Question

One of the emission lines of the hydrogen atom has a wavelength of 93.07 nm. (a) In what region of the elertromagnetic spectrum is this emission found? (b) Determine the initial and final values of \(n\) associated with this emission.

Short answer

(a) The emission line with a wavelength of 93.07 nm is in the ultraviolet region of the electromagnetic spectrum. (b) The initial and final values of n associated with this emission are \(n_1 = 2\) (final) and \(n_2 = 3\) (initial).

Step-by-step solution

Identify the region of the electromagnetic spectrum

The given wavelength is 93.07 nm. We can use this information to identify the region of the electromagnetic spectrum to which this emission belongs. The electromagnetic spectrum is divided into various regions, some of which are: - Ultraviolet (UV) region: 10 nm to 400 nm - Visible region: 400 nm to 700 nm - Infrared (IR) region: 700 nm to 1 mm The given wavelength of 93.07 nm falls in the ultraviolet (UV) region. Answer to (a): The emission line with a wavelength of 93.07 nm is in the ultraviolet region of the electromagnetic spectrum.

Use the Rydberg formula to relate wavelength to initial and final quantum numbers

The Rydberg formula for the hydrogen atom's emission spectra is given by: \[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] where \(\lambda\) is the wavelength, \(R_H\) is the Rydberg constant for hydrogen (approximately 1.097373 x \(10^7\) m\(^{-1}\)), \(n_1\) is the final principal quantum number, and \(n_2\) is the initial principal quantum number. It is important to note that \(n_2 > n_1\) for emission. Let's plug in the given value of the wavelength and solve for the combination of \(n_1\) and \(n_2\).

Arrange and solve the Rydberg formula for possible quantum numbers

First, we need to convert the wavelength from nm to meters: \(\lambda = 93.07 \times 10^{-9} \text{m}\) Now, plug in the value of the wavelength and the Rydberg constant into the Rydberg formula and solve for the difference of inverse squares of quantum numbers: \[\frac{1}{93.07 \times 10^{-9} \text{m}} = (1.097373 \times 10^7 \text{m}^{-1}) \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\] \[\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{93.07 \times 10^{-9} \text{m} \times 1.097373 \times 10^7 \text{m}^{-1}} \approx 0.122\] We will now create a table for values of \(n_1\) and \(n_2\) and their corresponding differences in inverse squares of quantum numbers to observe the possible combinations of quantum numbers. | \(n_1\) | \(n_2\) | \(\Delta(\frac{1}{n^2})\) |------|------|----------------| | 1 | 2 | 0.750 | | 1 | 3 | 0.889 | | 1 | 4 | 0.937 | | 1 | 5 | 0.960 | | 1 | 6 | 0.972 | | 2 | 3 | 0.139 | | 2 | 4 | 0.187 | | 2 | 5 | 0.210 | | 2 | 6 | 0.222 | The closest value to 0.122 in our table is 0.139, which corresponds to \(n_1 = 2\) and \(n_2 = 3\). Answer to (b): The initial and final values of n associated with this emission are \(n_1 = 2\) (final) and \(n_2 = 3\) (initial).

Key concepts

Ultraviolet Region

The hydrogen emission spectrum includes several regions across the electromagnetic spectrum. One of these regions is the ultraviolet (UV) region. The electromagnetic spectrum encompasses a wide range of wavelengths and frequencies. The UV region is situated just beyond the visible light spectrum and typically covers wavelengths from 10 nm to 400 nm. In the context of hydrogen emission, when an electron transitions from a higher energy level to a lower one, it can release energy as electromagnetic radiation, which often falls within the UV range.
Given a wavelength of 93.07 nm for a particular emission line of hydrogen, we can determine that this emission falls squarely within the ultraviolet region. Understanding where such emissions lie on the spectrum helps scientists identify the nature of these transitions and further comprehend the behavior of hydrogen and other elements at an atomic level.

Rydberg Formula

The Rydberg formula is a crucial tool for understanding hydrogen's emission spectrum. It mathematically describes the wavelengths of the spectral lines. The formula is:\[ \frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \]where:

• \( \lambda \) is the wavelength of the emitted light.
• \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.097373 \times 10^7 \text{m}^{-1} \).
• \( n_1 \) and \( n_2 \) are the final and initial principal quantum numbers, respectively.

For hydrogen emission, \( n_2 > n_1 \) since electrons move from higher to lower energy states. By inserting the given wavelength into this formula, we can find possible values for \( n_1 \) and \( n_2 \). This formula showcases the relationship between wavelength and quantum numbers, enabling the deduction of energy levels involved in electron transitions.

Quantum Numbers

Quantum numbers are essential for understanding the electronic structure of atoms, especially in regards to their emission spectra. There are four fundamental quantum numbers, but in this context, we're primarily concerned with the principal quantum number, \( n \).
The principal quantum number (\( n \)) indicates the energy level in which the electron resides. It is always a positive integer, and for hydrogen emissions, it allows us to determine the initial and final energy levels involved in an electron's transition. As an electron falls from a higher energy level (\( n_2 \)) to a lower one (\( n_1 \)), it emits energy in the form of light. This energy release can be calculated using the Rydberg formula, which connects the wavelength of the emission to these quantum numbers.
In our given problem, we used the Rydberg formula to establish that the emission involves a transition from \( n_2 = 3 \) to \( n_1 = 2 \). These values pinpoint the specific electron transitions responsible for the observed spectral line in the ultraviolet region.