If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of `He^(+)` is:

Options

A. `(5A)/(9)`
B. `(9A)/(5)`
C. `(36A)/(5)`
D. `(36A)/(7)`

Answer

D

Explanation

Shortest wavelength transition in Lyman series will be corresponding to
`n=oo rarrn=1" "(1)/(lambda)=R_(H)(1)^(2)[(1)/(1^(2))-(1)/(oo^(2))]," "lambda=(1)/(R_(H))`
Given wavelength for this transition is A therefore `(1)/(R_(H))=A" ...(i)"`
Longest wavelength transition for `He^(+)` in Pachen series `4rarr3`
`(1)/(lambda)=R_(H)(2)^(2)[(1)/(3^(2))-(1)(4^(2))]`
Putting value of `R_(H)` from equation (i) `lambda=(36A)/(7)`