For any given series of spectral lines of atomic hydrogen, let \Delta \mathop v\limits^\_ = \Delta \overline v _\max - \Delta \overline v _\min be the difference in maximum and minimum frequencies in cm–1. The ratio Lyman Balmer \Delta \overline v _Lyman \over \Delta \overline v _Balmer is :

Structure of Atom PYQ — JEE Main 2019 · 9 Apr · Shift 1 · Q19 with Solution

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Correct answer: A

\overline{\nu}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad n_2>n_1

where $n_1$ is fixed for the series.

Hence,

\Delta \overline{\nu}=\overline{\nu}_{\max}-\overline{\nu}_{\min}

with

\overline{\nu}_{\max}=R\left(\frac{1}{n_1^2}-0\right)=\frac{R}{n_1^2}

and

\overline{\nu}_{\min}=R\left(\frac{1}{n_1^2}-\frac{1}{(n_1+1)^2}\right).

Therefore,

\Delta \overline{\nu}=\frac{R}{(n_1+1)^2}.

\Delta \overline{\nu}_{\text{Lyman}}=\frac{R}{(1+1)^2}=\frac{R}{4}.

\Delta \overline{\nu}_{\text{Balmer}}=\frac{R}{(2+1)^2}=\frac{R}{9}.

\frac{\Delta \overline{\nu}_{\text{Lyman}}}{\Delta \overline{\nu}_{\text{Balmer}}} =\frac{R/4}{R/9}=\frac{9}{4}.

\boxed{9:4}

which is option A.

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Correct answer: A

\overline{\nu}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad n_2>n_1

where $n_1$ is fixed for the series.

Hence,

\Delta \overline{\nu}=\overline{\nu}_{\max}-\overline{\nu}_{\min}

with

\overline{\nu}_{\max}=R\left(\frac{1}{n_1^2}-0\right)=\frac{R}{n_1^2}

and

\overline{\nu}_{\min}=R\left(\frac{1}{n_1^2}-\frac{1}{(n_1+1)^2}\right).

Therefore,

\Delta \overline{\nu}=\frac{R}{(n_1+1)^2}.

\Delta \overline{\nu}_{\text{Lyman}}=\frac{R}{(1+1)^2}=\frac{R}{4}.

\Delta \overline{\nu}_{\text{Balmer}}=\frac{R}{(2+1)^2}=\frac{R}{9}.

\frac{\Delta \overline{\nu}_{\text{Lyman}}}{\Delta \overline{\nu}_{\text{Balmer}}} =\frac{R/4}{R/9}=\frac{9}{4}.

\boxed{9:4}

which is option A.