MNS 101 DATA SHEETS - DETACH FOR EASY REFERENCE
1
1
H
1.008
3
Li
6.94
11
Na
22.99
19
K
39.10
37
Rb
85.47
55
Cs
132.9
87
Fr
(223)
2
4
Be
9.012
12
Mg
24.31
3
4
20
21
22
Ca
Sc
Ti
40.08 44.96 47.88
38
39
40
Sr
Y
Zr
87.62 88.91 91.22
56
(57-71)
72
Ba
La-Lu
Hf
137.3
178.5
(89-103)
88
104
Ac-Lr
Ra
Rf
226
5
23
V
50.94
41
Nb
92.91
73
Ta
180.9
105
Db
6
24
Cr
52.00
42
Mo
95.94
74
W
183.9
106
Sg
7
25
Mn
54.94
43
Tc
(98)
75
Re
186.2
107
Bh
8
26
Fe
55.85
44
Ru
101.1
76
Os
190.2
108
Hs
9
27
Co
58.93
45
Rh
102.9
77
Ir
192.2
109
Mt
10
28
Ni
58.69
46
Pd
106.4
78
Pt
195.1
110
Ds
11
29
Cu
63.55
47
Ag
107.9
79
Au
197.0
111
Rg
12
30
Zn
65.38
48
Cd
112.4
80
Hg
200.6
112
Cn
13
5
B
10.81
13
Al
26.98
31
Ga
69.72
49
In
114.8
81
Tl
204.4
113
Nh
14
6
C
12.01
14
Si
28.09
32
Ge
72.59
50
Sn
118.7
82
Pb
207.2
114
Fl
15
7
N
14.01
15
P
30.97
33
As
74.92
51
Sb
121.8
83
Bi
209.0
115
Mc
16
8
O
16.00
16
S
32.07
34
Se
78.96
52
Te
127.6
84
Po
(209)
116
Lv
17
9
F
19.00
17
Cl
35.45
35
Br
79.90
53
I
126.9
85
At
(210)
117
Ts
18
2
He
4.003
10
Ne
20.18
18
Ar
39.95
36
Kr
83.80
54
Xe
131.3
86
Rn
(222)
118
Og
Conversion Factors and Constants:
1 Pa = 1 N m−2
1 mm = 10−3 m
e = 1.602×10−19 C
1 Pa = 1 J m−3
1 m = 10−6 m
me = 9.109×10−31 kg
1 MPa = 106 Pa
1 nm = 10−9 m
NA = 6.022×1023 mol−1
1 GPa = 109 Pa
1 pm = 10−12 m
R = 8.3145 J K−1 mol−1
1 mm2 = 10−6 m2
1 pm = 10−10 cm
0 °C = 273.15 K
Module 2:
f covalent = e−( 0.25)E
2
f covalent + fionic = 1
Module 3:
a0 =
a0 = 2r
c0  1.633a0
LD =
1
RD
4
r
3
a0 = 2 2r
a0 = 2r (HCP)
PD =
Acircle =  r 2
0.225 
k
LD =
BD = 3a0
0.414 
PF =
N cellVatom
Vcell
number of atoms centred on a direction vector
length of the direction vector
number of atoms centred on a plane
area of the plane
4
Vsphere =  r 3
3
r+
 0.414
r−
FD = 2a0
Asquare = a0 2
r+
 0.732
r−
Pf =
d=
N
i =1
r+
 0.732
r−
Mi
Vcell N A
Pf = 2r  LD
area of atoms on a plane
area of the plane
Vcube = a03
cell, i
1
Atriangle = bh
2
Module 4:
nv = ne
−
Qv
RT
 = ce
−
kd
b
d hkl =
a0
h +k +l
2
2
 r =  cos  cos 
2
F
A0
=
 y =  0 + Kd −1 2
Module 5:
S=
F
A0
e=
l − l0 l
=
l0
l0
 lf

%EL =  − 1  100%
 l0

 bend =
3FL
2wh 2
 =−
S = Ee
elateral
elongitudinal
Af 

%RA = 1 −
  100%
A0 

Ebend =
elateral =
 = S (1 + e )
d − d 0 d
=
d0
d0
1
Er =  y ey
2
 = ln (1 + e )
=
 A0 

 A
F
A
 = ln 
L3 F
4 wh3
Module 6:
P(V ) = e
 
− 

 0 
m
F (V ) = 1 − e
 
−  
 0 
m
AP =
Ww − Wd
 100%
Ww − Ws
TP =
−B
 100%

B=
Wd
Ww − Ws
Mn
N M
=
N
TP = AP + CP
Module 7:
m
M=
n
N = nN A
mmolecule
M
=
NA
Mw
DPw =
M0
N M
=
N M
i
Mn
DPn =
M0
Mw
2
i
i
i
i
i
i
 =  0e
i
i
i
Q
−t

 = 0 e RT
Module 8:
V = IR
=
RA
l
=
VA
Il
=
 total  T +  D
T =  RT (1 +  R T )
 n-type  n e e
 p-type  p e h
1

J = E
 D = b (1 − x ) x
J=
I
A
E =
V
l
 = n e e + p e  h
vd = eE
 = n e e
 = ni e ( e + h )
i