Question

caluclate the wavelength of second line in lyman series of hydrogen spectrum
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Answer 1

Answer:

do it by ur own ..............

Answer 2

For the first line in balmer series:

λ

1

=R(

2

2

1

−

3

2

1

)=

36

5R

For second balmer line:

4861

1

=R(

2

2

1

−

4

2

1

)=

16

3R

Divide both equations:

4861

λ

=

16

3R

×

5R

36

λ=4861×

20

27