CHEM 121 EXAM 2 REVIEW
1. rutherford's gold foil experiment setup
Answer - shot a beam of alpha particles at thin sheet of pure gold foil
- tracked the deflections of the alpha particles after hitting the gold foil
2. alpha particles
Answer positively charged, relatively massive
3. conclusions
Answer rutherford's experiment
- most particles went straight through (no deflection) => most of the atom is empty space
- small amount of particles had shallow deflections => small part of the atom is positively charged
- even smaller amount of particles "bounced back" (large deflection) => small part of the atom is more massive than
the alpha particle, thus is very dense
4. overall conclusion of rutherford's experiment
Answer atom is mostly empty space with a dense, positively-charged nucleus
5. moseley's x-ray experiment setup
,Answer bombarded atoms with energy and then measured x-ray frequencies emitted
6. moseley's experiment results
Answer - each element emits a unique x-ray frequency (unique, physical property)
- assigned an atomic number to each element - integer ranking of ordered mass
- when graphing x-ray frequency against atomic number, it formed a perfect curve (comparing the square root of x-ray
to atomic nmber => linear relationship)
7. moseley's experiment conclusionsAnswer - x-ray frequency is unique for each atom =>
corresponds to a unique, physical property of each element (there is a perfect relationship between x-ray frequency
and atomic number, so atomic number must also correspond to a unique physical property of the atom)
- atomic number are integers => implies we are counting something
8. moseley's main conclusionAnswer atomic number counts the number of protons (units of positive
charge) in the nucleus of an atom
9. coulomb's lawAnswer - quantifies the force between two objects
- PE is directly proportional to V(r) = q1*q2/r
10. potential energyAnswer - proportional to coulomb's law
,- will always be negative => large magnitudes of potential energy are lower numbers than small magnitudes of potential
energy
- lower potential energy => more stable state (electron experiencing greater coulombic attractions)
, 11. factors determining potential energy
Answer - core charge
- radius
- electron-electron repulsion
12. core charge
Answer - ettective nuclear charge felt by an electron
- core charge = # protons - # shielding electrons
- accounts for attraction and sum of all repulsions
13. ionization energy
Answer - amount of energy required to remove an electron from the atom
- when the electron is removed from the atom, then the coulombic attraction between the electron and the atom must
be zero
14. relationship between pe and ie
Answer - low pe corresponds to large ie
- high pe corresponds to small ie
1. rutherford's gold foil experiment setup
Answer - shot a beam of alpha particles at thin sheet of pure gold foil
- tracked the deflections of the alpha particles after hitting the gold foil
2. alpha particles
Answer positively charged, relatively massive
3. conclusions
Answer rutherford's experiment
- most particles went straight through (no deflection) => most of the atom is empty space
- small amount of particles had shallow deflections => small part of the atom is positively charged
- even smaller amount of particles "bounced back" (large deflection) => small part of the atom is more massive than
the alpha particle, thus is very dense
4. overall conclusion of rutherford's experiment
Answer atom is mostly empty space with a dense, positively-charged nucleus
5. moseley's x-ray experiment setup
,Answer bombarded atoms with energy and then measured x-ray frequencies emitted
6. moseley's experiment results
Answer - each element emits a unique x-ray frequency (unique, physical property)
- assigned an atomic number to each element - integer ranking of ordered mass
- when graphing x-ray frequency against atomic number, it formed a perfect curve (comparing the square root of x-ray
to atomic nmber => linear relationship)
7. moseley's experiment conclusionsAnswer - x-ray frequency is unique for each atom =>
corresponds to a unique, physical property of each element (there is a perfect relationship between x-ray frequency
and atomic number, so atomic number must also correspond to a unique physical property of the atom)
- atomic number are integers => implies we are counting something
8. moseley's main conclusionAnswer atomic number counts the number of protons (units of positive
charge) in the nucleus of an atom
9. coulomb's lawAnswer - quantifies the force between two objects
- PE is directly proportional to V(r) = q1*q2/r
10. potential energyAnswer - proportional to coulomb's law
,- will always be negative => large magnitudes of potential energy are lower numbers than small magnitudes of potential
energy
- lower potential energy => more stable state (electron experiencing greater coulombic attractions)
, 11. factors determining potential energy
Answer - core charge
- radius
- electron-electron repulsion
12. core charge
Answer - ettective nuclear charge felt by an electron
- core charge = # protons - # shielding electrons
- accounts for attraction and sum of all repulsions
13. ionization energy
Answer - amount of energy required to remove an electron from the atom
- when the electron is removed from the atom, then the coulombic attraction between the electron and the atom must
be zero
14. relationship between pe and ie
Answer - low pe corresponds to large ie
- high pe corresponds to small ie
