Physics

1. Bohr, 1913

2. atomos = "uncuttable"

3. 2A:B
B
Hydrogen
2A
Oxygen
Watter[H2O]

4. -
+
-
+
+
-
-
+ +
-
1897-J.J.Thomson
Helium gas
[-] Negative Charge
[+] Positive Charge
Unlike-sign charges attract.
e/me = - 1.758821011 C/kg
J.J. Thomson's experiment and the
charge-to-mass ratio of the electron

5. [+] Proton
[-] Electron
1907-Rutherford

6. Robert A. Millikan
-
-
-
Fg = mg
Fe = qE
1910-Millikan oil drop
e/me = - 1.758821011 C/kg
J.J. Thomson's experiment and the
charge-to-mass ratio of the electron
Newton's first law :
SF = 0 [Forces are Balanced]

7. 1913-Bohr
Proton
Electron

8. S
Proton
Electron
Neutron
+a
N

9. Mass spectrometry (MS) is an analytical technique that ionizes chemical species and sorts the ions based
on their mass-to-charge ratio. In simpler terms, a mass spectrum measures the masses within a sample.

10. The Classical/Solar system
Atomic Model is doomed
• Let’s consider atoms as a quasi sun/planet model (only
one planet so that it is just a two body problem.
• The force balance of circular orbits for an electron
“going around” a stationary nucleolus
where v is the tangential velocity of the electron. Circular motion is
accelerated, accelerated charges need to radiate energy off
according to Maxwell, loosing kinetic energy

11. Figure : A prism spectroscope can be used to observe emission
spectra (a). The emission spectra of mercury (b) and barium (c)
show characteristic lines.

12. Figure : This apparatus is used to produce the absorption
spectrum of sodium (a). The emission spectrum of sodium
consists of several distinct lines (b), whereas the absorption
spectrum of sodium is nearly continuous (c).

13. Balmer Series
• In 1885, Johann Balmer found an empirical formula for
wavelengths of the emission spectrum for hydrogen in nm within
the visible range
(where k = 3,4,5…)
• A good physical theory needs to make sense of this empirical result
There is a minimum wavelength corresponding
to a maximum frequency and energy of photons
657
nm
Red
486
Blue
434
Violet
410
397
365
nm
Series
limit

14. Rydberg Equation
• As more scientists discovered emission lines at infrared and
ultraviolet wavelengths, the Balmer series equation was extended
to the Rydberg equation, actually on the basis of this equation,
people went out looking for more lines :
(where RH = 1.096776 x 107 m-1)
Hydrogen Series of Spectral Lines
Discoverer (year) Wavelength n k
Lyman (1916) Ultraviolet 1 >1
Balmer (1885) Visible, Ultraviolet 2 >2
Paschen (1908) Infrared 3 >3
Brackett (1922) Infrared 4 >4
Pfund (1924) Infrared 5 >5
Can be applied to isotopes of hydrogen by modifying R slightly

 R
m
R
e
H

1
7
2
10
09737
.
1
2

 

 m
h
c
m
R e
a
M
m
M
m
e
e
+



Aside:

15. -
+
𝑣
𝑚𝑒
𝐹𝑐
𝐹𝑒
𝑟
𝐾𝐸 =
1
2
𝑚𝑒𝑣2
=
𝑒2
8𝜋𝜀0𝑟
𝑣 =
𝑒
4𝜋𝜀0𝑚𝑒𝑟

16. -
-
-
-
+
Coulomb’s potential
Energy
Radial Distance
𝐾𝐸 =
𝑒2
8𝜋𝜀0𝑟
0
Angular momentum is
quantized in units of h/2π

17. -
-
-
-
+
Energy
Radial Distance
0
𝑣 =
𝑒
4𝜋𝜀0𝑚𝑒𝑟
𝑛 = 1, 2, 3, …
𝑘 =
1
4𝜋𝜀0

18. +
Energy
Radial Distance
0
𝑟1 = 0.529 Å
𝑟2 = 2.116 Å
𝑟3 = 4.761 Å
𝑟4 = 8. 464 Å
𝑟5 = 13. 225 Å
𝑛 = 1
𝑛 = 2
𝑛 = 3
𝑛 = 4
𝐸4 = −0.850 𝑒𝑉
𝐸3 = −1.510 𝑒𝑉
𝐸2 = −3.400 𝑒𝑉
𝐸1 = −13.600 𝑒𝑉
𝑛 = 5 𝐸5 = −0.544 𝑒𝑉
-
-
-
-
-

19. +
Energy
Radial Distance
0
𝑛 = 1
𝑛 = 2
𝑛 = 3
𝑛 = 4
𝑛 = 5
-
-
-
-
ℎ = 4.135667 × 10−15𝑒𝑉 ∙ 𝑠

20. +
Energy
Radial Distance
0
𝑛 = 1
𝑛 = 2
𝑛 = 3
𝑛 = 4
𝑛 = 5
-
-
-
-
𝜐 =
𝑘𝑒2
2ℎ𝑎0
1
𝑛𝑓
2 −
1
𝑛𝑖
2
𝜆 =
𝑐
𝜐
=
2ℎ𝑎0
𝑘𝑒2
𝑛𝑖
2
𝑛𝑓
2
𝑛𝑓
2
− 𝑛𝑖
2 =
1
𝑅∞
𝑛𝑖
2
𝑛𝑓
2
𝑛𝑓
2
− 𝑛𝑖
2
𝑅∞ =
𝑘𝑒2
2ℎ𝑎0
= 1.097373 × 107
𝑚−1

21. + Radial Distance
0
𝑛 = 1
𝑛 = 2
𝑛 = 3
𝑛 = 4
𝐸3 = −1.510 𝑒𝑉
𝐸2 = −3.400 𝑒𝑉
𝐸1 = −13.600 𝑒𝑉
𝑛 = 5 𝐸5 = −0.544 𝑒𝑉
-
𝐸4 = −0.850 𝑒𝑉
-
-
-
∆𝐸5−1= 13.056 𝑒𝑉
∆𝐸4−1= 12.750 𝑒𝑉
∆𝐸3−1= 12.090 𝑒𝑉
∆𝐸2−1= 10.200 𝑒𝑉
∆𝐸5−1= 𝐸5 − 𝐸1
∆𝐸4−1= 𝐸4 − 𝐸1
∆𝐸3−1= 𝐸3 − 𝐸1
∆𝐸2−1= 𝐸2 − 𝐸1
𝜆5−1 = 94.9625 𝑛𝑚
𝜆4−1 = 97.2416 𝑛𝑚
𝜆3−1 = 102.550 𝑛𝑚
𝜆2−1 = 121.5521 𝑛𝑚
Photon
Energy
:
ℎ = 4.135667 × 10−15
𝑒𝑉 ∙ 𝑠
Energy

22. +
Energy
Radial Distance
0
𝑛 =2
𝑛 = 3
𝑛 = 4
𝑛 = 5
𝐸3 = −1.510 𝑒𝑉
𝐸2 = −3.400 𝑒𝑉
𝑛 = 6
𝐸5 = −0.544 𝑒𝑉
-
𝐸4 = −0.850 𝑒𝑉
-
-
-
∆𝐸6−2= 3.022 𝑒𝑉
∆𝐸5−2= 2.856 𝑒𝑉
∆𝐸4−2= 2.550 𝑒𝑉
∆𝐸3−2= 1.89 𝑒𝑉
∆𝐸6−2= 𝐸6 − 𝐸2
∆𝐸5−2= 𝐸5 − 𝐸2
∆𝐸4−2= 𝐸4 − 𝐸2
∆𝐸3−2= 𝐸3 − 𝐸2
𝜆6−2 = 410.2586 𝑛𝑚
𝜆5−2 = 434.1042 𝑛𝑚
𝜆4−2 = 486.1967 𝑛𝑚
𝜆3−2 = 655.9796 𝑛𝑚
ℎ = 4.135667 × 10−15
𝑒𝑉 ∙ 𝑠
𝐸6 = −0.378 𝑒𝑉
Photon
Energy
:

24. Rydberg constant : Error 0.05443 %
𝑅∞ = 1.097373 × 107𝑚−1
𝑅𝐻 = 1.096776 × 107
𝑚−1
+ -
M
m
M
m
e
e
e
+


 
 R
m
R
e
e
H

CM.
𝑟𝑀 𝑟𝑚𝑒
𝑟
𝑀 𝑚𝑒

25. -
+Z
𝑣
𝑚𝑒
𝐹𝑐
𝐹𝑒
𝑟
Atom number with Z > 1

26. +
Energy
Radial Distance
-
-
-
-
-
Radial Distance
Angular momentum is
quantized in units of h/2π
Duality wave
and particle

27. Thank for attention….End