If `n_(1)` and `n_(2)` are the boundary value principal quantum numbers of a portion fo spectrum of emission spectrum of `H` atom, determine the wavelength ( in metre ) corresponding to last line ( longest `lambda)` . Given `:n_(1) + n_(2) =7, n_(2)-n_(1)=3` and `R_(H)=1.097xx10^(7)m^(-1)` . ( Give your answer in multiple of `10^(-6))`

Options

A. 8
B. 4
C. 10
D. 15

Answer

2

Explanation

Here `n_(2)=5 & n_(1)=2`
So longest wavelength means least energy difference transition `i.e. n_(2)=5` to `n_(1)=4`
`(1)/(lambda)=R_(H)(1)^(2)((1)/((4)^(2))-(1)/((5)^(2)))`
So `lambda=4xx10^(-6)m`