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Question: What is the ratio of \( \Delta \nu = {\nu _{\max }} - {\nu _{\min }} \) for spectral lines correspon...

What is the ratio of Δν=νmax⁡−νmin⁡\Delta \nu = {\nu _{\max }} - {\nu _{\min }} for spectral lines corresponding to the Lyman and Balmer series of Hydrogen?
A. 9:4  9:4\;
B. 4:9  4:9\;
C. 5:7  5:7\;
D. 7:5  7:5\;

Explanation

Solution

Rydberg formula is used to estimate the wavelength of light arising from an electron in motion between energy levels of an atom. When an electron moves from an orbital with high energy to a lower energy state, a photon of light is created, and when the electron changes from low energy to a higher energy state, the atom absorbs a photon of light.
1λ=R[1n12−1n22]\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right] expression is found to be valid for all the lines in the hydrogen spectrum.

Complete answer:
A series of spectral lines of hydrogen atoms in different regions was recognized. These lines were grouped into five different series of lines, each named after the name of its discoverer. These are Lyman series, Balmer series, Paschen series, Brackett series and Pfund series.
When the electron jumps from all the energy levels higher than n=1n = 1 , i.e., n=2,3,4,5.....n = 2,3,4,5..... to n=1n = 1 energy level, the lines obtained fall in the ultraviolet region. These lines are called the Lyman series.
The lines obtained when the electron jumps to the second energy level (n=2)(n = 2) from higher energy levels (n=3,4,5,6..)(n = 3,4,5,6..) fall in the visible region. These are called the Balmer series.
According to Rydberg equation, 1λ=R[1n12−1n22]\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]
Since frequency is inversely proportional to wavelength the equation can be written as,
νc=R[1n12−1n22]\dfrac{\nu }{c} = R\left[ {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right]
For Lyman series: Δν=νmax⁡−νmin⁡\Delta \nu = {\nu _{\max }} - {\nu _{\min }}
Δν=cR[112−1∞2]−cR[1−122]\Delta \nu = cR\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{\infty ^2}}}} \right] - cR\left[ {1 - \dfrac{1}{{{2^2}}}} \right]
Δν=cR[1−34]−cR[14]\Delta \nu = cR\left[ {1 - \dfrac{3}{4}} \right] - cR\left[ {\dfrac{1}{4}} \right] …….(i)
For Balmer series: Δν=νmax⁡−νmin⁡\Delta \nu = {\nu _{\max }} - {\nu _{\min }}
Δν=cR[122−1∞2]−cR[122−132]\Delta \nu = cR\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{\infty ^2}}}} \right] - cR\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right]
Δν=cR4[1−59]\Delta \nu = \dfrac{{cR}}{4}\left[ {1 - \dfrac{5}{9}} \right]
Δν=cR9\Delta \nu = \dfrac{{cR}}{9} ….. (ii)(ii)
Comparing equation (i)(i) and (ii)(ii)
ΔνLymanΔνBalmer=cR4cR9=94\dfrac{{\Delta {\nu _{Lyman}}}}{{\Delta {\nu _{Balmer}}}} = \dfrac{{\dfrac{{cR}}{4}}}{{\dfrac{{cR}}{9}}} = \dfrac{9}{4}
The ratio of Δν=νmax⁡−νmin⁡\Delta \nu = {\nu _{\max }} - {\nu _{\min }} for spectral lines corresponding to the Lyman and Balmer series of hydrogen is 9:4  9:4\; .
Therefore the correct answer is option A.

Note:
The study of emission or absorption spectra is called spectroscopy.
The limiting line of any spectral series in the hydrogen spectrum is the line when n2{n_2} in the Rydberg equation is infinity i.e., n2=∞{n_2} = \infty . This line will have the shortest wavelength and largest wave number.