For emission line of atomic hydrogen from n i = 8 to n f = n , the plot of wave number v - against 1 n 2 will be: (The Rydberg constant, R H is in wave number unit)

According to Rydberg's equation, 1 λ = v - = R H Z 2 1 n 1 2 - 1 n 2 2 Z = 1 for H atom = R H × 1 2 ( 1 n f 2 − 1 8 2 ) v ¯ = R H n f 2 − R H 64 slope, m = + R H The curve is linear

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ChemistryAtomic StructureSpectrumMedium2 minPYQ_2019

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For emission line of atomic hydrogen from $n_{i} = 8$ to $n_{f} = n,$ the plot of wave number $(\bar{v})$ against $(\frac{1}{n^{2}})$ will be: (The Rydberg constant, $R_{H}$ is in wave number unit)

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Answer:

Solution:

According to Rydberg's equation,

$\frac{1}{λ} = \bar{v} = R_{H} Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

$Z = 1$ for $H$ atom

$= R_{H} \times 1^{2} (\frac{1}{n_{f}^{2}} - \frac{1}{8^{2}})$

$\bar{v} = \frac{R_{H}}{n_{f}^{2}} - \frac{R_{H}}{64}$

slope, $m =$ $+ R_{H}$

The curve is linear

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Spectrum

⏱ 2mℹ️ Source: PYQ_2019

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