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))`

Answer

4

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`