The wave leagth of the first line of lyman series of hydrogen is identical to that second line of balmer series for some hydrogen like ion `X` The `IF_(2)` for X is

Options

A. `-54.4 eV`
B. `-328 eV~`
C. `-13.6 eV`
D. `-3.8 eV`

Explanation

Gives `lambda(lyman H_(2) = lambda (Balmer , X)`
i.e `R_(H)xx 1^(2)((1)/(1^(2)) - (1)/(2^(2))) = R_(2)X^(2) ((1)/(2^(2)) - (1)/(3^(2)))`
Second line Balmer `rArr n = 3`
`rArr X = 2` Thus
`IE_(2) = - 13.6(z^(2))/(n^(2)) = - 13.6 xx (2^(2))/(n^(2)) = -13.6 eV`