability of detecting a photon at a particular point is directly proportional to the square of the light-wave amplitude function at that point:

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Prob(δx,x) = P(x)δx

Probability DensityProbability DensityWe can define the probability density P(x) such that

In one dimension, probability density has SI units of m–1. Thus the probability density multiplied by a length yields a dimensionless probability.NOTE: P(x) itself is not a probability. You must multiply the probability density by a length to find an actual probability. The photon probability density is directly proportional to the square of the light-wave amplitude: