For any given series of spectral lines of atomic hydrogen, let $\Delta \overline{\mathrm{V}}=\Delta \overline{\mathrm{V}}_{\text {max }}-\Delta \overline{\mathrm{V}}_{\text {min }}$ be the difference in maximum and minimum frequencies in $\mathrm{cm}^{-1}$. the ration $\Delta \overline{\mathrm{V}}_{\text {Lymann }} / \Delta \overline{\mathrm{V}}_{\text {Balmer }}$ is

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Atomic Structure

For any given series of spectral lines of atomic hydrogen, let

Δ \overline{V} = Δ \overline{V}_{max} - Δ \overline{V}_{min}

be the difference in maximum and minimum frequencies in

cm^{- 1}

. the ration

Δ \overline{V}_{Lymann} /Δ \overline{V}_{Balmer}

Answer: (C)

For a series the spread in wavenumber runs from the series limit down to the first line. In the Lyman series (

n_{1} = 1

) the largest is

R_{H}

and the smallest is

R_{H} (1 - \frac{1}{4}) = \frac{3 R _{H}}{4}

, so

Δ \overset{ν}{ˉ}_{L y man} = \frac{R _{H}}{4}

. In the Balmer series (

n_{1} = 2

) the largest is

\frac{R _{H}}{4}

and the smallest is

R_{H} (\frac{1}{4} - \frac{1}{9}) = \frac{5 R _{H}}{36}

, so

Δ \overset{ν}{ˉ}_{B a l m er} = \frac{R _{H}}{9}

. The ratio is

\frac{R _{H} /4}{R _{H} /9} = \frac{9}{4}

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option