Five lines in the \(\mathrm{H}\) atom spectrum have these wavelengths \((\) in \(\dot{A}):(a) 1212.7 ;(b) 4340.5 ;(c) 4861.3 ;(d) 6562.8 ;(e) 10,938\) Three lines result from transitions to \(n_{\text {finul }}=2\) (visible series). The other two result from transitions in different series, one with \(n_{\text {final }}=1\) and the other with \(n_{\text {final }}=3\). Identify \(n_{\text {initial }}\) for each line.

The Balmer series is part of the hydrogen atom spectrum that corresponds to electronic transitions where the final state is the second energy level (_{final} = 2). These lines fall in the visible spectrum, making them observable with the naked eye. Johann Balmer first formulated an empirical equation to describe these lines. Using the Rydberg formula, we can express it as: \[ \bar{u} = R \left( \frac{1}{2^2} - \frac{1}{n_{initial}^2} \right) \] where \( R \) is the Rydberg constant, having a value of 109,737 cm^{-1}.
To identify specific transitions for the Balmer series in our exercise, we convert the wavelengths (given in Å) to wavenumbers and match them to known lines in the series. Lines (b), (c), and (d) correspond to this series.

Named after Theodore Lyman, the Lyman series includes transitions where the final state is the ground state (_{final} = 1) of the hydrogen atom. These lines are in the ultraviolet region and are not visible to the naked eye. The Rydberg formula for the Lyman series is: \[ \bar{u} = R \left( \frac{1}{1^2} - \frac{1}{n_{initial}^2} \right) \] By solving for \( n_{initial} \), we can identify which transition corresponds to a given wavelength. For our problem, line (a) falls within the Lyman series.

The Paschen series, discovered by Friedrich Paschen, involves electronic transitions where the final state is the third energy level (_{final} = 3). These lines are in the infrared region, beyond the range of visible light. The Rydberg formula for the Paschen series is: \[ \bar{u} = R \left( \frac{1}{3^2} - \frac{1}{n_{initial}^2} \right) \] By solving for \( n_{initial} \), we determine that line (e) belongs to the Paschen series.

The Rydberg formula is crucial for calculating the wavelengths of spectral lines in many hydrogen-like elements. It is given by: \[ \bar{u} = R \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right) \] where \( \bar{u} \) is the wavenumber, \( R \) is the Rydberg constant (109,737 cm^{-1}), \( n_{final} \) is the final energy level, and \( n_{initial} \) is the initial energy level. Using this formula, we can identify the series and specific transitions in a hydrogen atom's spectrum.

To find the wavenumber (), we first need to convert the wavelength from Ångströms (Å) to centimeters (cm). The conversion factor is 1 Å = 10^{-8} cm. The calculation is as follows:
Line (a): 1212.7 Å = 1.2127 x 10^{-5} cm
Line (b): 4340.5 Å = 4.3405 x 10^{-5} cm
Line (c): 4861.3 Å = 4.8613 x 10^{-5} cm
Line (d): 6562.8 Å = 6.5628 x 10^{-5} cm
Line (e): 10,938 Å = 1.0938 x 10^{-4} cm
The wavenumber is then given by:
\( \bar{u} = \frac{1}{\text{Wavelength in cm}} \)
For example, the wavenumber for line (a) is:
\( \bar{u} = \frac{1}{1.2127 x 10^{-5}} \, cm^{-1} = 8.2472 x 10^{4} \, cm^{-1} \)}