By calculating 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.

Options

A. Balmer series, visible region.
B. Lyman series, UV-region
C. Paschen series, IR-region.
D. Brackett series, IR-region.

Answer

A

Explanation

Radius of the nth orbit of H like species,
`r_n = (52.9(n^2))/z` pm
`r_1 = 1.3225 nm = 1322.5` pm
`= (52.9n_1^2)/z`
`r_2 = 211.6 " pm " = (52.9n_2^2)/z`
`r_1/r_2 = (1322.5)/(211.6)=(n_1^2)/(n_2^2)`
`(n_1^2)/(n_2^2) = 6.25 " or " n_1/n_2 =2.5`
`n_1 = 5, n_2 = 2`, so, the transition (emission transition) is from 5th orbit to 2nd orbit and it belongs to Balmer series.
`barv= 1/lambda = 1.09677 xx 10^7 (1/(n_1^2)-1/(n_2^2))m^(-1)`
`1/lambda = 1.09677 xx 10^7 (1/2^2 -1/5^2) m^(-1)`
`1/lambda = 1.09677 xx 10^7 xx 21/100 = 2.303 xx 10^6 m^(-1)`
`lambda =0.434xx10^(-6) m =434xx10^(-9)m =434 nm`
It belongs to the visible region.
Note: In the above Rydberg formula if `n_1 = 1, 2... " then " n_2 = n_1 + 1, n_1 +2`...For Balmer series, the spectral region is visible.