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

For a 1 electron system, radius of n-th orbit=`(52.9n^(2))/(Z)`pm
The radius of the orbit from which the transition of the electron occurs `r_(1)=1.3225` nm=1322.5 pm=`(52.9n_(1)^(2))/(Z)`
The radius of the orbit to which the electron is added.
`r_(2)=211.6"pm"=(52.9n_(2)^(2))/(Z)`
So, `(r_(1))/(r_(2))=(1322.5)/(211.6)=(n_(1)^(2))/(n_(2)^(2)) or, (n_(1))/(n_(2))=2.5`
When `n_(1)=5 and n_(2)=2`, the equation obtained for `n_(1) and n_(2)` is satisfied. thus, the transition occurs from n=5 to n=2 and belongs to Balmer series.
`therefore`Wave number `(overline(v))=109677xx((1)/(2^(2))-(1)/(5^(2)))=2.3xx10^(4)cm^(-1)`
and wavelength `(lamda)=(1)/(overline(v))=(1)/(2.3xx10^(4))cm`
`=4.35xx10^(-5)cm=435nm`
Thus, it lies in the visible region.