Question

The shortest wavelength of transition in Paschen series of `He^(+)` ion in nanometer (nm) is `(1/(R_(H))=91.12nm)`.

Answer 1

Correct Answer - `205.02`
For shortest wavelength `n_(2)=oo`
`1/(lamda_("min"))=4xxR_(H)[1/(3^(2))-1/(oo^(2))]` ltrbgt `1/(lamda_("min"))=4/9 R_(H)`
`lamda_("min")=9/4xx1/(R_(H))nm`
`=205.02nm`