Question

The Hα-line in Balmer series of the hydrogen spectrum has a wavelength 6563 Å From this calculate the wavelength for the first line of the Lyman series (Lyα). [Ans: 1215 Å]

Answer 1

Answer:

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Answer 2

Answer:

1215 Å

Explanation:

Wavelength = 6563 Å

The wavelength of the spectral lines in Hydrogen atom are given by ,

1/λ = 1/R (1/nf²−1/ni²)

where R is the Rydberg constant .

For the first member of the Balmer series ,nf=2 and ni=3

Therefore ,

1/λ = 1/R(1/2²−1/3²)

= 5R/36 --- 1

For the first member of the Lyman series nf=1 and ni=2

Therefore,

1/λ = 1/R( 1/1²−1/2²)

= 3R/4 --- (2)

Now from the equations 1 and 2

= λ1/λ1' =5R/36 × 4/3R

= 5/27

Therefore, λ1' = 5/27×6563

= 1215

Thus, the wavelength for the first line of the Lyman series is 1215 Å.