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What is the ratio of wavelength of II line of Balmer series and I line of Lyman series


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For II line of Balmer series $n_1=2,n_2=4$
$\therefore \large\frac{1}{\lambda_{2B}}=$$R_H\times\big[\large\frac{1}{2^2}-\frac{1}{4^2}\big]$
$\Rightarrow R_H\times \large\frac{3}{16}$-----(1)
For I line of Balmer series $n_1=1,n_2=2$
$\therefore \large\frac{1}{\lambda_{1L}}$$=R_H\times \big[\large\frac{1}{1^2}-\frac{1}{2^2}\big]$
$\Rightarrow R_H \times \large\frac{3}{4}$-------(2)
By equ (1) & (2)
$\large\frac{\lambda_{2B}}{\lambda_{1L}}=\frac{R_H\times 3\times 16}{4\times 3\times R_H}$$=4$
Hence (d) is the correct answer.
answered Jan 17, 2014 by sreemathi.v

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