Quantum Theory and Atomic Models Slide 2 / 83 The Electron - - PowerPoint PPT Presentation

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Quantum Theory and Atomic Models

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The Electron

· Streams of negatively charged particles were found to emanate from cathode tubes. · J. J. Thompson is credited with their discovery (1897).

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Discovery and Properties of the Electron

It was found that these rays could be deflected by electric

• r magnetic fields. By adjusting those fields the charge to

mass ratio of the unknown "ray" was found.

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FE FB v

Discovery and Properties of the Electron

First, they found the velocity of the particle by adjusting the magnetic field electric forces so that they cancelled out; the "ray" traveled in a straight line. ΣF= ma FB - FE = 0 qvB = qE v = E/B Since they could measure E and B, they could calculate v.

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Discovery and Properties of the Electron

ΣF= ma FB = ma qvB = mv2/r qB = mv/r q/m = v/Br q/m = 1.76 x 1011 C/kg

FB v

r

Then they turned off the electric field and the particle moved in a circular path. They measured the radius of the circle, by seeing where the particle struck the tube, and then determined the charge to mass ratio: q/m.

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1 Which one of the following is not true concerning cathode rays?

A They originate from the negative electrode. B They travel in straight lines in the absence of electric or magnetic fields. C They impart a negative charge to metals exposed to them. D They are made up of electrons. E The characteristics of cathode rays depend on the material from which they are emitted.

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Millikan Oil Drop Experiment

Once the charge/mass ratio

• f the electron was known,

determination of either the charge or the mass of an electron would yield the

• ther.

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The mass of each droplet was estimated by its size. The electric field was adjusted so the drop fell with constant velocity. The data showed that the charge was always an integral multiple of a smallest charge, e. That must be the charge of one electron.

Discovery and Properties of the Electron

FE

mg

ΣF= ma FE - mg = 0 qE = mg q = mg/E q was always an integer multiple of the same number, which was given the symbol "e"

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2 Which of these could be the charge of an object?

A 0.80 x 10-19 C B 2.0 x 10-19 C C 3.2 x 10-19 C D 4.0 x 10-19 C

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3 The charge on an electron was determined in the __________.

A cathode ray tube, by J. J. Thompson B Rutherford gold foil experiment C Millikan oil drop experiment D Dalton atomic theory E atomic theory of matter

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Blackbody Radiation

All objects emit electromagnetic radiation which depends on their temperature: thermal radiation. A black body absorbs all electromagnetic radiation (light) that falls on it. Because no light is reflected or transmitted, the

• bject appears black when it is
• cold. However, black bodies emit

a temperature-dependent spectrum termed blackbody radiation.

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Blackbody Radiation

At normal temperatures, we are not aware of this radiation. But as objects become hotter, we can feel the infrared radiation or heat. At even hotter temperatures,

• bjects glow red and at still

hotter temperatures, object can glow white hot such at the filament in a light bulb.

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This figure shows blackbody radiation curves for three different temperatures. The wavelength at the peak, #p, is related to the temperature by: #pT = 2.90 x 10-3 m-K Classical physics couldn't explain the shape of these spectra.

Blackbody Radiation

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4 Which of the following colors would indicate the hotest temperature? A Black B Red C Yellow D Blue

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· The wave nature of light could not explain the way an

• bject glows depending on its temperature: its

spectrum. · Max Planck explained it by assuming that atoms only emit radiation in quantum amounts...in steps given by the formula: E = hf where h is Planck’s constant and f is the frequency of the light h = 6.6 ´ 10-34 J-s

Planck’s Quantum Hypothesis

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Planck’s Quantum Hypothesis

Planck didn't believe this was a real...it just worked. It was like working from the answers in the book...getting something that works, but having no idea why. It didn't make sense that atoms could

• nly have steps of energy. Why

couldn't they have any energy? Planck thought a "real" solution would eventually be found...but this

• ne worked for some reason.

Which brings us to our next mystery...

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When light strikes a metal, electrons sometimes fly off. Classical physics couldn't explain some specific features about how the effect works. So Einstein used Planck's idea to solve it.

The Photoelectric Effect

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If atoms can only emit light in packets of specific sizes; maybe light itself travels as packets of energy given by Planck's formula. He called these tiny packets,

• r particles, of light, photons.

The Photoelectric Effect

E = hf

where h is Planck’s constant h = 6.6 ´ 10-34 J-s

• r

4.14 x 10-15 eV-s

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5 What is the energy (in nJ) of a photon with a frequency of 5 x 1022 Hz?

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The maximum kinetic energy of these photons can be measured using a variable voltage source and reversing the terminals so that the electrode C is negative and P is

• positive. If the voltage is

increased, there is a point when the current reaches zero. This is called the stopping voltage, V0, and it is given by: KEmax = eV0

The Photoelectric Effect

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We said earlier that when light strikes a metal, electrons sometimes fly off. Since electrons are held in the metal by attractive forces, some minimum energy, W0, called the work function, is required just to get an electron free from the metal. The input energy of the photon will equal the kinetic energy of the ejected electron plus the energy required to free the electron. hf = KE + W0

The Photoelectric Effect

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• A. H. Compton scattered short-wavelength light from various

materials and discovered that the scattered light had a slightly lower frequency than the incident light, which indicated a loss

• f energy. He applied the laws of conservation of momentum

and energy and found that the predicted results corresponded with the experimental results.

The Compton Effect

Incident photon (#) Electron at rest Electron after collision Scattered photon (#') # #

A single photon wavelength, #, strikes an electron in some material, knocking it out of its

• atom. The scattered photon

has less energy since it gave some to the electron and thus has a wavelength of , #'.

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The Compton Effect

Incident photon (#) Electron at rest Electron after collision Scattered photon (#') # #

The momentum of a photon is given by: p = E/c Since E = hf, p = hf/c = h/# Using conservation of momentum: Where m0 is the rest mass of the electron.

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Photon Interactions

• 1. The Compton Effect - the photon can be scattered by an

electron and lose energy in the process.

• 2. The photoelectric effect - a photon can knock an electron
• ut of an atom and disappear in the process.
• 3. The photon can knock and atomic electron to a higher

energy state if the energy of the photon is not sufficient to knock it out of the atom.

• 4. Pair production - a photon can produce an electron and a

positron and disappear in the process. (The inverse of pair production can occur if an electron collides with a positron. This is called annihilation.)

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Photon Theory of Light

This particle theory of light assumes that an electron absorbs a single photon and made specific predictions that proved

• true. For instance, the kinetic energy of escaping electrons
• vs. frequency of light shown below:

This shows clear agreement with the photon theory, and not with wave theory.

This shows that light is made of particles, photons; light is not a wave.

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Wave-Particle Duality; the Principle of Complementarity

Earlier we proved that light is a wave. Now we've proven that light is a particle. Which is it? This question has no answer; we must accept the dual wave-particle nature of light. While we cannot imagine something that is a wave and is a particle at the same time; that turns out to be the case for light.

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6 The ratio of energy to frequency for a given photon gives

A its amplitude. B its velocity. C Planck's constant. D its work function.

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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7 What is a photon?

A an electron in an excited state B a small packet of electromagnetic energy that has particle-like properties C

• ne form of a nucleon, one of the

particles that makes up the nucleus D an electron that has been made electrically neutral

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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8 The energy of a photon depends on

A its amplitude. B its velocity. C its frequency. D none of the given answers

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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9 Which color of light has the lowest energy photons?

A red B yellow C green D blue

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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10 The photoelectric effect is explainable assuming

A that light has a wave nature. B that light has a particle nature. C that light has a wave nature and a particle nature. D none of the above

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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11 The energy of a photon that has a frequency 110 GHz is

A 1.1 × 10-20 J B 1.4 × 10-22 J C 7.3 × 10-23 J D 1.3 × 10-25 J

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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12 The frequency of a photon that has an energy of 3.7 x 10-18 J is

A 5.6 × 1015 Hz B 1.8 × 10-16 Hz C 2.5 × 10-15 J D 5.4 × 10-8 J E 2.5 × 1015 J

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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13 The energy of a photon that has a wavelength of 12.3 nm is

A 1.51 × 10-17 J B 4.42 × 10-23 J C 1.99 × 10-25 J D 2.72 × 10-50 J E 1.62 × 10-17 J

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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14 If the wavelength of a photon is halved, by what factor does its energy change?

A 4 B 2 C 1/4 D 1/2

E = hf c = lf c = 3.00 ´ 108 m/s h = 6.6 ´ 10-34 J-s

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Energy, Mass, and Momentum of a Photon

Clearly, a photon must travel at the speed of light, since it is light. Special Relativity tell us two things from this: The mass of a photon is zero. The momentum of a photon depends on its wavelength. m = 0 p = hf / c and since c = lf p = h/l This last equation turned out to have huge implications.

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Wave Nature of Matter

de Broglie asked, "If light can behave like a wave or a particle, can matter also behave like a wave?" Amazingly, it does! de Broglie combined p = h/l with p = mv to get The wavelength of matter l = h/(mv) This wavelength is really small for normal objects, so it had never been noticed before. But it has a dramatic impact on the structure of atoms.

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Wave Nature of Matter

Electron wavelengths are often about 10-10 m, about the size

• f an atom, so the wave character of electrons is important.

In fact, the two-slit experiment that showed that light was a wave, has been replicated with electrons with the same result...electrons are particles and waves. Electrons fired one at a time towards two slits show the same interference pattern when they land on a distant screen. The "electron wave" must go through both slits at the same time...which is something we can't imagine a single particle doing...but it does.

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These photos show electrons being fired one at a time through two slits. Each exposure was made after a slightly longer time. The same pattern emerges as was found by light. Each individual electron must behave like a wave and pass through both slits. But each electron must be a particle when it strikes the film, or its wouldn't make one dot

• n the film, it would be spread out.

This one picture shows that matter acts like both a wave and a particle.

The most amazing experiment ever!!!

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15 What is the wavelength of a 0.25 kg ball traveling at 20 m/s?

l = h/(mv) h = 6.6 ´ 10-34 J-s

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16 What is the wavelength of a 80 kg person running 4.0 m/s?

l = h/(mv) h = 6.6 ´ 10-34 J-s

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17 What is the wavelength of the matter wave associated with an electron (me = 9.1 x 10-31kg) moving with a speed of 2.5 × 107 m/s?

A

0.29 nm

B

0.36 nm C 0.48 nm D 0.56 nm

l = h/(mv) h = 6.6 ´ 10-34 J-s

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18 What is the wavelength of the matter wave associated with an electron (me = 9.1 x 10-31kg) moving with a speed of 1.5 × 106 m/s?

A

0.29 nm

B

0.36 nm C 0.48 nm D 0.56 nm

l = h/(mv) h = 6.6 ´ 10-34 J-s

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The Atom, circa 1900

· The prevailing theory was that

• f the “plum pudding” model,

put forward by Thompson. · It featured a positive sphere of matter with negative electrons imbedded in it.

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Discovery of the Nucleus

Ernest Rutherford shot a particles at a thin sheet of gold foil and observed the pattern of scatter of the particles.

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Discovery of the Nucleus

While most particles went straight through, as expected, some bounced back...which was totally unexpected.

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Early Models of the Atom

The only way to account for that was to assume all the positive charge was contained within a tiny volume. A small very dense nucleus must lie within a mostly empty atom. Now we know that the radius of the nucleus is 1 / 10,000 that of the atom.

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The Nuclear Atom

Since some particles were deflected at large angles, Thompson’s model could not be correct.

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Early Models of the Atom

Rutherford's experiment showed that the positively charged nucleus must be very small compared to the rest of the atom. Then I remember two or three days later Geiger coming to me in great excitement and saying "We have been able to get some of the alpha-particles coming backward …" It was quite the most incredible event that ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.

• Rutherford

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The Nuclear Atom

· Rutherford postulated a very small, dense nucleus with the electrons around the outside of the atom. · Most of the volume of the atom is empty space.

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The Nuclear Atom

If an atom were magnified to be the size of a gymnasium (about 100 m across), the proton would be about the size of a ping pong ball (1 cm across), the electrons would be too small to see, and all the rest would be just emptiness...not filled with air (like a gym), but nothing.

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19 The gold foil experiment performed in Rutherford's lab __________.

A confirmed the plum-pudding model of the atom B led to the discovery of the atomic nucleus C was the basis for Thomson's model

• f the atom

D utilized the deflection of beta particles by gold foil E proved the law of multiple proportions

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20 In the Rutherford nuclear- atom model, __________.

A

the heavy subatomic particles reside in the nucleus

B

the principal subatomic particles all have essentially the same mass C the light subatomic particles reside in the nucleus D mass is spread essentially uniformly throughout the atom

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Subatomic Particles

· Protons were discovered by Rutherford in 1919. · Neutrons were discovered by James Chadwick in 1932. · Protons and electrons are the only particles that have charge. · Protons and neutrons have essentially the same mass. · The mass of an electron is extremely small.

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The Problem with the Nuclear Atom

· The nucleus of an atom is small, 1/10,000 the size of the atom. · The electrons are outside the nucleus, moving freely within the vast empty atom · The nucleus is positive; the electron is negative · There is an electric force, FE = kq1q2/r2, pulling the electron towards the nucleus · There is no other force acting on the electron; there is nothing to support it; it experiences a net force towards the nucleus · Why don't the electrons fall in...why doesn't the atom collapse into its nucleus?

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The Problem with the Nuclear Atom

Based on the equations #F = ma and x = x0 + vot +1/2at2 All atoms would collapse in about 10-10 s Earth would collapse to less than a mile across in less than a billionth of a second. The universe as we know it would end.

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The Problem with the Nuclear Atom

Perhaps electrons orbit the nucleus...like planets orbit the sun. But then they would constantly be accelerating as they travel in a circle: a = v2/r But it was known that an accelerating charge radiates electromagnetic energy...light. All the kinetic energy would radiate away in about that same billionth of a second...then it would fall into the nucleus. All the atoms in the universe would still collapse.

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The Problem with the Nuclear Atom

Classical physics failed to explain how atoms could exist. A new approach was needed. The next step led to the Bohr model of the atom, which was a semi-classical explanation of atoms. It would be an important transition to modern quantum theory. An important clue was found in the spectra of gas discharge tubes.

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Atomic Spectra

A very thin gas heated in a discharge tube emits light

• nly at characteristic frequencies.

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Atomic Spectra

An atomic spectrum is a line spectrum – only certain frequencies appear. If white light passes through such a gas, it absorbs at those same frequencies.

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Atomic Spectra

Why don't atoms radiate, or absorb, all frequencies of light? Why do they radiate light at only very specific frequencies, and not at others?

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Atomic Spectra

The wavelengths of electrons emitted from hydrogen have a regular pattern: Balmer series Lyman series Paschen series

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Atomic Spectra: Key to the Structure of the Atom

A portion of the complete spectrum of hydrogen is shown

• here. The lines cannot be explained by the Rutherford theory.

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The Bohr Atom

Bohr proposed that electrons could orbit the nucleus, like planets orbit the sun...but only in certain specific orbits. He then said that in these orbits, they wouldn't radiate energy, as would be expected normally of an accelerating charge. These stable orbits would somehow violate that rule. Each orbit would correspond to a different energy level for the electron.

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The Bohr Atom

These possible energy states for atomic electrons were quantized – only certain values were possible. The spectrum could be explained as transitions from one level to another. Electrons would only radiate when they moved between orbits, not when they stayed in one orbit.

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The Bohr Atom

As long as an electron was in an orbit given by the below formula, it would not emit electromagnetic radiation. The observed spectrum of the hydrogen atom is predicted successfully by transitions between these orbits.

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The Bohr Atom

An electron is held in orbit by the Coulomb force:

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The Bohr Atom

Using the Coulomb force, we can calculate the radii of the orbits. These matched the sizes of known atoms very well.

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The Bohr Atom

The radii of the orbits of a hydrogen atom are given by the below formula, with the smallest

• rbit,

rn = n2r1, (for hydrogen, Z = 1) Z r1 = 0.53 x 10-10m. n = 1, 2, 3, 4, .... Notice that the orbits grow in size as the square of n, so they get much larger as n increases.

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21 The radius of the orbit for the third excited state (n=4)

• f hydrogen is ______r1.

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22 The radius of the orbit for the fifth excited state (n=6)

• f hydrogen is ____ x 10-10m.

r1 = 0.50 x 10-10m

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The Bohr Atom

Using the Coulomb force, he calculated the energy of each orbit. For hydrogen he arrived at this result: n = 1, 2, 3, 4, .... E = -13.6 eV n2 Notice that the energy levels are all negative,

• therwise the electron would be free of the atom.

The levels get closer together, and closer to zero, as n increases.

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The Electron Volt (eV)

In atomic physics, energies are so low that it's awkward to use Joules (J). A smaller unit of energy is the electron-volt (eV). It's value equals the potenial energy of an electron in a region of space whose voltage (V) is 1.0 volt. UE = qV UE = (1.6 x 10-19C)(1.0V) UE = 1.6 x 10-19J # 1.0 eV

1.0 eV = 1.6 x 10-19J

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The Electron Volt (eV)

It's also convenient to convert Plank's constant to units of eV-s rather than J-s.

1.0 eV = 1.6 x 10-19J

h = 6.63 x 10-34 J-s h = (6.63 x 10-34 J-s) h = 4.14 x 10-15 eV-s 1.6 x 10-19J 1.0 eV )

(

h = 4.14 x 10-15 eV-s

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23 What is the energy of the second excited state (n=3)

• f hydrogen?

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24 What is the energy of the fifth excited state (n=6) of hydrogen?

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The Bohr Atom

The lowest energy level is called the ground state; the

• thers are excited states.

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de Broglie’s Hypothesis Applied to Atoms

Scientists didn't like the lack of an explanation for why electrons didn't radiate when in those orbits. But de Broglie's wave theory of matter explained it very well. As long as the wavelength on an electron in orbit was the same as the circumference of the orbit, it would not radiate. This approach yields the same relation that Bohr had proposed. In addition, it makes more reasonable the fact that the electrons do not radiate, as one would otherwise expect from an accelerating charge.

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de Broglie’s Hypothesis Applied to Atoms

These are circular standing waves for n = 2, 3, and 5.

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Quantum Physics

While a big step forward, Bohr's model only worked for atoms that had one electron, like hydrogen or certain ionized atoms. It failed for all atoms other than hydrogen. The idea that the electron was a particle in orbit around the nucleus, but with wavelike properties that only allowed certain orbits, worked only for hydrogen. Semi-classical explanations failed except for hydrogen. It turned out that only a lucky chance let it work even in that case.

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Quantum Mechanics

Our goal was to explain why electrons in an atom don't fall into the nucleus. An electron, as a charged particle, would fall in because of Newton's Second Law. #F = ma But electrons, in atoms, aren't particles, they're waves. Waves don't follow Newton's Second Law. Schrodinger had to invent a new equation for wave mechanics. H# = E#

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Quantum Mechanics

H# = E# The simplicity of this equation is deceptive. Here's what it looks like when expanded for one type of problem. It's only solved for general cases in advanced university courses. However, computers have been used to exactly solve it for many specific cases: atoms, molecules, etc.

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Quantum Mechanics

Those solutions will allow us to understand how the microscopic world works: atoms, the periodic table, molecules, chemical bonds, etc. Quantum mechanics is very different from classical physics– you can predict what a lot of electrons will do on average, but have no idea what any individual electron will do. In Chemistry, you'll be using the solutions to Schrodinger's Equation equations, and the physics you've learned this year, to explored the nature of matter.