Question

Calculate the wave length
length of the first
the series limit for the lyman series of
Hydrogen?​

Answer 1

Answer:

For Lyman series, n

1

=1.

For shortest wavelength in Lyman series (i.e., series limit), the energy difference in two states showing transition should be maximum, i.e., n

2

=∞.

So,

λ

1

=R

H

[

1

2

1

−

∞

2

1

]=R

H

λ=

109678

1

=9.117×10

−6

cm

=911.7

A

˚

For longest wavelength in Lyman series (i.e., first line), the energy difference in two states showing transition should be minimum, i.e., n

2

=2.

So,

λ

1

=R

H

[

1

2

1

−

2

2

1

]=

4

3

R

H

or λ=

3

4

×

R

H

1

=

3×109678

4

=1215.7×10

−8

cm

=1215.7

A

˚

.

Explanation:

hope this helps .......