3121
Pass
①
the
What's
wavelength
of
3*108=2
C- 2.ir
2
A
3×108
②
( in
light
that
)
nm
2.34
•
✗
000001282
=
.
1282
What
the
is
complete
Iss
sodium Wa)
Is
(c)
'
2s
'
"
( Fl)
Is
'
Oxygen
10)
Is
'
2s
'
'
2s
Quantum
:
( N)
#
describes
n=
3s
frequency
2.34×10
of
"
s
in
nm
configuration
for the
following
elements
'
"
2ps
Zp
"
Numbers
Quantum
Principle
'
Zp
Zp
2s
Fluorine
electron
a
10/4
mis
Carbon
has
the
1,2-3,4
energy
.
-
level
of
an
electron
inside
(1--0)
.
an
atom
→
n= ,
n=2
n=3
Angular
Momentum :
#
Quantum
This
"
*
Sphere
btw
Isn l
-
1=0
1=1
d -71=2
f
sub level
-
d
1=2
-
p
the
-
1=1
Relationship
→
of
-
1=0
→
shape
P
s
s
the
Il )
E-
☒
describes
number
-71=3
let
n
f-
1=3
within
an
energy
level
Magnetic
Quantum
#
This
:
describes
(Me)
Numbers
orbital
the
2
3
soblevel
°
0
a-
a
Me
l
I
wtin
-1
-2
0
,
-1
,
-3 , -2
,
,
,
I
0
1
,
,
2
-1/0,1/2,3
1ms)
Electron Spin :
ms=±tz
Electron
Configuration
Sub Level
/
Is
/
⇐
/
ZP
2s
/
←
/
/
4p
4s
5s
←
6s
←
*
/
7p
2
P
holds
6
☐
holds
10
F
holds
/
Sf
I
3
$
/
Gd
Gp
/
7s
/
/
,
holds
Subleuel
subkvel
4f
sd
Sp
/
}
4cL
/
/
←
3d
3p
3s
←
/
5
14
7-
soba "
Sublevel
③
Energy
The
is
of
J
E-
-
photon
a
that
has
(6.626×10-34)
=
-
C- =hV
C-
=
h=
C-
1.207×10
=
frequency
a
( 1.821
✗
1.821×1016
of
s
10K )
"
j
energ
Planck 's
Constant
(G. 626
×
10-34
VI. frequency
3④
The
is
energy
-
C-
=
E-
De
of
photon
a
2.432
Broglie Wavelength
hlmv
a
frequency
of
3.671
✗
1016s
i.
(6.626×10-34)
=
has
✗ to
of
(3.671+1016)
-17
a
Particle
:
Be round
to
decimal
z
places
-1
-1