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

Assuming that the involved species is a H atom, the radius of the nth orbit is given by
`r_(n)=0.529 n^2 A=52.9n^2 pm`
Suppose the orbits involved in the transition are `n_1 and n_2`
`r_(n_(1))=1.3225 nm=1322.5 pm =52.9n_(1)^2`
`n_(1)=((1322.5)/(52.9))^(1//2)=5`
`r_(n_(2))=211.6pm=52.9n_(2)^2`
`n_(2)=((211.6)/(52.9))^(1//2)=2`
Thus, `n_(1)=5 and n_(2)=2` i.e, the transition is from 5th orbit to 2nd orbit. This transition belongs to Balmer series. The wave number `barv` of this transition is given by
`barv=1.09679 xx 10^(7) (1/5^2-1/2^2)m^(-1) (therefore R=1.096679 xx 10^(7) m^(-1))`
`=1.09679 xx 10^(7) xx (21)/(100) m^(-1)`
`=2.303 xx 10^(6) m^(-1)`
Wavelength of the transition
`1/barv=1/(2.303 xx 10^(6)) =4.34 xx 10^(-7)m`
=434 nm
This lines lies in the visible region.