For any given series of spectral lines of atomic hydrogen, let∆v—= v—max —v—min be the difference in maximum and minimum frequencies in cm–1. The ratio of ∆ν-lyman to ∆ν-balmer

For Lyman series: ν-max=RH 112-1∝2=RH ν-mini.=RH 112-122=3RH4 for lyman series, ∆ν-lyman=RH4 similarly, ∆ν-balmer=RH9

For any given series of spectral lines of atomic hydrogen, let∆v—= v—max —v—min be the difference in maximum and minimum frequencies in cm–1. The ratio of ∆ν-lyman to ∆ν-balmer

For Lyman series: ν-max=RH 112-1∝2=RH ν-mini.=RH 112-122=3RH4 for lyman series, ∆ν-lyman=RH4 similarly, ∆ν-balmer=RH9

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For any given series of spectral lines of atomic hydrogen, let $∆$ $\overset{—}{v}$ = $\overset{—}{v}$ _max $—$ $\overset{—}{v}$ _min be the difference in maximum and minimum frequencies in cm^–1. The ratio of $∆ {\bar{ν}}_{lyman} to ∆ {\bar{ν}}_{balmer}$

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4:1

9:4

5:4

27:5

answer is B.

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Detailed Solution

For Lyman series:

${\bar{ν}}_{\max} = R_{H} (\frac{1}{1^{2}} - \frac{1}{\propto^{2}}) = R_{H} {\bar{ν}}_{mini .} = R_{H} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) = \frac{3 R_{H}}{4} for lyman series, ∆ {\bar{ν}}_{lyman} = \frac{R_{H}}{4} similarly, ∆ {\bar{ν}}_{balmer} = \frac{R_{H}}{9}$