Reporting Category 2: Atomic Structure and Nuclear Chemistry

9 STAAR QUESTIONS*indicates readiness standards

*6A Experiments and Conclusions that Developed Atomic Theory

John Dalton's Atomic Theory and Evidence theory:

1. All matter consist of tiny particles called atoms.

2. Atoms are indestructible and unchangeable. Atoms of an element cannot be created, destroyed, broken into smaller parts or transformed into atoms of another element.

3. All atoms of the same element are identical. Elements are characterized by the mass of their atoms. All atoms of the same element have identical weights.

4. Atoms combine in new ways during a chemical change. When elements react, their atoms combine in simple, whole- number rations.

Explanation:Dalton's theory explained that if all matter consists of indestructible atoms, then any chemical reaction simply changes the attachments between atoms but does not destroy the atoms itself. All of the original atoms are still present a the end, they have merely changed partners. Thus the total mass after this chemical change remains the same as it was before the change took place.

Dalton states that if a water molecule consists on one atom of oxygen and two atoms of hydrogen, and if an oxygen atom has a mass 16 times that of a hydrogen atom, then the mass ratio nof these two elements in water must always be 8.0 to 1.0. The two elements can never be found in water in any other mass ratio.

Law of Multiple Proportions: When two elements form several compounds, the mass ration in one compound will be a small whole-number multiple of the mass ration in another. (empirical and molecular formulas)

6.B -understand the electromagnetic spectrum and the mathematical relationships between energy, frequency, and wavelength of light;

Frequency is the number of cycles per second and is expressed in Hertz (Hz).

E = hvEnergy = (planck's constant)(frequency)

increased energy = smaller wavelength = greater frequencydecreased energy = increased wavelength = decreased frequency

Example: The red light from a helium-neon laser has a wavelength of 633 nm . What is the energy of one photon?

Two equations are needed:

E = hf where

E = energyh = Planck's constant = 6.626 x 10-34 J·sf = frequency

and

c = λf where

c = speed of light = 3 x 10-8 m/secλ = wavelengthf = frequency

f = c/λ

E = hνE = hc/λE = 6.626 x 10-34 J·s x 3 x 10-8 m/sec/ (633 nm x 10-9 m/1 nm)E = 1.988 x 10-25 J·m/6.33 x 10-7 m E = 3.14 x -19 J

The energy of a single photon of red light from a helium-neon laser is 3.14 x -19 J.

Example: The Aurora Borealis is a night display in the Northern latitudes caused by ionizing radiation interacting with the Earth's magnetic field and the upper atmosphere. The distinctive green color is caused by the interaction of the radiation with oxygen and has a frequency of 5.38 x 1014 Hz. What is the wavelength of this light?

6.C - calculate the wavelength, frequency, and energy of light using Planks constant and the speed of light

The speed of light, c, is equal to the product of the wavelength, & lamda;, and the frequency, f.

Therefore

λ = c/f

λ = 3 x 108 m/sec/(5.38 x 1014 Hz)λ = 5.576 x 10-7 m

1 nm = 10-9 mλ = 557.6 nm

Answer:

The wavelength of the green light is 5.576 x 10-7 m or 557.6 nm.

*6.E Express the arrangement of electrons in atoms through electron configurations and Lewis valence electron dot

structures

The p- sublevel is 3 peanut-shaped orbitals sitting on 3 different axis. Each orbital can hold 2 electrons for a maximum of 6 electrons

The s-sublevel is shaped like a sphere. There is only one orbital for a total of 2 electrons

The d sublevel has 5 orbitals each holding 2 electrons for a maximum of 10 electrons.

The f sublevel has 7 orbitals each holding 2 electrons for a maximum of 14 electrons.

Wave characteristics

• Frequency is the number of cycles per second and is expressed in Hertz (Hz).

E = hvEnergy = (planck's constant)(frequency)

increased energy = smaller wavelength = greater frequency

•decreased energy = increased wavelength = decreased frequency

Longer wavelength = low energy = less harmfulShorter wavelength = higher energy = hazardous

3 rules for filling orbitals

• Aufbau Principle –electrons fill orbitals starting with lowest n and moving upwards;

• Pauli Exclusion Principle –no two electrons can fill one orbital with the same spin;

• Hund's Rule –for degenerate orbitals, electrons fill each orbital singly before any orbital gets a second electron.

Electron Configuration of SodiumNoble gas configuration of Chlorine

Lewis Dot Diagrams• A Lewis Dot Structure is like a simplified electron energy level model.• The Lewis structure contains the element symbol with dots representing

the valence electrons.• The electrons are place around the element symbol, clockwise or

counterclockwise, and then grouped in pairs as more electrons are added.

*12A describe the characteristics of alpha, beta, and gamma radiation;

Alpha DecayHelium nucleus is given off. Atomic mass of parent nucleus drops by 4.Atomic number of parent nucleus drops by 2.Creates a new atom (daughter nucleus) and an alpha particle (helium nucleus).

Beta decay

Gives off an electron (beta particle) and a neutrino.A neutron ahs become a proton.The atomic number increases by one but the mass remains the same

Gamma Decay

*12.B Describe radioactive decay process in terms of balanced nuclear equations

TO BE ANNOUNCED…..

J. J. Thomson and the Discovery of the Electron Experiment and Conclusions

Cathode Ray Experiment: A cathode ray is a small glass tube with a cathode (negatively charged metal plate) and an anode ( a positively charged metal plate) at opposite ends. By separating the cathode and anode by a short distance, the cathode ray tube can generate what are known as "cathode rays" - rays of electricity that flowed from cathode to anode. Thomson wanted to know what these rays were and whether or not they had any mass or charge.

J.J. Thomson pointed a substance known as "phosphor" so he could see exactly where the cathode rays hit because the cathode rays make the phosphor glow.

The cathode ray traveled away from the negative cathode and towards the positive anode

Experiment Observation

He placed a positively charged metal plate on one side of the cathode ray tube and negatively charged plate on the other side.

The flow of the cathode rays passing through the hole in the anode was bent upward towards the positive metal plate and away from the negative metal plate

ConclusionCathode ray may be negatively charged.

Using the "opposites attract, like repel" rule, Thomson argued that if the cathode rays were attracted to the positively charged metal plate and repelled from the negatively charged metal plate. They themselves must have a negative charge

J. J. Thomson and the Discovery of the Electron Experiment and Conclusions

Based on what was learned about electrons, two other inferences were made about atomic structure.1. Because atoms are electrically neutral, they must contain a positive charge to balance the negative electrons.

2. Because electrons have so much less mass than atoms, atoms must contain other particles that account for most of their mass. Final conclusion: Plum Pudding Model

Ernest Rutherford's Experiment and Conclusions

The scientists bombarded a thin piece of gold foil with fast-moving alpha particles, which are positively charged particles with about four times the mass of a hydrogen atom. Alpha particles are very dense positively charged radiation. The alpha particles are emitted from a radioactive ore housed in a lead box with one opening for the fast traveling particles to leave the box. The alpha particles are shoot at a piece of gold foil that Rutherford thought was not very dense because the model of the atom at the time was the Plum pudding model. In the plum pudding model the mass is spread out over the entire volume of the atom. Rutherford thought shooting the alpha particles at the very this sheet of gold foil was like shooting a 16 inch shell at a piece of tissue paper. He thought the high speed very dense alpha particle would go through the gold leaf.

• Hypothesis: If the plum pudding model of the atom is correct, atoms have no concentration of mass or charge (atoms are 'soft' targets

ExperimentGold Leaf Scattering Experiment1. fire massive alpha particles at the atoms in thin metal foil2. alpha partilce should pass like bullets straight through soft plum pudding atoms

ObservationsMost of the alpha particles went straight through the gold leaf. But to his surprise a few alpha particles ricocheted.

ConclusionsMost of the alpha particles went straight through the gold leaf because most of the atom was empty space. All the mass and positive charges occupy about 1/10,000 the volume (this is the nucleus). The electrons are in the empty space outside the center of the very dense positive charged area called the nucleus. The nucleus has to be extremely dense for the alpha particles to deflected

Neils Bohr Observations, Conclusions, and Model

Observation: Light emitted from samples of atom exposed to energy when passed through a prism emit distinct wavelengths of light line spectrum instead of emitting all the colors of light. Each element emits the same line spectrum and are just as characteristic to that element as finger prints are to people.

Hypothesis: If energized atoms emit only discrete wavelengths, maybe electrons can have only discrete energies.

Conclusion: Bohr's Theory 1. electrons occupy orbits at discrete distances from the nucleus called energy levels. 2. Electrons in a non-energized atom occupy the lowest energy orbit (closest to nucleus is called the round state) 3. Electrons in energized atoms absorb just enough energy to move from a lower energy orbital to a higher energy orbital (this is called the excited state) 4. Electrons do not remian excited state and return to ground state releasing energy that corresponds to all the possible difference in energy of the allowed orbits in the atom.

1. circular orbit - the electron travels around the nucleus2. energy of an orbital is directly proportional to the distance from the nucleus3. quantized - limited number of countable allowed energy levels.4. absorption is low energy to high energy5. emission is high energy to low energy

6-D -use isotopic composition to calculate average atomic mass of an element

An element has three isotopes. Given the abundances and relative masses, calculate the average atomic mass and determine (from the periodic table) which element it is.Abundances Relative masses0.005% 234.040947 amu0.720% 235.043924 amu99.275% 238.050784 amu

Average Atomic Mass = 0.00005(234.040947 amu) + 0.00720(235.043924 amu) + 0.99275(238.050784 amu)