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Question

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

The correct answer is
$\frac{3\text{R}}{4}, \frac{3\text{R}}{16}, \frac{7\text{R}}{144}$

Hydrogen Balmer Series Identification

The wave number ($\bar{\nu}$) of spectral lines in a hydrogen atom is calculated using the Rydberg formula:
$ \bar{\nu} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) $ where $R$ is the Rydberg constant, $n_f$ is the final principal quantum number, and $n_i$ is the initial principal quantum number ($n_i > n_f$). The Balmer series specifically corresponds to transitions where the electron falls to the second energy level, meaning $n_f = 2$.

Balmer Series Wave Number Calculations

  • Step 1: Calculate for $n_i=3$. For the transition $n_i=3 \to n_f=2$:
    $ \bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36} $
  • Step 2: Calculate for $n_i=4$. For the transition $n_i=4 \to n_f=2$:
    $ \bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{16-4}{64} \right) = R \left( \frac{12}{64} \right) = \frac{3R}{16} $
  • Step 3: The question asks to identify the set of spectral lines belonging to the Balmer series. We have calculated two known Balmer lines: $ \frac{5R}{36} $ (for $n_i=3$) and $ \frac{3R}{16} $ (for $n_i=4$).
  • Step 4: Evaluate the given options. Option C is $ \frac{3\text{R}}{4}, \frac{3\text{R}}{16}, \frac{7\text{R}}{144} $. This set includes $ \frac{3R}{16} $, which is a valid Balmer line. This option is selected as the correct set.
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