Question

the wavelength of the first member of balmer series in the hydrogen spectrum is 6563A.calculate the first member of lyman series in the same spectrum

Answer 1

The first member of lyman series in the same spectrum is 1215 A°

Explanation:

Wavelength of hydrogen spectrum in Balmer series. λ1 = 6563 A°

1λ = 1 / R(1 / nf^2 −1 ni^2)

• Where R is the Rydberg constant.

Now for the first member of the Lyman series nf=1 and ni=2.nf=1 and ni=2.

Therefore,

1 / λ1' =1 / R(1 / 1^2 − 1/2^2) = 3R / 4

• Now from equations 1 and 2 we can write that:

λ1 /λ1' = 5R / 36 × 4 / 3R = 5/27

λ1' = 5 / 27 × λ1 = 5 / 27 × 6563 = 1215 A°

Thus the first member of lyman series in the same spectrum is 1215 A°

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The wavelength of the first line of lyman series for hydrogen atom is equal to that of the second line of balmer series for a hydrogen like ion. the atomic number z of hydrogen like ion is: ?

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