If the series limit wavelength of the Lyman series for hydrogen atom is `912 Å`, then the series limit wavelength for the Balmer series for the hydrogen atom is

Options

A. 912 Å
B. `912xx2 Å`
C. `912 xx 4 Å`
D. `912//2 Å`

Answer

C

Explanation

For series limit wavelength of Lyman series,
`p=1` and `n=oo`
`1/lambda_(L)=R(1/p^(2)-1/n^(2))=R(1/1-1/oo)`
`:. lambda_(L)=1/R` ...(1)
For series limit wavelength of Balmer series,
`p=2` and `n=oo`
`1/lambda_(B)=R(1/p^(2)-1/n^(2))=R(1/4-1/oo)`
`:. lambda_(B)=4/R` ...(2)
From eq. (1) and (2) we have,
`lambda_(B)=4 lambda_(L)=4xx912 Å`