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( 29
( 29
( 29
m
10
21
.
1
10
64
.
1
10
00
.
3
10
626
.
6
E
hc
7
J
18

s
m
8
s
J
34


×
=
×
×
×
=
=
•
λ
Solve:
Calculate the wavelength of light emitted when the hydrogen
electron transitions from
n
= 2 to
n
= 1
E=hc/
λ,
E
n
= 2.18 x 10
18
J (1/
n
2
)
Concept Plan:
Relationships:
n
i
= 2, n
f
= 1
λ,
m
Given:
Find:

=
2
H
1
R
E
n
n
i
,
n
f
∆
E
atom
E
photon
λ
E
c
h
•
=
∆
E
atom
= E
photon
J
10
64
.
1
2
1
1
1
J
10
18
.
2
E
18
2
2
18
atom


×

=

×

=
∆
E
photon
= (1.64 x 10
18
J) =
1.64 x 10
18
J
the unit is correct, the wavelength is in the UV, which is
appropriate because more energy than 3→2 (in the visible)
Check:
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Probability & Radial Distribution
Functions
•
ψ
2
is the probability density
the probability of finding an electron at a particular point in
space
for
s
orbital maximum at the nucleus?
decreases as you move away from the nucleus
•
the Radial Distribution function represents the total
probability at a certain distance from the nucleus
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This note was uploaded on 05/20/2008 for the course CHEM 115 taught by Professor Larkin during the Spring '08 term at Bloomsburg.
 Spring '08
 LARKIN
 Electron, pH

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