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

Calculate the wavelength for the emission transition if it starts from

1.	Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer» Solution :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