Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. 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.