Expression for frequency of limited radiation of hydrogen atoms
Answers
Answer:
Electronic transitions in the hydrogen atom are given by the Rydberg equation:
Δ
E
=
−
R
H
(
1
n
2
f
−
1
n
2
i
)
where:
R
H
=
2.18
×
10
−
18
J/H atom
is the Rydberg constant. Yes, it is the ionization energy of hydrogen atom as well!
h
=
6.626
×
10
−
34
J
⋅
s
is Planck's constant.
n
k
is the principal quantum number for the
k
th state. Thus,
n
f
indicates the destination energy level, and
n
i
indicates the initial energy level.
So,
n
=
4
→
1
implies
n
f
=
1
and
n
i
=
4
. This means...
Δ
E
=
−
2.18
×
10
−
18
J/H atom
×
(
1
1
2
−
1
4
2
)
which should be negative with respect to the system! (why?)
...and we do indeed get a negative number...
Δ
E
=
−
2.044
×
10
−
18
J/H atom
The energy released from the
H
atom due to the relaxation process contains photons, which each have energy
|
Δ
E
|
=
E
photon
=
h
ν
where
ν
is the frequency in
s
−
1
.
E
photon
=
+
2.044
×
10
−
18
J/H atom
=
h
ν
and the frequency is therefore:
ν
=
E
photon
h
=
+
2.044
×
10
−
18
J
H atom
×
1 H atom
photon
6.626
×
10
−
34
J
⋅
s
=
3.084
×
10
15
s
−
1
−−−−−−−−−−−−−−