Light Notes & the Quantum Mechanical Model



Light Notes & the Quantum Mechanical Model

Electromagnetic radiation is a form of energy that travels in waves. The electromagnetic radiation (EMR) that travels from the sun contains many different types of radiation. The radiation stretches from radio waves to gamma rays. There is a very narrow band of the electromagnetic spectrum from 400 to 700 nm that is the visible light area of the EMR spectrum. When the wavelength is multiplied by the frequency the answer is always a constant 3.00 x 10 8 m/s which we call the speed of light constant.

“Red Martians invade Venus using Xray guns” will help learn parts of spectrum.

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This relationship can be summarized in the relationship : c = λ υ

c = speed of light 3.00 x 10 8 m/s

λ = (lambda) wavelength

υ = (nu) frequency

The types of problems you will see would be problems where you were given the frequency or the wavelength. Traps in the problem would be having the wavelength in nm or micrometers or kilometers instead of meters. Frequency traps would be things like kHZ or MHz which when used in the problem would give the wrong answers.

Ex. What is the frequency of EMR with a wavelength of 4.21 x 10 -6 m?

The wavelength is in m so use the formula, plug in and solve. Remember to show work or you will only get 1 point for the answer!!!!!!!!

c = λ υ

3.00 x 10 8 m/s = 4.21 x 10-6 m x υ

3.00 x 10 8 m/s = 4.21 x 10-6 m x υ

4.21 x 10 -6 m 4.21 x 10 -6 m

υ ’ 7.13 x 10 13 1/s

Ex. What is the wavelength of EMR spectrum that has a frequency of 4.32 MHz?

c = λ υ

What is the trap in the problem? MHz!

|4.32 MHz |1000000Hz |= 4320000 Hz (1/s) |

| |1 MHz | |

Now that you missed the trap work the problem.

3.00 x 10 8 m/s = λ x 43200000 1/s (Yes you could use 4.32 x 10 6 1/s instead)

3.00 x 10 8 m/s = λ x 4320000 1/s

4320000 1/s 4320000 1/s

λ = 69.4 m

Ex. The laser in a compact disc (CD) uses light with a wavelength of 780nm. What is the frequency of this light?

c = λ υ

Τrap: 780nm

|780 nm |1 m |= 7.8 x 10 -7 m |

| |1000000000 nm | |

c = λ υ

3.00 x 10 8 m/s = 7.8 x 10 -7 m x υ

3.00 x 10 8 m/s 7.8 x 10 -7 m

υ ’ 3.8 x 10 14 1/s

All waves have certain characteristics in common. These include wavelength, amplitude, crest, trough, and origin. The wavelength is the distance from any point on a wave to the same point on the next wave. This is generally thought of as the distance from crest to crest or trough to trough but in reality it is from any point on one wave to the same point on the next wave. Amplitude is the distance a wave rises or falls from the origin point. (For sound waves a large amplitude is a loud sound; for light a large amplitude is bright and a small amplitude is dim.) The crest is the highest point on a wave and the trough is the lowest point on a wave.

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When waves have long wavelengths they have lower frequencies and vice versa. See diagram below:

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Light is considered a form of energy and matter is considered to be particles. Funny thing though there is a wave-particle duality in nature. The smaller the particle the more it acts like a wave and the larger the particle the more it acts like matter. A physicist de Broglie came up with a great equation that allows you to calculate the wavelength of something. The formula is:

λ = h

m υ

λ ’ wavelength, h = Planck’s constant 6.626 x 10 -34 J*s, m = mass in kg , υ = frequency.

“n automobile moving at 25 m/s and having a mass of 910 kg would have a wavelength of 2.9 x 10 -38 m a wavelength that can’t be detected with even the most sensitive instrument. An electron on the other hand moving at the same speed has a wavelength of 2.9 x 10 -5 m. You won’t calculate wavelength!

As physicists studied matter and light more and more discoveries were made. Max Planck and Einstein made the connection that heated objects absorb or lose energy only in specific amounts called quanta.

Planck was able to make the connection that

Energy of a quantum of light = Planck’s constant x frequency

E = h x υ

E = energy of light

h = Planck’s constant 6.626 x 10 -34 J*s

υ= frequency

Example problem: What is the energy of a photon red light having a frequency of

4.48 x 1014 Hz?

E = h x υ

E = 6.626 x 10 -34 J*s x 4.48 x 10 14 1/s

Ε ’ 2.97 x 10 -19 J

Einstein came up with the photoelectric effect.

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For the photoelectric effect to occur the threshold frequency must be met. Think of breaking a window….a ping pong ball could hit a window hundreds of times without damage but a baseball hitting the window one time will break it because the force is enough to break the window…in the photoelectric effect a beam of light with the correct frequency will eject an electron. If enough electron are ejected a current is produced and the solar cell will run a device. (calculator, etc.)

When the atom was first realized by Democritus and then experimentally proved by Dalton, modified by Thomson, then Rutherford, Bohr changed the atom by giving electrons places to be and definite locations. The Bohr atom told us that electrons are in the ground state they gain a quanta of energy jump to the excited state. The jump upward isn’t as easy to detect as the return from the excited state. When electrons return from the excited state to the ground state they must lose that extra energy or they can’t go home again. The extra energy is transformed into light which we call a photon. If you remember the flame test lab we added energy to the metal cation and recorded the characteristic color. The unknown was identified by what color the flame turned as the electrons were returning to the ground state. Because we were still adding heat the electrons continued jumping back and forth giving plenty of time to determine the unknown.

When electrons jump and return because of their gain in energy you can have phosphorescence, fluorescence or other types of energy changes. Fluorescence occurs when electrons are struck by UV light and jump up but hesitate in the blue range of visible light before the electrons return to their ground state. We think of fluorescent as bright. Tide and other detergents have been adding brighteners to their detergents that make clothes appear brighter. Remember in class with the black light the white shirts and when you smiled. As soon as you turn off the UV however the glowing stops.

Phosphorescence is different the electrons jump up but return to the ground state in tiny steps (like a pinball bouncing around). This phenomenon is what makes your glow in the dark toys possible.

When Bohr studied the hydrogen atom he realized that electrons dropped from a higher energy orbit to a lower energy that the energy emitted was corresponding to the energy differences between the two energy levels.

ρE = E higher energy orbit – E lower energy orbit = E photon = h υ

It turns out that humans can only see electrons that are returning to the 2nd energy level. See diagram below:

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Electrons returning to the 1st energy level are ultraviolet. Electrons returning to 3rd energy level are in the infrared we can’t see them but snakes, insects, fish etc. seem to be able to.

Electrons obviously must have some type of arrangement in the electron cloud because the valence electrons control the chemical behavior of an atom. There is a direct relationship between the arrangement of the electrons in the electron cloud and the position of an element on the periodic table.

How are electrons arranged in the electron cloud? The first thing you need to realize is that the cloud is divided into energy levels, sublevels, orbitals, and the all important spin.

When you are placing electrons in the electron cloud arrangement follow three simple rules: Aufbau Principle, Hund’s Rule, and Pauli Exclusion Principle.

Aufbau principle: states that electrons fill from the lowest energy sublevel and move upward.

Hund’s Rule: states that 1 electron must occupy each orbital before any orbital will contain two. (Room mate rule)

Pauli Exclusion Principle: States that an orbital will contain a maximum of 2 electrons with opposite spins. They may contain 0, 1 or 2 electrons but never more than 2.

Summary of electron cloud information you must know:

|Energy Level |# of sublevels |Sublevels |# of orbitals |# of electrons |

| 1 | 1 | s | 1 | 2 |

| 2 | 2 | s & p | 1, 3 =4 | 8 |

| 3 | 3 |s, p, d | 1, 3, 5 = 9 | 18 |

| 4 | 4 |s,p, d , f | 1,3,5,7 =16 | 32 |

| 5 –7 (like 4) | 4 |s, p, d, f | 1,3,5, 7 = 16 | 32 |

Remember you can calculate the number of electrons present in an energy level with the formula 2n2. n = energy levels 1, 2, 3, or 4

Ex. How many electrons are found on the third energy level?

2(3) 2 = 18

What if we were writing the electron filling diagram for Yttrium which has 39 electrons.

4d η ___ ___ ____ ___

5s ηι

4p ηι ηι ηι

3d ηι ηι ηι ηι ηι

4s ηι

3p ηι ηι ηι

3s ηι

2p ηι ηι ηι

2s ηι

1s ηι

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