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May 2, 2024
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Naomi Uiyoshioria 009 Quantum mechanics I 03/14/2024 Table 1. Substance peak Wavelength color Brightness H 4 690 510 410 400 Dim red Dim Blue Blue Violet Dull Dull Bright Dull He 7 700 690 585 400 450 600 400 Red Dim red yellow blue Dim blue Dim yellow Dim Violet Bright Dull Bright Bright Dull Dull dull Ne 8 450 520 630 570 680 585 630 420 Dim blue Dim green Orange Dim yellow Red Dim purple Yellow Dim orange Dull Dull Bright Dull Bright Dull Bright dull Table. 2 Substance Flame color Flame brightness LiCl Bright red bright NaCl Orange bright CuCl2 green/blue light SrCl2 Red bright BaCl2 green light Unknown red bright
Table 3. Solution Absorbed Absorbance color color CuSo4 700-800 Red/orange Light blue CoCl2 410-500 blue/green Light red/pink KMnO4 600-600 Yellow Purple/ violet Table 4 Wavelength (nm) Wavelength (m) Wavelength (m-1) ni ni2 1/ni 657.80 489.90 6.578 x 10-7 4.849 x 10-7 1520218.912 206280.883 3 4 9 16 0.111 0.0625 436.50 413.50 4.365 x 10-7 4.135 x 10-7 229060.745 2418379.686 5 6 25 36 0.04 0.0278 Figure 1. shows the relationship between 1/n
2 and 1
λ trendline is
y= - 1E+07x + 3E +06 the R
2 value is 0.9995
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wwwwwwww
mmmm
10
10
-16
1024
T
10-16
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10
10
-14
1022
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400
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1020
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10-14 10-12
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20
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10-10
10
X-rays
18
-10
10
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10¹
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1018
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TI
X rays UV
10-8
1014
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Visible spectrum
IR
10¹4
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10¹2
10"
500
600
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T
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1010
10-6 104 10-²
12
Microwave
T
10
108
Microwave FM
10
10⁰
10
|AM
Radio waves
700
10⁰°
10⁰
10°
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s
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Long radio waves
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6.
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&
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tab
2.
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Wavenumbers (cm-¹)
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D
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There is no such thing
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OTrue
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&n(x) =
2
sin
NTX
L
n = 1, 2, 3,...
ws(x) = √777 sin (374)
2
2
(esa(z))² = sin³² (37) - (sin (7))*
(√3(x))²
=
7 Å
ΤΑ 7A
(13(x))² =
0.286 Å-¹
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1:51 PM Sun Oct 9
-1/2
5) For a particle on a ring with the wavefunction: () = π
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-1/2
Y(0) = IT
21T
K=-ħ²d²
21d²
2 cos ($)
= kdo
T
(17¹/2)²
211
- f = "Cos (0) . # de
TỈ Cos (O). t d. TỈ Cosco) do
=
21
0
21
cos() calculate the average
cos(0). -ħ² d² do
21 do²
10:52
@ 79%
+ :
5
9
+
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1-2 u
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horva
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BossMabay
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A
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matter
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energy
amplitude
O O O O
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²6°
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|| THz
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DD
F1
F2
F3
F4
F5
F6
F7
F8
F9
F10
F11
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1
2
3
4
5
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