The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen like ion x. Calculate energies of the first four levels of x.

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`