Daphne Fry

2022-05-15

Lines in one spectral series can overlap lines in another. Use the Rydberg equation to see if the range of wavelengths in the n₁=1 series overlaps the range in the n₁=2 series.

Maeve Holloway

The Rydberg equation is given by:
$\frac{1}{\lambda }={R}_{H}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)$
Where,
$\lambda$ - wavelength
${R}_{H}$ - Rydberg constant for hydrogen atom $=1.096×{10}^{7}{m}^{-1}$
${n}_{1}=$ principal quantum number for lower energy level
${n}_{2}=$ principal quantum number for higher energy level
The n = 1 series corresponds to the Lyman series of spectral lines.
The n = 2 series corresponds to the Balmer series of spectral lines.
The n = 3 series corresponds to the Paschen series of spectral lines.
The n = 4 series corresponds to the Brackett series of spectral lines.
The n = 5 series corresponds to the Pfund series of spectral lines.
The Lyman series of lines lie in the UV-region of the electromagnetic spectrum.
The Balmer series of lines lie in the visible region of the electromagnetic spectrum
For Lyman and Balmer series, we will take one lowest and one highest energy transitions to see if they overlap.
For Lyman series, let's take n1 = 1 and n2 = 2.
Thus,
$\frac{1}{\lambda }=1.096×{10}^{7}\left(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}}\right)=8.220×{10}^{6}{m}^{-1}$
and n1 = 1 and n2 = infinity.
$\frac{1}{\lambda }=1.096×{10}^{7}\left(\frac{1}{{1}^{2}}-\frac{1}{\mathrm{\infty }}\right)=1.096×{10}^{6}{m}^{-1}$
Thus, the range of wavelengths of the two series shows that the lines in these two series do not overlap.

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