COMEDK Exam 2022 Solved Paper

Wavelength in Balmer series for hydrogen are given by

1

λ

=RH(

1

22

−

1

n2

)
=RH(

1

4

−

1

n2

),n=3,4,5.....
The second line in Balmer series corresponds to n=4
Hence,

1

λ2

=RH(

1

4

−

1

16

)=

3RH

16

or, λ2=

16

3RH

The wavelength of the first line (n=2) in Lyman series is

1

λ1

=RH(1−

1

22

)=RH(1−

1

4

)=

3RH

4

or, λ1=

4

3RH

∴

λ1

λ2

=

4

3RH

×

3RH

16

=

1

4

⇒ λ1=

λ2

4

=

486.4

4

=121.6nm