Question

Calculate the FREQUECY of the last line of lyman series in. hydrogen spectrum.

Answer 1

For last line of lyman series n1 = 1 , n2 = infinte

1/lamda = RZ^2 (1/n1^2 - 1/n2^2)

1 / lamda = R (1/1 - 0)
= R
= 109700cm^-1

v = C/lamda
= C * R
= 3 * 10^8 * 109700
= 3.29 *10^15 / sec

Answer 2

Answer:
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
−−−−−−−−−−