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 particles
`=(0.529n^(2))/Z Å=(52.9n^(2))/Z p m`
`r_(1)=1.3225 nm=1322.5 p m =52.9 n_(1)^(2)`
`r_(2)=211.6 p m=(52.9n_(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)=2.5`
`:.` If `n_(2)=2, n_(1)=5`. Thus the transition is from `5th` orbit to `2nd` orbit. It belongs to Balmer series.
`bar(v)=1.097xx10^(7)m^(-1)(1/2^(2)-1/5^(2))`
`=1.097xx21/100xx10^(7) m^(-1)`
or `lambda=1/bar(v)=100/(1.097xx21xx10^(7))m`
`=434xx10^(-9)m=434 nm`
It lies in the visible region.