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1
Atomic Model
2
Line Spectra of Atomic Gases
Neon (Ne)
Mercury (Hg)
Solar absorption spectrum
Emission spectra of gases of atoms are composed of discrete lines (i.e., the wavelengths of the emitted light from the atoms are discrete). Why?
3
Lyman series
Balmer series
Paschen series
2 , 3 , 4 , 1 1
1 1 2 2
n n
R
3 , 4 , 5 , 1 2
1 1 2 2
n n
R
4 , 5 , 6 , 1 3
1 1 2 2
n n
R
Rydberg’s description for the Line
Spectrum of Hydrogen
Wavelength,
Lyman series Balmer series Paschen series
where R = 1.097 x 10^7 m-1^ is the Rydberg constant.
4
Line Spectrum of Hydrogen
By E = hc/the formulae figured out by Rydberg for in the hydrogen spectrum means that the energy of the emitted photons are of the form:
E = hcR(1/n 22 – 1/n 12 ), where n 1 > n 2 are integers.
If the emissions are due to the hydrogen atoms falling from a higher-energy state to a lower-energy state, it makes sense to think that the energy of a hydrogen atom takes on discrete values equal to hcR/n 2 , where n is a number characterizing the energy state of a hydrogen atom. How does this energy quantization (E~1/n^2 , n=1,2,3,…) dependence come from?
5
Observed Line Spectra of Hydrogen
E (^) i Ef hf
Ef Ei
6
Energy of an Atom
r
kZe E
r
kZe mv
E
2
2 2 2 1
KU
r
kZe mv r
kZe r
mv (^) 2 2 2
2 2
Centripetal force
Electric force
r
+Ze
F
v
e-
But this doesn’t lead to quantization of E. To proceed, we need some hypotheses.
7
Bohr’s Model of Atom
De Broglie suggested standing particle waves as an explanation for Bohr’s angular momentum assumption.
Bohr hypothesized that the angular momentum of electrons in an atom is quantized:
nh
mvr n n = 1, 2, 3, …
r
nucleus n
n mv
nh p
h
2 r n n
which leads to Bohr’s hypothesis given above.
8
Bohr’s Model for Hydrogen-like Atoms
By using
nh mvr (^) n (p. 7),
r
kZe E 2
2
and
1 , 2 , 3 ,
2 2
2 2
2 2 4 ^
n n
Z h
mke E (^) n
the energy levels of an atom with atomic no. Z:
. J 2
2 2 18 1018 n
Z
. eV 2
2 13 6 n
Z
En n = 1, 2, 3, …
r
kZe mv
2 (^2)
(p. 6), Bohr derived an expression for
9
Li 2+^ is a lithium atom ( Z=3 ) with only one electron. What is the ionization energy of Li 2+^ in the ground state? (n.b. Ionization energy is the minimum energy required to just free an electron from its atom.)
Ionization Energy of Li 2+
- –(13.6 eV)(3 2 /1^2 )
- +(13.6 eV)(3^2 /1^2 )
- -(13.6 eV)(3^2 /2^2 )
- +(13.6 eV)(3^2 /2^2 )
- Cannot be determined (^10)
Ionization Energy of Li 2+
. eV . eV 122 eV
2 2
2
n
Z
E n
Ionization energy 122 eV
Ionization energy is the minimum energy required to just free an electron from its atom. So the final state of the electron should have zero total energy (i.e., K = U = 0.) Thus, the ionization energy is 0 – E n = - E n. Because an Li 2+^ ion has only one orbiting electron, it is like a hydrogen atom with a nuclear charge of +3e. We can use Bohr’s model to find the ground state energy, E 1 :
11
The full quantum mechanical description of atoms actually involves solving the Schrödinger’s equation - named after Erwin Schrödinger who shared the Nobel Prize in 1933 for his contribution leading to quantum mechanics. The solution to Schrödinger’s equation reveals that four different quantum numbers are required to describe the electronic state of an atom.
1.The principal quantum number n.
Four Quantum Numbers of an Atom
This number determines the total energy of the atom and can have only integer values. It is the same n that appears in Bohr’s Energy Levels.
n 1 , 2 , 3 , 12
Four Quantum Numbers of an Atom
2. The orbital quantum number l****.
This number determines the orbital angular momentum, L , of the electron by:
1 , 2 , 3 ,, n 1
h L
19
Filling the electrons – Shell Model
The table shows some examples of electronic configurations of elements successfully predicted by the Shell Model. These are what would give the lowest possible energy the atoms can have and so is called the ground-state. If the electronic configuration of an atom deviates from that of the ground- state, the atom is in an excited state. De- excitation of an atom leads to emissions.
