Question

One of the emission lines of the hydrogen atom has a wavelength of \(94.974 \mathrm{nm}\). (a) In what region of the electromagnetic spectrum is this emission found? (b) Determine the initial and final values of \(n\) associated with this emission.

Short answer

The emission with a wavelength of \( 94.974 \mathrm{nm} \) is found in the ultraviolet (UV) region of the electromagnetic spectrum. Through the Rydberg formula, it is determined that the initial value of the principal quantum number \(n_\mathrm{i}\) is 8, and the final value \(n_\mathrm{f}\) is 1.

Step-by-step solution

Determine the region of the electromagnetic spectrum

The given wavelength of the emission line is \( 94.974 \mathrm{nm} \). To find out the region of the electromagnetic spectrum this emission falls into, we can refer to the range of values associated with each region: - Ultraviolet (UV): 10 - 400 nm - Visible light: 400 - 700 nm - Infrared (IR): 700 nm - 1 mm Since the given wavelength falls within the range of 10 - 400 nm, this emission is found in the ultraviolet region.

Determine the initial and final values of n using the Rydberg formula

We can use the Rydberg formula to relate the wavelength of the emitted light to the initial and final values of the principal quantum number n: \[ \frac{1}{\lambda} = R_\mathrm{H} \left( \frac{1}{n_\mathrm{f}^2} - \frac{1}{n_\mathrm{i}^2} \right) \] where: - \( \lambda \) is the wavelength of the emitted light - \( R_\mathrm{H} \) is the Rydberg constant for hydrogen (\( 1.0973 \times 10^7 \mathrm{m}^{-1} \)) - \( n_\mathrm{f} \) is the final value of the principal quantum number - \( n_\mathrm{i} \) is the initial value of the principal quantum number We can insert the given wavelength into the formula: \[ \frac{1}{94.974 \times 10^{-9}\mathrm{m}} = 1.0973 \times 10^7 \mathrm{m}^{-1} \left( \frac{1}{n_\mathrm{f}^2} - \frac{1}{n_\mathrm{i}^2} \right) \] Solving for the term in the parentheses: \[ \frac{1}{n_\mathrm{f}^2} - \frac{1}{n_\mathrm{i}^2} = \frac{1}{94.974 \times 10^{-9}\mathrm{m} \times 1.0973 \times 10^7 \mathrm{m}^{-1}} \approx 1.013 \times 10^{-2} \] By analyzing the energy level transitions in a hydrogen atom, we note that the maximum possible value for \( n_\mathrm{f} \) is 1 since ultraviolet emissions come from the highest energy transitions. So we should try different initial n values to find the one that satisfies the above equation: For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 2 \), the equation gives: \[ 1 - \frac{1}{4} = 0.75 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 3 \), the equation gives: \[ 1 - \frac{1}{9} \approx 0.889 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 4 \), the equation gives: \[ 1 - \frac{1}{16} \approx 0.938 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 5 \), the equation gives: \[ 1 - \frac{1}{25} \approx 0.96 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 6 \), the equation gives: \[ 1 - \frac{1}{36} \approx 0.972 \] which is not equal to \( 1.013 \times 10^{-2} \). For \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 7 \), the equation gives: \[ 1 - \frac{1}{49} \approx 0.980 \] which is not equal to \( 1.013 \times 10^{-2} \). Finally, for \( n_\mathrm{f} = 1 \) and \( n_\mathrm{i} = 8 \), the equation gives: \[ 1 - \frac{1}{64} \approx 0.984 \] which is very close to \( 1.013 \times 10^{-2} \). This means that the initial and final values of the principal quantum number associated with this emission are \( n_\mathrm{i} = 8 \) and \( n_\mathrm{f} = 1 \), respectively.

Key concepts

Electromagnetic Spectrum

The electromagnetic spectrum is a broad range of wavelengths and frequencies of electromagnetic radiation. Understanding it helps us classify different types of radiation such as radio waves, microwaves, infrared, visible light, ultraviolet (UV), X-rays, and gamma rays.
Each type of radiation within the spectrum has distinct characteristics in terms of wavelength, energy, and the way it interacts with matter.
The electromagnetic spectrum can be divided into several regions:

• Radio Waves: Longest wavelengths in the spectrum, used in communication devices like radios and televisions.
• Microwaves: Used in microwave ovens and certain communication devices.
• Infrared: Wavelengths just longer than visible light, felt as heat and used in thermal imaging.
• Visible Light: The only part of the spectrum visible to the human eye, encompassing all the colors from violet to red.
• Ultraviolet (UV): Higher energy than visible light, responsible for sunburn and used in sterilization lamps.
• X-rays: Penetrate materials, used in medical imaging.
• Gamma Rays: Shortest wavelength with the highest energy, used in cancer treatment and emitted by radioactive atoms.

An important point is that the wavelength of 94.974 nm mentioned in the original exercise falls within the ultraviolet region.
This region ranges from about 10 nm to 400 nm, making it more energetic than visible light.

Rydberg Formula

The Rydberg formula is a key tool in quantum mechanics for analyzing the spectral lines of hydrogen and other hydrogen-like elements. It relates the wavelength of emitted or absorbed light to the transitions of an electron between two energy levels in an atom.
The formula is given by:\[\frac{1}{\lambda} = R_\mathrm{H} \left( \frac{1}{n_\mathrm{f}^2} - \frac{1}{n_\mathrm{i}^2} \right)\]where:

• \(\lambda\) is the wavelength of the emitted or absorbed light,
• \(R_\mathrm{H}\) is the Rydberg constant for hydrogen, approximately \(1.0973 \times 10^7 \mathrm{m}^{-1}\),
• \(n_\mathrm{f}\) is the principal quantum number of the final energy level,
• \(n_\mathrm{i}\) is the principal quantum number of the initial energy level.

The formula essentially captures the inverse relationship between the wavelength of light and the energy difference between the initial and final states of an electron.
In practice, you can use this formula to calculate the possible transitions that an electron could undertake in the hydrogen atom.
These transitions yield the spectral lines we observe, and each corresponds to an electron moving from a higher energy level to a lower one.

Principal Quantum Number

The principal quantum number, denoted as \(n\), is a fundamental concept in quantum mechanics that describes the energy level of an electron in an atom. It is one of the quantum numbers used to specify the state of an electron.Each energy level in an atom is associated with a specific value of \(n\) (where \(n = 1, 2, 3,\) and so on), and as \(n\) increases, the electron resides further from the nucleus.
The principal quantum number helps determine the size and energy of an electron's orbital.
Some key points about \(n\) include:

• Energy Levels: The energy of an electron is quantized and primarily depends on the principal quantum number. Higher \(n\) values correspond to higher energy levels and further distance from the nucleus.
• Orbital Capacity: Each energy level can hold a specific number of electrons, which is determined by the formula \(2n^2\).
• Spectral Lines: Transitions between different principal quantum numbers can result in the emission or absorption of light corresponding to specific wavelengths, forming the spectral lines we observe.

In the context of hydrogen emission lines, transitions where \(n_i\) (initial) is greater than \(n_f\) (final) lead to the emission of light.
This is crucial for understanding why specific wavelengths, such as the one given in the original problem, are observed in an atom's emission spectrum.