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 `= (0.529 n^(2))/(Z) Å = (52.9 n^(2))/(Z)` 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)) = 2.5`
`:.` If `n_(1) = 2, n_(1) = 5`. Thus, the transition is from 5th orbit to 2nd orbit. it belongs to Balmer series.
`bar(v) = 1.097 xx 10^(7) m^(-1) ((1)/(2^(2)) - (1)/(5^(2))) = 1.097 xx (21)/(100) xx 10^(7) m^(-1)`
or `lamda = (1)/(v) = (100)/(1.097 xx 21 xx 10^(7)) m = 434 xx 10^(-9) m = 434 nm`
It lies in the visible region