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Wave nature is most evident with particles that are confined to a small region of space, for ex.
e

bound to a nucleus. deBroglie relation;
p
=
h
l
KE of particle =
E
=
1
2
mv
2
=
p
2
/2
m
Thus,
λ
=
h
p
=
h
nmE
=
h
2
2
m
1
E
;
h
2
2
m
=150
eV
2
Wavelength of E = 10eV is:
=
150
eV
(
)
1
10
eV
~ 4
; approx. size of atom =
/2
p
~ 0.6
In order for an atom to exist it much have some
KE (be orbiting the nucleus) otherwise it would collapse onto the proton.
E
total
=
E
=
E
k
+
E
p
;
E
k
=
1
2
mv
2
=
L
2
2
mr
2
;
E
p
= 
1
4
pe
o
e
2
r
Where
L
=
mvr
angular momentum ; minus sign
a
attractive force ;
E
o
= 8.85
10
 12
C
2
Nm
2
Orbits of the
e

are made up of integral # of
’s. Circumference:
2
pr
=
nl
=
n
h
p
n
2
p
h
p
; angular momentum:
L
=
mvr
=
pr
=
n
h
2
p
Quantized in
units of h!
Only certain modes or states of an
e

can exist; boundary conditions will only permit certain
’s to exist and for bound
e

only discrete energy levels exist. H
atoms;
min energy we set
dE dr
= 0
;
dE
dr
= 
L
2
mr
3
+
1
4
pe
o
e
2
r
2
= 0
;

1
r
2
E
k
+
E
p
(
)
= 0
E
k
= 
1
2
E
p
so that
=
ε
o
h
2
pme
2
n
2
Radius has discrete values & orbits.
E
min
=
1
2
E
p
= 
1
8
pe
o
e
2
r
min
= 
me
4
8
e
o
h
2
1
n
2
; For n=1
E
min
= 
me
4
8
e
o
h
2
=  2.12
10
 18
J
=  13.6
eV
(ionization energy for H)
r
min
=
e
o
h
2
pme
2
= 5.3
10
 11
m
= 0.53
= a
o
Bohr radius; E(in eV)
=

13.6
eV
n
2
n=1,2,3… If we apply UV light such that hv>13.6eV (
=
91
nm
) we can free the
e

from the nucleus (ionization).
∆
E
n
=
E
initial

E
final
=13.6
eV
1
n
2
inital

1
n
2
final
Thermal Equilibrium and Population Inversion
Boltzmann Distribution
N
i
=
N
o
e

Ei

Eo
(
)
kbT
i=1,2,3…
N
i
=# atoms/mol per unit volume in the
i
th
state,
N
o
=# atoms/mol per unit volume in the ground state &
k
b
=1.4
10
 23
J
/
K
= 8.7
10
 5
eV
/
K
is Boltzmann’s constant.
k
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This note was uploaded on 03/03/2008 for the course A&EP 110 taught by Professor Gaeta during the Fall '05 term at Cornell.
 Fall '05
 GAETA

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