The wave numbers of three spectral lines of H | vedclass

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Question

(A) For the Balmer series, the wave number $\bar{v}$ is given by the formula: $\bar{v} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$, where $n = 3

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$ \frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100} $

Vedclass · Answer

(A) For the Balmer series, the wave number $\bar{v}$ is given by the formula: $\bar{v} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$, where $n = 3, 4, 5, \dots$ For $n = 3$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36}$ For $n = 4$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left[ \frac{4-1}{16} \right] = \frac{3R}{16}$ For $n = 5$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{25} \right] = R \left[ \frac{25-4}{100} \right] = \frac{21R}{100}$ Thus, the set of spectral lines is $\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100}$.

The wave numbers of three spectral lines of $H$ atom are considered. Identify the set of spectral lines belonging to the Balmer series. ($R =$ Rydberg constant)

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