Calculate the wavelength for the emission transition if it starts from the orbit having radius `1.3225 nm` ends at `211.6 p m`. Name the series to which this transition belongs and the region of the spectrum.

Explanation

Radius of nth orbit of H like species = `(0.529 n^2)/(Z)Å`
or `= (52.9 n^2)/(Z) p m`
`r_1 = 1.3225 nm = 1322.5 pm = (52.9 n_1^2)/(Z)`
`r_2 = 211.6 p m = (52.9 n_2^2)/(Z)`
`:. (r_1)/(r_2) = (1322.5)/(211.6) = (n_1^2)/(n_2^2)`
`:. (n_1^2)/(n_2^2) = 6.25 " or " (n_1)/(n_2) = 2.5`
If `n_2 = 2, n_1 = 5`. This transition corresponds to transition from 5th orbit to 2nd orbit. Hence, it belongs to Balmer series.
Now , `bar(v) = 1.097 xx 10^(7) m^(-1) (1/(2^2) - 1/(5^2))`
`= 1.097 xx 10^(7) xx 21/100 m^(-1)`
`:. lambda = 1/v = 1/(23.037 xx 10^5 m^(-1))`
`= 434 xx 10^(-9) m " or " 434 nm`.
It lies in the visible region.