EE403/503 Introduction to Plasma Processing
Examples solutions
Test 1: Basic Atomic Theory
Example 1. How fast is an Electron and a Hydrogen atom at room temperature?
Electrons:
me = 9.1 x 10-31 Kg
k = 1.38 x 10-23 J/K (Boltzmann’s Constant)
T = 293K
3
1
kT  mv 2
2
2
3kT
v
m
v  1.2  10 5 m / s
Hydrogen:
mH = 1.66 x 10-27 Kg
k = 1.38 x 10-23 J/K (Boltzmann’s Constant)
T = 293K
3kT
m
 2703m / s
v rms 
vrms
Example 2. In the vacuum tube shown in the figure below, determine the location,
velocity and energy of e, H and H+ at the time when the electron arrives at one of
the electrodes. All three particles are initially at rest.
me = 9.1 x 10-31 Kg
k = 1.38 x 10-23 J/K (Boltzmann’s Constant)
T = 293K
Time for electron arrives at the anode t
1
x  at 2 (x = distance travel)
2
= Electric Field)
a
F
me
F  qE
(E
dV 2
 1
dx 2
qE 1.6  10 19  1
a

 1.76  1011 m / s 2
31
me
9.1  10
1
1

x  at 2
(1)  (1.76  1011 )t 2
2
2
6
t  3.37  10 s
E

Electron:
Location : at the anode
Velocity:
v  at  1.76  1011  3.37  10 6
v  5.93  10 5 m / s
1
1
Energy:
E  mv 2  (9.11  10 31 )(5.93  10 5 ) 2
2
2
19
E  1.6  10 J  1.0eV
Ion:
qE 1.6  10 19  1

 9.58  10 7 m / s 2
me
1.67  10 27
1
1
x  at 2  (9.58  10 7 )(3.37  10 6 )
2
2
3
x  0.5  10 m = 0.5 mm from the center toward cathode
v  at  9.58  10 7  3.37  10 6
v  323m / s
1
1
E  mv 2  (1.67  10  27 )(323) 2
2
2
23
E  8.7  10 J  5  10 4 eV
a
Location:
Velocity:
Energy:
Atom:
qE 0  E

0
me
me
x  0 at the center
v  0 at rest
E0
a
Location:
Velocity:
Energy:
Example 3. Using the Bohr’s model, write a program to:
(1) Calculate the five lowest energy states including the ground state
of hydrogen atoms in eV.
(2) Calculate the ionization potential of hydrogen atoms in eV.
(3) Find the energy states of hydrogen atoms that can absorb
photons with wavelengths in the visible range (400-700 nm).
(1):
1 

En  13.59921  2 
 n 
n=1
E1 = 0
eV 
n=2
n=3
n=4
n=5
(2):
(3):
E2 = 10.19 (eV)
E1 = 12.08 (eV)
E1 = 12.74 (eV)
E1 = 13.05 (eV)
1
1
 2]
2
n1 n2
RH = 13.6 eV
n1 = 1
Eionization  RH  13.6eV
E  RH [
E photon  E  hv 
hc
(6.63  10 34 )(3  10 8 )
  400nm

E photon  4.97  10 19  3.1eV
  700nm
E photon  2.84  10 19  1.6eV
Series
Lyman
Balmer
Paschen


n2 = 
n
E (eV)
 (nm)
12
13
14
15
10.20
12.09
12.75
13.06
121.32
102.37
97.06
94.78
23
24
25
1.89
2.55
2.86
655.15
485.29
433.30
34
35
0.66
0.97
1871.85
1279.58
Visi