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

The radius of the `n^(th)` orbit of hydrogen-like particles is given by,
`r = (0.529n^(2))/(Z) Å`
`r = (52.9n^(2))/(Z) "pm"`
For radius `(r_(1)) = 1.3225 nm`
`= 1.32225 xx 10^(-9) m`
`= 1322.25 xx 10^(-12) m`
`= 1322.25 "pm"`
`n_(1)^(2) = (r_(1)Z)/(52.9)`
`n_(1)^(2) = (1322.25Z)/(52.9)`
Similarly,
`n_(2)^(2) = (211.6Z)/(52.9)`
`(n_(1)^(2))/(n_(2)^(2)) = (1322.5)/(211.6)`
`(n_(1)^(2))/(n_(2)^(2)) = 6.25`

`(n_(1))/(n_(2)) = 2.5`
`(n_(1))/(n_(2)) = 25/10 = 5/2`
`rArr n_(1) =5` and `n_(2) = 2`
Thus, the transition is from the `5^(th)` orbit to the `2^(nd)` orbit. It belongs to the Balmer series.
Wave number `(barv)` for the transition is given by,
`1.097 xx 10^(7) m^(-1)((1)/(2^(2)) - (1)/(5^(2)))`
`= 1.097 xx 10^(7) m^(-1) ((21)/(100))`
`= 2.303 xx 10^(6) m^(-1)`
`:.` Wavelength `(lambda)` associated with the emission transition is given by,
`lambda = (1)/(barv)`
`= (1)/(2.303 xx 10^(6) m^(-1))`
`= 0.434 xx 10^(-6) m`
`= 0.434 xx 10^(-6) m`
`lambda = 434 nm`
`rArr` This transition belongs to Balmer series and comes in the visible region of the spectrum.