Question

The Lyman series of emission lines of the hydrogen atom are those for which \(n_{f}=1 .\) (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. (b) Calculate the wavelengths of the first three lines in the Lyman series- those for which \(n_{i}=2,3,\) and 4 .

Short answer

The Lyman series of hydrogen are emission lines observed in the UV region of the electromagnetic spectrum. Using the Rydberg formula, we can calculate the wavelengths of the first three lines in the series (ni=2, 3, and 4) as approximately \(1.215 \times 10^{-7}m\), \(9.724 \times 10^{-8}m\), and \(9.574 \times 10^{-8}m\), respectively.

Step-by-step solution

Understand the Lyman series of hydrogen and its relation to the electromagnetic spectrum.

The Lyman series of emission lines of the hydrogen atom are those for which nf = 1. It means that electrons are transitioning from higher energy levels to the lowest energy level (ground state). The Lyman series emissions are found in the UV region of the electromagnetic spectrum since the transitions correspond to large energy changes, which result in shorter wavelengths.

Use the Rydberg formula to calculate the wavelength of the Lyman series.

The wavelengths of the Lyman series can be determined using the Rydberg formula: \[ \frac{1}{\lambda} = R_H\Big(\frac{1}{n_{f}^2} - \frac{1}{n_{i}^2}\Big) \] Where: - λ is the wavelength of the emission line - \(R_H\) is the Rydberg constant, approximately equal to \(1.097 \times 10^7 m^{-1}\) - nf is the final energy level, which should be 1 in the case of the Lyman series - ni is the initial energy level. We can now calculate the wavelengths of the first three lines in the series (those for which ni = 2, 3, and 4).

Calculate the wavelengths for ni = 2, 3, and 4.

Step 1: For \(n_i = 2\): \[ \frac{1}{\lambda_{1}} = R_H\Big(\frac{1}{1^2} - \frac{1}{2^2}\Big) \] \[ \lambda_{1} = \frac{1}{R_H\left(\frac{3}{4}\right)} \] \[ \lambda_{1} = \frac{1}{(1.097 \times 10^7)(\frac{3}{4})} \] \[ \lambda_{1} = 1.215 \times 10^{-7}m \] Step 2: For \(n_i = 3\): \[ \frac{1}{\lambda_{2}} = R_H\Big(\frac{1}{1^2} - \frac{1}{3^2}\Big) \] \[ \lambda_{2} = \frac{1}{R_H\left(\frac{8}{9}\right)} \] \[ \lambda_{2} = \frac{1}{(1.097 \times 10^7)(\frac{8}{9})} \] \[ \lambda_{2} = 9.724 \times 10^{-8}m \] Step 3: For \(n_i = 4\): \[ \frac{1}{\lambda_{3}} = R_H \Big(\frac{1}{1^2} - \frac{1}{4^2} \Big) \] \[ \lambda_{3} = \frac{1}{R_H\left(\frac{15}{16}\right)} \] \[ \lambda_{3} = \frac{1}{(1.097 \times 10^7)(\frac{15}{16})} \] \[ \lambda_{3} = 9.574 \times 10^{-8}m \] In conclusion, the first three lines in the Lyman series (ni=2, 3, and 4) have wavelengths of approximately \(1.215 \times 10^{-7}m\), \(9.724 \times 10^{-8}m\), and \(9.574 \times 10^{-8}m\), respectively. These lines are observed in the UV region of the electromagnetic spectrum.

Key concepts

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, which include everything from radio waves to gamma rays. Each type of radiation or wave on the spectrum is characterized by its wavelength and frequency. At one end of the spectrum, we find long wavelengths and low frequencies, typical for radio waves. On the other end, the short wavelengths and high frequencies belong to gamma rays.
The Lyman series of the hydrogen atom is particularly interesting as it falls within the ultraviolet (UV) region of the spectrum. This is because the transitions from higher energy levels to the ground state in a hydrogen atom involve large energy releases, resulting in these short wavelengths. Hence, when observing the Lyman series, one is essentially looking at ultraviolet light, beyond the violet portion of visible light, which is invisible to the naked eye. This transition in the hydrogen atom's energy levels provides us with insights into atomic structure and behavior.

Rydberg Formula

The Rydberg formula is a mathematical equation used to predict the wavelengths of spectral lines emitted by a hydrogen atom. It plays a crucial role in understanding atomic spectra, particularly for hydrogen. The formula is:\[ \frac{1}{\lambda} = R_H\left(\frac{1}{n_{f}^2} - \frac{1}{n_{i}^2}\right) \]In this equation:

• \( \lambda \) is the wavelength of the emitted light.
• \( R_H \) is the Rydberg constant, approximately \(1.097 \times 10^7 m^{-1}\).
• \( n_f \) and \( n_i \) are the final and initial energy levels of the electron, respectively.

For the Lyman series, \( n_f \) is always 1 because the electron is transitioning to the ground state. By plugging different values of \( n_i \) (such as 2, 3, 4, etc.), we can calculate the specific wavelengths, which correspond to different emission lines within the Lyman series. This formula serves as an essential tool in quantum mechanics and helps in understanding the transition processes in atoms.

Hydrogen Atom Emission Lines

The emission lines of a hydrogen atom are unique signatures produced when an electron transitions between energy levels. In a hydrogen atom, these lines appear as a set of discrete wavelengths on the electromagnetic spectrum. They are a direct indication of the quantized nature of energy levels in atoms. The Lyman series specifically refers to emissions that result when electrons move from higher energy levels (\(n_i\)) down to the first energy level (\(n_f = 1\)). These emissions typically output light in the ultraviolet region, as these transitions involve relatively large energy differences.
In the provided exercise, calculating the wavelengths for these emissions involves understanding which quantum jumps correspond to the series. The primary lines in the Lyman series for \(n_i = 2, 3,\) and \(4\) have been calculated using the Rydberg formula. Each of these lines is identified by a distinct wavelength, facilitating their study and application in understanding the electromagnetic properties of atoms.