Question

The frequency of limiting line in balmer series of hydrogen atom

Answer 1

Answer:
f
=
8.225
×
10
14
l
Hz

Explanation:
The Balmer series corresponds to all electron transitions from a higher energy level to
n
=
2
.

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The wavelength is given by the Rydberg formula

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
∣
a
a
1
λ
=
R
(
1
n
2
1
−
1
n
2
2
)
a
a
∣
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−


where

R
=
the Rydberg constant and
n
1
and
n
2
are the energy levels such that
n
2
>
n
1

Since
f
λ
=
c

we can re-write the equation as

1
λ
=
f
c
=
R
(
1
n
2
1
−
1
n
2
2
)

or

f
=
c
R
(
1
n
2
1
−
1
n
2
2
)
=
R
′
(
1
n
2
1
−
1
n
2
2
)

where
R
′
is the Rydberg constant expressed in energy units (
3.290
×
10
15
l
Hz
).

In this problem,
n
1
=
2
, and the frequency of the limiting line is reached
as
n
→
∞
.

Thus,

f
=
lim
n
→
∞

R
′
(
1
4
−
1
n
2
2
)
=
R
′
(
0.25
−
0
)
=
0.25
R
′

=
0.25
×
3.290
×
10
15
l
Hz
=
8.225
×
10
14
l
Hz