Inorganic Chemistry 5e Gary Miessler, Paul Fischer
Inorganic Chemistry 5e Gary Miessler, Paul Fischer
Exam (elaborations)
Inorganic Chemistry 5th Edition By Gary Miessler, Paul Fischer, Donald Tarr (Solution Manual)
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Inorganic Chemistry 5e Gary Miessler, Paul Fischer
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Inorganic Chemistry 5e Gary Miessler, Paul Fischer
Inorganic Chemistry 5e Gary Miessler, Paul Fischer, Donald Tarr (Solution Manual)
Inorganic Chemistry 5e Gary Miessler, Paul Fischer, Donald Tarr (Solution Manual)
The energy of the electromagnetic radiation emitted in this transition is too low for
humans to see, in the infrared region of the spectrum.
2.6
E 102823.8530211 cm 1 97492.221701 cm 1 5331.6313201 cm 1
E hc (6.626 1034 J s)(2.998 108 m s)(100cm m)(5331.6313201 cm 1 )
E 1.059 1019 J
This is the same difference found via the Balmer equation in Problem 2.5 for
a transition from n 4 to n 3. The Balmer equation does work well for hydrogen.
2.7 a. We begin by symbolically determining the ratio of these Rydberg constants:
2 2 He + (Z He+ )2 e4 2 2 H (Z H )2 e4
RHe + RH
(4 0 )2 h2 (4 0 )2 h2
RHe + He+ (Z He+ )2 4 He+
(since Z 2 for He )
RH H (Z H ) 2 H
The reduced masses are required (in terms of atomic mass units):
1 1 1 1 1
H me m proton 5.4857990946 10 m 1.007276466812 mu
4
u
1
1823.88 mu1; H 5.482813 104 mu
H
1 1 1 1 1
He me mHe2 5.4857990946 10 m 4.001506179125 mu
4
u
1
1823.13 mu1; He + 5.485047 10 4 mu
He +
c. The energy difference is first converted to Joules:
E hc (6.626 1034 J s)(2.998 108 m s)(100cm m)(329179.76197 cm 1 )
E 6.5391 1018 J
The Rydberg equation is applied, affording nearly the identical Rydberg constant for He+:
1 1 1 1
E RHe + ( 2 2 ) 6.5391 1018 J RHe + ( )
n n 1 4
l h
RHe + 8.7188 1018 J
h2 2
2.8 a. – E ; A sin rx B cos sx
8 2 m x 2
= Ar cos rx – Bs sin sx
x
2
= –Ar 2 sin rx – Bs 2 cos sx
x 2
h2
–
8 m
– Ar
2
2
sin rx – Bs2 cos sx E A sin rx + B cos sx
If this is true, then the coefficients of the sine and cosine terms must be independently
equal:
h 2 Ar 2 h 2 Bs 2
EA ; = EB
8 2 m 8 2 m
8 2 mE 2
r 2 s2 ; r s 2mE
h 2
h
b. A sin rx ; when x 0, A sin 0 0
when x a, A sin ra 0
n
ra n ; r
a
r 2 h 2 n 2 2 h 2 n 2h 2
c. E
8 2 m a 2 8 2 m 8ma 2
a a a
x
2 n a nx nx
d. dx
2 2
A sin dx A
2
sin 2 d
a n a a
0 0 0
a
a 2 1 nx 1 2nx
A – sin 1
n 2 a 4 a 0
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