Question

Calculate the wavelength and frequency of light emitted when an electron changes from \(n=3\) to \(n=1\) in the H atom. In what region of the spectrum is this radiation found?

Short answer

The radiation has a wavelength of \(1.025 \times 10^{-7}\,\text{m}\), frequency \(2.93 \times 10^{15}\,\text{Hz}\), and is in the UV spectrum.

Step-by-step solution

Understanding the Exercise

To solve this problem, we need to calculate the wavelength and frequency of light emitted during the electronic transition from the third energy level \(n=3\) to the first energy level \(n=1\) in a hydrogen atom. This involves using the Rydberg formula to find the wavelength.

The Rydberg Formula

The Rydberg formula for wavelength is given by \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \), where \( R_H \) is the Rydberg constant \( R_H = 1.097 \times 10^7 \text{ m}^{-1} \), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

Applying the Rydberg Formula

Given that \( n_1 = 1 \) and \( n_2 = 3 \), substitute these values into the formula: \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \].Calculate each component inside the parenthesis.

Simplifying the Expression

Calculate \( \frac{1}{1^2} - \frac{1}{3^2} = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \). Now plug this back into the formula: \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{8}{9} \].

Calculating the Wavelength

Calculate \( \frac{8}{9} = 0.8889 \), so \[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.8889 = 9.748 \times 10^6 \text{ m}^{-1} \]. Then, \( \lambda = \frac{1}{9.748 \times 10^6} \approx 1.025 \times 10^{-7} \text{ m} \).

Calculating the Frequency

Use the speed of light equation \( c = \lambda u \) to find the frequency, where \( c = 3.00 \times 10^8 \text{ m/s} \). Rearrange to find frequency: \( u = \frac{c}{\lambda} \). Substitute \( \lambda \approx 1.025 \times 10^{-7} \text{ m}\), so \( u = \frac{3.00 \times 10^8}{1.025 \times 10^{-7}} \approx 2.93 \times 10^{15} \text{ Hz} \).

Determining the Spectrum Region

The wavelength \( \approx 1.025 \times 10^{-7} \text{ m} \) corresponds to the ultraviolet (UV) region of the electromagnetic spectrum.

Key concepts

Electron Transition

When an electron transitions between energy levels in an atom, it either absorbs or emits energy in the form of light. This transition is crucial in understanding how light is produced. In the context of a hydrogen atom, these energy levels are quantized, meaning electrons can only exist at specific energy states. The change from a higher energy level (n=3) to a lower one (n=1) releases energy, a process which emits light. This emission occurs because the electron is losing energy, moving to a state with less energy. Electron transitions are fundamental to many fields, including chemistry and physics, as they explain the behavior of atoms when interacting with light.

Hydrogen Atom

The hydrogen atom is the simplest atom consisting of one proton and one electron. Its simplicity makes it an excellent model for studying atomic structure and behavior. In the Bohr model of the hydrogen atom, which is often used for these calculations, it is depicted as a small nucleus surrounded by electron orbits. Each orbit corresponds to a different energy level. The electron can jump between these levels, resulting in absorption or emission of light. This model helps in understanding the discrete line spectra observed experimentally when light is emitted by the hydrogen atom.

Wavelength Calculation

Calculating the wavelength of light emitted from an electron transition in a hydrogen atom involves using the Rydberg formula. This formula is essential for determining the wavelengths of spectral lines. To find the wavelength \( \lambda \), we use the equation:

• \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)

where \( R_H \) is the Rydberg constant, \( n_1 \) is the final energy level, and \( n_2 \) is the initial energy level. Substituting the values for a transition from \( n=3 \) to \( n=1 \), we find the wavelength comes out to approximately \( 1.025 \times 10^{-7} \text{ m} \). This value corresponds to a specific color or type of light in the electromagnetic spectrum.

Frequency Calculation

The frequency of light \( u \) is directly related to its wavelength \( \lambda \) through the speed of light \( c \). The relationship is expressed by the equation:

• \( u = \frac{c}{\lambda} \)

In this equation, \( c \) is the speed of light, approximately \( 3.00 \times 10^8 \text{ m/s} \). By plugging in the known wavelength, we can calculate the frequency. For the given electron transition, the frequency computes to approximately \( 2.93 \times 10^{15} \text{ Hz} \). This frequency is a key factor in identifying the type of light emitted, such as ultraviolet or visible light.

Ultraviolet Spectrum

The ultraviolet (UV) spectrum is a part of the electromagnetic spectrum that ranges from about 10 nm to 400 nm in wavelength. It sits between visible light and X-rays. UV light is not visible to the human eye, but it can be detected with special instruments. The UV spectrum is divided into several regions, including UV-A, UV-B, and UV-C, which differ in their wavelength ranges and biological effects. In the case of the hydrogen atom transition from \( n=3 \) to \( n=1 \), the emitted light falls into the ultraviolet region, making it a part of the Lyman series. These UV emissions are important for studying various physical phenomena and are used in applications ranging from sterilization to fluorescence.