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.

Answer

`- 54.4 e V, - 13.6 e V, - 6.04 e , - 3.52 e V ,54.4 e V`

Explanation

We know that
`(1)/(lambda) = RZ^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
For first line of Lyman series in hydrogen atom,
`(1)/(lambda_(1)) = R ((1)/(1^(2)) - (1)/(2^(2))) = (3 R)/(4)` (i)
For second line of Balmer series of hydrogen-like ion `X`
`(1)/(lambda) = Z^(2) R ((1)/(2^(2)) - (1)/(4^(2))) = (3 Z^(2) R)/(16)` (ii)
given that `lambda_(1) = lambda_(2)`
`:. (3 R)/(4) = (3 Z^(2) R)/(16) or Z = 2`
Thus, the ion `X` is singly ionized helium atom . Energy of `n^(th)` state of ion `X` is given by
`E_(S) = - (13.4)/(n^(2)) xx Z^(2)`
`:. E_(X)_(1) = - (13.4 xx 4)/(1) = - 54.4 e V`
`E_(X)_(2) = - (13.4 xx 4)/(4) = - 13.6 e V`
`E_(X)_(3) = - (13.4 xx 4)/(9) = - 6.04 e V`
`E_(X)_(4) = - (13.4 xx 4)/(16) = - 3.52 e V`