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Answers
All you have to do here is to use the Rydberg formula for a hydrogen atom
1
λ
e
=
R
⋅
(
1
n
2
1
−
1
n
2
2
)
Here
λ
e
is the wavelength of the emitted photon (in a vacuum)
R
is the Rydberg constant, equal to
1.097
⋅
10
7
m
−
1
n
1
represents the principal quantum number of the orbital that is lower in energy
n
2
represents the principal quantum number of the orbital that is higher in energy
Notice that you need to have
n
1
<
n
2
in order to avoid getting a negative value for the wavelength of the emitted photon.
So, convert the wavelength from nanometers to meters
434
nm
⋅
1 m
10
9
nm
=
4.34
⋅
10
−
7
m
Rearrange the Rydberg formula to isolate the unknown variables
1
n
2
1
−
1
n
2
2
=
1
λ
e
⋅
1
R
Plug in your values to find
1
n
2
1
−
1
n
2
2
=
1
4.34
⋅
10
−
7
m
⋅
1.097
⋅
10
7
m
−
1
1
n
2
1
−
1
n
2
2
=
0.21
This will be equivalent to
n
2
2
−
n
2
1
(
n
1
⋅
n
2
)
2
=
21
100
At this point, you have two equations with two unknowns, since
{
n
2
2
−
n
2
1
=
21
n
1
⋅
n
2
=
√
100
Use the second equation to write
n
1
=
10
n
2
(*)
Plug this into the first equation to get
n
2
2
−
(
10
n
2
)
2
=
21
n
2
2
−
100
n
2
2
=
21
This is equivalent to
n
4
2
−
100
=
21
⋅
n
2
2
n
4
2
−
21
n
2
2
−
100
=
0
If you take
n
2
2
=
x
, you will have
x
2
−
21
x
−
100
=
0
Use the quadratic equation to get
x
1
,
2
=
−
(
−
21
)
±
√
(
−
21
)
2
−
4
⋅
1
⋅
(
−
100
)
2
⋅
1
x
1
,
2
=
21
±
√
841
2
x
1
,
2
=
21
±
29
2
⇒
{
x
1
=
21
−
29
2
=
−
4
x
2
=
21
+
29
2
=
25
Since we know that
n
2
2
=
x
you can only use the positive solution here, so
n
2
2
=
25
⇒
n
2
=
5
According to equation
(*)
, you will have
n
1
=
10
5
=
2
This means that your transition is taking place from
n
=
5
→
n
=
2
which is part of the Balmer series.