Wavelength of the first line of Lyman series in
Hatom is (RH = Rydberg constant)
Answers
Answer:
2 Answers
Tony
Aug 18, 2017
121.6
nm
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.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
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
−−−−−−−−−−
