Physical Chemistry 20130501 week 5 Wendesday May 1, 2013 page 1
experiments that show wave property of light:
1. diffraction
2. Interference
particle property of light:
1. photoelectric effect
E=hν
E is particle concept
ν is wave property
Niels Bohr (1913)
applied quantization to hydrogen atom
1
1 1
=R ( 2 - 2 )
λ
n1 n2
R is Rydberg constant = 1.096776*105cm-1
n1 =1,2,3…
n2 =2,3,4…
n2 >n1
cm-1 is wavenumber(unit of energy)
see spectrum handout
Spectral lines at first were empirical.
Bohr’s postulates:
1. Energy of the hydrogen atom is quantized. energy levels E1, E2, E3,… with E1<E2<E3<E4…
Each energy level is called a stationary state or allowed energy. Why doesn’t electron fall into nucleus?
2. An electron in a stationary state does not radiate electromagnetic energy (radiation).
3. Emission line spectrum results from the transition between upper stationary state and a lower
stationary state.
4. The electron in a hydrogen atom stationary state moves in a circular path around the nucleus.
5. The angular momentum of the electron is quantized. (from deBroglie’s theory)
4 and 5 are false.
h
mvr=n ( )
2π
n=1,2,3…for quantization
focus on
1. eq A is a difference formula
2. eq A is energy formula
v is speed
1
1 1
=R ( 2 - 2 )
λ
n1 n2
m is mass of electron
equation A
r is radius
Planck’s quantum of energy is incorporated
E=hν=
hc
1 1
=hc (R ( 2 - 2 ))
λ
n1 n2
energy formula
A hydrogen atom has one electron, which orbits the nucleus at radius r. It’s attracted to the nucleus by
the electric force. The electron is under acceleration, otherwise it would go in a straight line and not be
bound to the atom.
Coulomb’s Law: in cgs units for convenience
F=
ze2
r2
e is charge of electron times charge of proton
centripetal force: F=ma=m
v2
r
ze2 mv 2
ze2
2
=
⟹mv
=
r2
r
r
Etot=KE+PE
z is atomic number
r is radius
v2
is acceleration
r
equation I
kinetic energy + potential energy
1
-ze2
Etot = mv 2 +
2
r
equation II
the-sign due to proton and electron having opposite charges
Etot =
1 ze2 -ze2 1 ze2
+
=2 r
r
2 r
1
1
Etot = mv 2 +(-mv 2 )=- mv 2
2
2
h
mvr=n ( )
2π
ħ=
equation III
equation IV
h
2π
from I, (mvr)v=ze2, substitute into v
n(
v=
2πze2
nh
eq V
h
) v=ze2
2π
speed of electron in Hydrogen atom
v=9.4*107cm/s for n=1, the first orbit
-2mπ2 z 2 e4
Etot =
n2 h2
everything on right side except n is constant
-2mπ2 z 2 e4
=2.18*10-11 erg
h2
n=1,2,3…
for n=1:
E=-2.18*10-18J=-13.6eV
for n=2:
E=-5.4*10-19J=-3.40eV
for n=3:
E=-2.42*10-19J=-1.51eV
For n↑, E gets less negative. As n→∞, E→0, in which case it would be no longer bound to the nucleus
(free electron).
1eV=1.60*10-19J=8066cm-1
eV is electron volts
mvr=n (
v=
h
)
2π
nh
2πmr
Substitute for v in equation I:
mh 2 ze2
m(
) =
2πmr
r
rn =
n2 h2
4π2 mze2
h2
=a =5.29*10-9 cm=.529Å=Bohr ' s radius
4π2 mze2 0
rn=(.529Å)n2
hν=En2 -En1 =
-2mπ2 z 2 e4 2mπ2 z 2 e4 2mπ2 z 2 e4 1
1
+
=
(- 2 + 2 ) divide by h:
2 2
2 2
2 2
n2 h
n1 h
n2 h
n2 n1
ν=
2π2 mz 2 e4 1
1
(- 2 + 2 )
h3
n2 n1
ν=
c
so:
λ
1 2π2 mz 2 e4 1
1
=
(- 2 + 2 )
3
λ
ch
n2 n1
1
1 1
=R ( 2 - 2 )
λ
n1 n2
1 ν
=
λ c
2π2 mz 2 e4
are all constants
ch3
R is Rydberg constant
That formula is correct, even though it was derived based on two false assumptions.
Heisenberg threw in a monkey wrench…