# HYDROGEN ATOM VISIBLE SPECTRUM by klutzfu51

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• pg 1
```									Fig. 4-9, p. 125
HYDROGEN ATOM VISIBLE SPECTRUM

This grouping of lines has a name. It is called the Balmer Series.
(1885, Johann Balmer )
Notice how the spacing increases as the wavelength gets longer
This will become important later.
•     the energy levels are found to follow the
empirical formulas given by Balmer and
Rydberg.
⎛ Z 2 ⎞
E = − 2 . 178 x 10 − 18 J ⎜ 2 ⎟
⎝ n ⎠
Z = atomic number (1 for H)     n = integer (2, 3, ….., ….)
• R = -2.178 x 10-18 J is the Rydberg Constant
• We can use the equation to predict what ΔE is
for any two energy levels
ΔE = E final − E initial
⎛ 1 ⎞               ⎛ 1 ⎞
−18                 −18
ΔE = −2.178x10 J⎜ 2 ⎟ − (−2.178x10 J)⎜ 2 ⎟
⎜n ⎟
⎝ final ⎠           ⎝ n initial ⎠

⎛ 1        1 ⎞
ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟
⎜n
⎝ final n initial ⎟
⎠
• H Atom Energy level diagram Visible Spectrum
Energy
0                                       Note: As n goes to
n=∞
-R/25                            5         Infinity energy levels
-R/16                            4          get closer and closer
-R/9                            3         together.Looks more
Continuous
-R/4                            2
Balmer                                 R
En i = −        2
ni
R
En f = −        2
nf
⎛ 1  1 ⎞
ΔE = E n f   − En i = − R 2 − 2 ⎟
⎜
⎜n   ni ⎟
⎝  f    ⎠

-R                             1
Fig. 4-13, p. 129
⎛ Z2 ⎞
E = −2.178x10−18 J⎜ 2 ⎟
⎝n ⎠

• Energy levels get closer together
as n increases

• at n = infinity, E = 0
Bohr Model of the Atom
• Postulates:
1. The coulomb force on a planetary electron provides
the centripetal acceleration required for a
dynamically stable circular orbit.

2. The only permissible orbits are those in the discrete
set for which the angular momentum
of the electron equals an integer
n times h, where h = h / 2π
Note Quantization of Angular Momentum
Postulates Part II
3. An electron moving in one of these
stable orbits does not radiate.
4. Emission or absorption of radiation
occurs only when an electron makes a
transition from one orbit to another.
The Bohr Hydrogen Atom
• We can use Bohr’s postulates to derive the
energy levels in Hydrogen
• Postulate 1 implies:
Z=1

2      2
mv  ke
= 2
r  r

From the second postulate we can write
mvr = nh
Radius of the Hydrogen Atom
• We get the equation
2 2
n h     2
r=       2
= n a0
mke
Which if evaluated comes out as:

a0 = 5.29 10-11 m ≈ 0.53Å
Energy Levels in the Bohr Atom
• The kinetic energy of an electron in circular motion is:
2        2
1 2    ke   ke
K = mv = r 2 =
2      2r    2r
The potential energy in a Coulomb Potential
2
ke
V =−
r
The total energy is then
2        2          2
ke   ke     ke
E = K +V = −    +     =−
r    2r     2r
Energy Levels Part II
• The energy of a given level depends on
the radius of the electron:   Actually “Reduced Mass
m m /(m + m )
e p   e   p
2          2 4
ke            mk e      w0
En =      2
=−     2 2
=− 2
2n a0          2h n    n
2.178 eV
w0 = 13.60 x 10 -18 Joules
Fig. 4-13, p. 129
• Example: At what wavelength will emission
from
n = 4 to n = 1 for the H atom be observed?
⎛ 1        1 ⎞
ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟
⎜n
⎝ final n initial ⎟
⎠
1        4
−18    ⎛  1⎞
ΔE = −2.178x10                   J⎜1− ⎟ = −2.04 x10−18 J
⎝ 16 ⎠

hc
ΔE = 2.04 x10   −18
J=                     λ = 9.74 x10−8 m = 97.4nm
λ
Alternative Units
• 1 electron volt = 1.602 x 10-19 joules

• 2.18 x 10 – 18 joules/ 1.602 x 10-19 joules/eV

= 13.6 eV
• We can use the equation to predict the
Photon Energy Connecting
any two energy levels

ΔE = E final − E initial
⎛ 1 ⎞               ⎛ 1 ⎞
−18                 −18
ΔE = −2.178x10 J⎜ 2 ⎟ − (−2.178x10 J)⎜ 2 ⎟
⎜n ⎟
⎝ final ⎠           ⎝ n initial ⎠

⎛ 1          1 ⎞
Ephoton =   ΔE = −2.178x10−18 J⎜ 2 − 2 ⎟
⎜n
⎝ final   n initial ⎟
⎠
⎛ 1     1 ⎞ eV
Ephoton = 13.6⎜ 2 − 2 ⎟
⎝n m ⎠
HYDROGEN ATOM
IONIZATION
Energy

= R/n2 = 2.179874113 x 10
-18 J
0.00
Fig. 4-12, p. 129
Extension to Higher Z
• The Bohr model can be extended to any
single electron system….must keep track of Z
(atomic number).

⎛ Z2 ⎞   Z = atomic number
E = −2.178x10−18 J⎜ 2 ⎟
⎝n ⎠     n = integer (1, 2, ….)

• Examples: He+ (Z = 2), Li+2 (Z = 3), etc.
Extension to Higher Z (cont.)

• Example: At what wavelength will emission
from
n = 4 to n = 1 for the He+ atom be observed?
⎛ 1        1 ⎞
ΔE = −2.178x10−18 J (Z 2 ) 2 − 2 ⎟
⎜
⎜n
⎝ final n initial ⎟
⎠
2    1      4
⎛  1⎞
ΔE = −2.178x10     −18
J (4)⎜1− ⎟ = −8.16x10−18 J
⎝ 16 ⎠
hc
−18
ΔE = 8.16x10     J=                λ = 2.43x10−8 m = 24.3nm
λ
λ H > λ He   +
Albert Einstein explained the
photoelectric effect

Explanation involved
light having particle-like
behavior.
–
Fig. 4-14, p. 132
EM Radiation Has Wave
Properties

λν = c
EM Waves Have Particle Properties

hc
E = hν or E=
λ
=
h = Planck’ constant 6.626x 10-34 J ⋅ s
s
Momentum P = h/λ
Next Advance in Understanding Atomic Structure Came
From Asking “If EM Waves Have Particle Properties-Do
Other particles than the photon Have Wave
Properties” ?

• yes- Particles also exhibit duality
De Broglie Associated a Wave Length
With Particles

Planck’s Constant   Mass of particle

λ = h/(mV)
Velocity of particle
Wavelength of an electron traveling 1.24 x 107 meters/sec
0.041 x speed of light
λ = h/(mV)
=         (6.626 x 10-34 Joule sec )
(9.11 x 10-31 kilograms x 1.24 x 107 meter/sec)
= 5.87 x 10-11 meters
comparable to interatomic distances

Wavelength of a baseball traveling 92.5 miles/hour
1.4 x 10-7 speed of light
λ = h/(mV) =       (6.626 x 10-34 Joule sec )
(0.149 kilograms x 41.3 meter/sec)
= 1.08 x 10-34 meters
Much smaller than baseball or distance to home plate
Bohr’s Model was improved upon
in the 1920’s
“WAVE MECHANICAL” or
“QUANTUM MECHANICAL” Model.

An “Operator Equation” – performing the operation
indicated by H on the wave function leads to a simple
answer , the energy times the wave function

They govern the motion of the electrons

IN PARTICULAR

IF CORRECT THEN OUR JOB
IS TO DEFINE H
AND THE WAVE FUNCTION
H Must Contain the Interaction terms
of the electrons
•   Schrödinger proposed a Wave Equation to describe how
electrons move in atoms
•   It has a form   HΨ=ΕΨ
•   Where H is the Hamiltonian “Operator” *
Ψ is the Wave Function       E is the energy of the system

*Simple Example of Operator equation:
In the equation 2a = b, the “operation” is multiply by 2.
Operating on a gives b.

Operations can be more complicated
STANDING WAVES

Fig. 4-17, p. 135
STANDING WAVES

Fig. 4-18, p. 135
Particle in a One Dimensional Box

Fig. 4-24, p. 146
H

GENERAL SOLUTION OF THIS DIFFERENTIAL EQUATION
General
Solution

AT ENDS OF BOX     DEMAND Ψ = 0 WHY?

TRUE WHEN   INTEGER

OR

0   L

SO
If Wave Function squared is probability then integrated over all space must equal 1

SO

We Can Now Plot These Solutions For n =, 1.2, 3 etc.
1

2

3

Fig. 4-24, p. 146
Fig. 4-24, p. 146

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