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 the orbit of H like species = `(0.529)/(Z) n^(2) Ã…=(52.9)/(Z) n^(2)`pm
`r_(1)=1.3225 nm = 1322.5 "pm" =(52.9 n_(1)^(2))/(Z)`
`r_(2)=211.6 "pm" =(52.9 n_(2)^(2))/(Z)`
`(r_(1))/(r_(2))=(1322.5)/(211.6)=(n_(1)^(2))/(n_(2)^(2))`
Or `" " (n_(1)^(2))/(n_(2)^(2))=6.25 or (n_(1))/(n_(2))=(6.25)^(1//2)=2.5`
Thus , if `n_(1)=2, n_(2)=5`. This transition corresponds to transition from 5th orbit to 2nd orbit. This means that the transition belongs to Balmer series.
Now, `" "` Wave number `(bar(v))=(1.097xx10^(7)m^(-1))xx((1)/(2^(2))-(1)/(5^(2))) , =1.097 xx 10^(7)xx(21)/(100)m^(-1)`
Wave `(lambda)=(1)/(bar(v))=(1)/((23.037xx10^(5)m^(-1)))=434xx10^(-9) m or 434 nm`
It lies in the visible region of light.