Name the series of lines observed in the emission spectrum of hydrogen in
which the minimum value of n2 is 2. Calculate the wavelength of the third line
of this series. [RH = 109677 cm-1]
Answers
Answer:
Explanation:
Explanation:
1
λ
=
R
(
1
(
n
1
)
2
−
1
(
n
2
)
2
)
⋅
Z
2
where,
R = Rydbergs constant (Also written is
R
H
)
Z = atomic number
Since the question is asking for
1
s
t
line of Lyman series therefore
n
1
=
1
n
2
=
2
since the electron is de-exited from
1
(
st
)
exited state (i.e
n
=
2
) to ground state (i.e
n
=
1
) for first line of Lyman series.
enter image source here
Therefore plugging in the values
1
λ
=
R
(
1
(
1
)
2
−
1
(
2
)
2
)
⋅
1
2
Since the atomic number of Hydrogen is 1.
By doing the math, we get the wavelength as
λ
=
4
3
⋅
912
.
A
since
1
R
=
912
.
A
therefore
λ
=
1216
.
A
or
λ
=
121.6
nm
Answer link
Ernest Z. · Truong-Son N.
Aug 18, 2017
λ
=
121.569 nm
Explanation:
The first emission line in the Lyman series corresponds to the electron dropping from
n
=
2
to
n
=
1
.
Lyman 1
Lyman 1
(Adapted from Tes)
The wavelength is given by the Rydberg formula
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
∣
a
a
1
λ
=
−
R
(
1
n
2
f
−
1
n
2
i
)
a
a
∣
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−
where
R
=
the Rydberg constant (
109 677 cm
-1
) and
n
i
and
n
f
are the initial and final energy levels
For a positive wavelength, we set the initial as
n
=
1
and final as
n
=
2
for an absorption instead.
1
λ
=
−
109 677 cm
-1
×
(
1
2
2
−
1
1
2
)
=
109 677
×
10
7
l
m
-1
(
1
4
−
1
1
)
=
109 677 cm
-1
×
1
−
4
4
×
1
=
−
109 677 cm
-1
×
(
−
3
4
)
=
82 257.8 cm
-1
λ
=
1
82 257.8 cm
-1
=
1.215 69
×
10
-5
l
cm
=
1.215 69
×
10
-7
l
m
=
121.569 nmExplanation:
1
λ
=
R
(
1
(
n
1
)
2
−
1
(
n
2
)
2
)
⋅
Z
2
where,
R = Rydbergs constant (Also written is
R
H
)
Z = atomic number
Since the question is asking for
1
s
t
line of Lyman series therefore
n
1
=
1
n
2
=
2
since the electron is de-exited from
1
(
st
)
exited state (i.e
n
=
2
) to ground state (i.e
n
=
1
) for first line of Lyman series.
enter image source here
Therefore plugging in the values
1
λ
=
R
(
1
(
1
)
2
−
1
(
2
)
2
)
⋅
1
2
Since the atomic number of Hydrogen is 1.
By doing the math, we get the wavelength as
λ
=
4
3
⋅
912
.
A
since
1
R
=
912
.
A
therefore
λ
=
1216
.
A
or
λ
=
121.6
nm
Answer link
Ernest Z. · Truong-Son N.
Aug 18, 2017
λ
=
121.569 nm
Explanation:
The first emission line in the Lyman series corresponds to the electron dropping from
n
=
2
to
n
=
1
.
Lyman 1
Lyman 1
(Adapted from Tes)
The wavelength is given by the Rydberg formula
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
∣
a
a
1
λ
=
−
R
(
1
n
2
f
−
1
n
2
i
)
a
a
∣
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−
where
R
=
the Rydberg constant (
109 677 cm
-1
) and
n
i
and
n
f
are the initial and final energy levels
For a positive wavelength, we set the initial as
n
=
1
and final as
n
=
2
for an absorption instead.
1
λ
=
−
109 677 cm
-1
×
(
1
2
2
−
1
1
2
)
=
109 677
×
10
7
l
m
-1
(
1
4
−
1
1
)
=
109 677 cm
-1
×
1
−
4
4
×
1
=
−
109 677 cm
-1
×
(
−
3
4
)
=
82 257.8 cm
-1
λ
=
1
82 257.8 cm
-1
=
1.215 69
×
10
-5
l
cm
=
1.215 69
×
10
-7
l
m
=
121.569 nmExplanation:
1
λ
=
R
(
1
(
n
1
)
2
−
1
(
n
2
)
2
)
⋅
Z
2
where,
R = Rydbergs constant (Also written is
R
H
)
Z = atomic number
Since the question is asking for
1
s
t
line of Lyman series therefore
n
1
=
1
n
2
=
2
since the electron is de-exited from
1
(
st
)
exited state (i.e
n
=
2
) to ground state (i.e
n
=
1
) for first line of Lyman series.
enter image source here
Therefore plugging in the values
1
λ
=
R
(
1
(
1
)
2
−
1
(
2
)
2
)
⋅
1
2
Since the atomic number of Hydrogen is 1.
By doing the math, we get the wavelength as
λ
=
4
3
⋅
912
.
A
since
1
R
=
912
.
A
therefore
λ
=
1216
.
A
or
λ
=
121.6
nm
Answer link
Ernest Z. · Truong-Son N.
Aug 18, 2017
λ
=
121.569 nm
Explanation:
The first emission line in the Lyman series corresponds to the electron dropping from
n
=
2
to
n
=
1
.
Lyman 1
Lyman 1
(Adapted from Tes)
The wavelength is given by the Rydberg formula
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
∣
a
a
1
λ
=
−
R
(
1
n
2
f
−
1
n
2
i
)
a
a
∣
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−
where
R
=
the Rydberg constant (
109 677 cm
-1
) and
n
i
and
n
f
are the initial and final energy levels
For a positive wavelength, we set the initial as
n
=
1
and final as
n
=
2
for an absorption instead.
1
λ
=
−
109 677 cm
-1
×
(
1
2
2
−
1
1
2
)
=
109 677
×
10
7
l
m
-1
(
1
4
−
1
1
)
=
109 677 cm
-1
×
1
−
4
4
×
1
=
−
109 677 cm
-1
×
(
−
3
4
)
=
82 257.8 cm
-1
λ
=
1
82 257.8 cm
-1
=
1.215 69
×
10
-5
l
cm
=
1.215 69
×
10
-7
l
m
=
121.569 nm
- Explanation:
- Name the series of lines observed in the emission spectrum of hydrogen in
- which the minimum value of n2 is 2. Calculate the wavelength of the third line
- of this series. [RH = 109677 cm-1]