#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdfCEM 141 141
- chapter
- chapter
2.pdf##################################################################
2.pdfCEM 141 - chapter 2.pdf###########################################
CEM 141 - chapter 2
_____ light has:
high frequency
high energy
short wavelength - ANS-UV
_____ light has:
low frequency
low energy
long wavelength - ANS-IR
___________ nuclei become more stable when fused together.
____________ nuclei break apart to form lighter, more stable nuclei
* Both reactions release E - ANS-light
heavy
- 1-100 nm
- have diff. properties than bulk materials
- gold nanoparticles interact w/light different depending on the size of the particle.
-properties depend on surface area to size ratio. IMF's are more important in understanding the
properties of nanomaterials. This is the case for many biomolecules which fall into the nano size
range - ANS-nanoparticles
- 13.8 billion years ago
-rapid expansion of the Universe - ANS-the big bang
- absorption of a photon will promote an electron to a higher energy level
- it immediately falls back down --emitting a photon (the metal shines)
- the metal interacts with light of many wavelengths, so the metal appears white or colorless
(silverly). - ANS-properties of metals
- each energy level has a quantum number
- the higher the number the higher the energy
- energy levels are NOT ORBITS (they represent energy only, NOT the distance from the
electron to the nucleus)
- electrons transition between energy levels by absorbing or emitting photons. - ANS-energy
diagrams
- electrons move in orbits arounds nucleus
- these orbits have definite energies and are at definite distances from the nucleus. so the
energies of electrons in atoms are quantized.
#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdf####################################################################################C
141 - chapter 2.pdfCEM 141 - chapter 2.pdf###########################################
,#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdfCEM 141 141
- chapter
- chapter
2.pdf##################################################################
2.pdfCEM 141 - chapter 2.pdf###########################################
- explained emission and absorption spectra by invoking discrete energy levels - characterized
by QUANTUM NUMBERS (n)
- photons of electromagnetic energy are emitted or absorbed by atoms as electrons move from
one energy level to another
- the energy of the photons corresponds to the DIFFERENCE in energy levels of the electrons. -
ANS-niel bohr's model of the atom
- energy is transferred as a particle (photon) that has a definable energy. E = hv
- one photon ejects one electron
- if the photon does not have enough energy - then no electron is ejected - ANS-e/m radiation
as a PARTICLE - p.e. effect
- involve rearrangements (sharing, donating, or accepting) of valence electrons.
- the identity of the element undergoing a chemical reaction doesn't change EVER -
ANS-chemical reactions
- when short wavelength (high frequency) light shines on a metal surface electrons are emitted -
creating a current.
- if the light intensity (brightness) is increased--more electrons are emitted (more light is shining
on the metal)
- theres a THRESHOLD FREQUENCY below which no electrons are ejected - no matter how
bright (intense) the light - ANS-evidence for photoelectric effect
-adding 2 nuclei together - ANS-fusion
-core electrons 'cancel out' the positive charge from the same number of protons
-each electron in the valence shell feels the effect of the protons that are left
ex) Carbon
[He} 2s^2 2p^2
* 2 core, 4 valence electrons ----(6 protons)
ENC = 6 (protons in nucleus) - 2(protons screened by core electrons) = 4
* each valence electron is attracted by 4 protons - ANS-effective nuclear charge
-electrons circling the nucleus like planets around the sun.
- this DOES NOT explain atomic absorption and emission spectra - ANS-rutherford's model of
the atom
-involve nucleus
#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdf####################################################################################C
141 - chapter 2.pdfCEM 141 - chapter 2.pdf###########################################
,#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdfCEM 141 141
- chapter
- chapter
2.pdf##################################################################
2.pdfCEM 141 - chapter 2.pdf###########################################
- often result in change in element (since element is defined by the number of protons) -
ANS-nuclear reactions
-Nuclear Charge: the greater the nuclear charge, the greater the electrostatic forces of
attraction.
-Distance between the nucleus and the valence electrons. Less number of energy shells means
the electrostatic forces of attraction are greater - ANS-
-nuclear reactions are accompanied by changes in mass (unlike chemical reactions)
- the mass is converted into E (mass - E equivalence): E = mc^2
- E is the "binding E", which is ?????????????? - ANS-the energy released when mass is
converted to energy
(electrons are waves and can therefore combine constructively and destructively)
- in this theory, n (# of) atomic orbitals are combined to give n MO's
- bonding MO's are (typically) stabilizing
------electrons in bonding orbitals make the species more stable
- antibonding MO's are of higher E destabilizing (cancelling out)
------electrons in antibonding orbitals make the species less stable. - ANS-Molecular orbit (MO)
theory
(Hz, s^-1, 1/s) the # of wavefronts per second - ANS-frequency
(m) the distance from peak to peak - ANS-wavelength
(when the protons and neutrons fuse together in the nucleus, some of that mass is converted
into E) - ANS-
*matter doesn't behave the same on a small scale
*electrons have wave properties - ANS-
*typically, the # of neutrons > the # of protons for HEAVIER elements (strong nuclear force
overcome p+ <--> p+ repulsion.
*have 'magic numbers' of nucleons (protons or neutrons or both)
-----2,8,20, 50, 82, 126
(there are quantized states in the nucleus too)
*have less than 92 protons - ANS-stable nuclei
# protons im nucleus is increasing, core electrons are staying the same --- across a row - ANS-
1 amu = ?? - ANS-1.6606 x 10^-27 kg
1 J = ???? - ANS-1 kg m^2s^-2
#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdf####################################################################################C
141 - chapter 2.pdfCEM 141 - chapter 2.pdf###########################################
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdfCEM 141 141
- chapter
- chapter
2.pdf##################################################################
2.pdfCEM 141 - chapter 2.pdf###########################################
CEM 141 - chapter 2
_____ light has:
high frequency
high energy
short wavelength - ANS-UV
_____ light has:
low frequency
low energy
long wavelength - ANS-IR
___________ nuclei become more stable when fused together.
____________ nuclei break apart to form lighter, more stable nuclei
* Both reactions release E - ANS-light
heavy
- 1-100 nm
- have diff. properties than bulk materials
- gold nanoparticles interact w/light different depending on the size of the particle.
-properties depend on surface area to size ratio. IMF's are more important in understanding the
properties of nanomaterials. This is the case for many biomolecules which fall into the nano size
range - ANS-nanoparticles
- 13.8 billion years ago
-rapid expansion of the Universe - ANS-the big bang
- absorption of a photon will promote an electron to a higher energy level
- it immediately falls back down --emitting a photon (the metal shines)
- the metal interacts with light of many wavelengths, so the metal appears white or colorless
(silverly). - ANS-properties of metals
- each energy level has a quantum number
- the higher the number the higher the energy
- energy levels are NOT ORBITS (they represent energy only, NOT the distance from the
electron to the nucleus)
- electrons transition between energy levels by absorbing or emitting photons. - ANS-energy
diagrams
- electrons move in orbits arounds nucleus
- these orbits have definite energies and are at definite distances from the nucleus. so the
energies of electrons in atoms are quantized.
#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdf####################################################################################C
141 - chapter 2.pdfCEM 141 - chapter 2.pdf###########################################
,#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdfCEM 141 141
- chapter
- chapter
2.pdf##################################################################
2.pdfCEM 141 - chapter 2.pdf###########################################
- explained emission and absorption spectra by invoking discrete energy levels - characterized
by QUANTUM NUMBERS (n)
- photons of electromagnetic energy are emitted or absorbed by atoms as electrons move from
one energy level to another
- the energy of the photons corresponds to the DIFFERENCE in energy levels of the electrons. -
ANS-niel bohr's model of the atom
- energy is transferred as a particle (photon) that has a definable energy. E = hv
- one photon ejects one electron
- if the photon does not have enough energy - then no electron is ejected - ANS-e/m radiation
as a PARTICLE - p.e. effect
- involve rearrangements (sharing, donating, or accepting) of valence electrons.
- the identity of the element undergoing a chemical reaction doesn't change EVER -
ANS-chemical reactions
- when short wavelength (high frequency) light shines on a metal surface electrons are emitted -
creating a current.
- if the light intensity (brightness) is increased--more electrons are emitted (more light is shining
on the metal)
- theres a THRESHOLD FREQUENCY below which no electrons are ejected - no matter how
bright (intense) the light - ANS-evidence for photoelectric effect
-adding 2 nuclei together - ANS-fusion
-core electrons 'cancel out' the positive charge from the same number of protons
-each electron in the valence shell feels the effect of the protons that are left
ex) Carbon
[He} 2s^2 2p^2
* 2 core, 4 valence electrons ----(6 protons)
ENC = 6 (protons in nucleus) - 2(protons screened by core electrons) = 4
* each valence electron is attracted by 4 protons - ANS-effective nuclear charge
-electrons circling the nucleus like planets around the sun.
- this DOES NOT explain atomic absorption and emission spectra - ANS-rutherford's model of
the atom
-involve nucleus
#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdf####################################################################################C
141 - chapter 2.pdfCEM 141 - chapter 2.pdf###########################################
,#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdfCEM 141 141
- chapter
- chapter
2.pdf##################################################################
2.pdfCEM 141 - chapter 2.pdf###########################################
- often result in change in element (since element is defined by the number of protons) -
ANS-nuclear reactions
-Nuclear Charge: the greater the nuclear charge, the greater the electrostatic forces of
attraction.
-Distance between the nucleus and the valence electrons. Less number of energy shells means
the electrostatic forces of attraction are greater - ANS-
-nuclear reactions are accompanied by changes in mass (unlike chemical reactions)
- the mass is converted into E (mass - E equivalence): E = mc^2
- E is the "binding E", which is ?????????????? - ANS-the energy released when mass is
converted to energy
(electrons are waves and can therefore combine constructively and destructively)
- in this theory, n (# of) atomic orbitals are combined to give n MO's
- bonding MO's are (typically) stabilizing
------electrons in bonding orbitals make the species more stable
- antibonding MO's are of higher E destabilizing (cancelling out)
------electrons in antibonding orbitals make the species less stable. - ANS-Molecular orbit (MO)
theory
(Hz, s^-1, 1/s) the # of wavefronts per second - ANS-frequency
(m) the distance from peak to peak - ANS-wavelength
(when the protons and neutrons fuse together in the nucleus, some of that mass is converted
into E) - ANS-
*matter doesn't behave the same on a small scale
*electrons have wave properties - ANS-
*typically, the # of neutrons > the # of protons for HEAVIER elements (strong nuclear force
overcome p+ <--> p+ repulsion.
*have 'magic numbers' of nucleons (protons or neutrons or both)
-----2,8,20, 50, 82, 126
(there are quantized states in the nucleus too)
*have less than 92 protons - ANS-stable nuclei
# protons im nucleus is increasing, core electrons are staying the same --- across a row - ANS-
1 amu = ?? - ANS-1.6606 x 10^-27 kg
1 J = ???? - ANS-1 kg m^2s^-2
#############################CEM
######################################################################################!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Q
CEM 141 - chapter 2.pdf####################################################################################C
141 - chapter 2.pdfCEM 141 - chapter 2.pdf###########################################
