Question

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electromagnetic radiation is emitted in each transition? a. \(n=4 \rightarrow n=3\) b. \(n=5 \rightarrow n=4\) c. \(n=5 \rightarrow n=3\)

Short answer

a. Transition \(n=4 \rightarrow n=3\): The emitted light has a wavelength of approximately \(1.97 \times 10^{-6}\,m\), corresponding to the infrared region of the spectrum. b. Transition \(n=5 \rightarrow n=4\): The emitted light has a wavelength of approximately \(4.01 \times 10^{-7}\,m\), corresponding to the visible light region of the spectrum. c. Transition \(n=5 \rightarrow n=3\): The emitted light has a wavelength of approximately \(6.55 \times 10^{-7}\,m\), corresponding to the visible light region of the spectrum.

Step-by-step solution

Rydberg Formula

We will use the Rydberg formula to calculate the wavelength, which 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 for hydrogen \((=1.097373 \times 10^7\,\text{m}^{-1})\), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level.

Transition - a. \(n=4 \rightarrow n=3\)

For the first transition \(n_1=3\) and \(n_2=4\). Substitute these values into the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{3^2} - \frac{1}{4^2}\right). \]Now, solve for the wavelength \(\lambda\): \[\lambda = \frac{1}{R_H \left(\frac{1}{3^2} - \frac{1}{4^2}\right)} \approx 1.97 \times 10^{-6}\,\text{m}.\]This wavelength corresponds to infrared radiation in the electromagnetic spectrum.

Transition - b. \(n=5 \rightarrow n=4\)

For the second transition \(n_1=4\) and \(n_2=5\). Substitute these values into the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{4^2} - \frac{1}{5^2}\right). \]Now, solve for the wavelength \(\lambda\): \[\lambda = \frac{1}{R_H \left(\frac{1}{4^2} - \frac{1}{5^2}\right)} \approx 4.01 \times 10^{-7}\,\text{m}.\]This wavelength corresponds to visible light in the electromagnetic spectrum.

Transition - c. \(n=5 \rightarrow n=3\)

For the third transition \(n_1=3\) and \(n_2=5\). Substitute these values into the Rydberg formula: \[\frac{1}{\lambda} = R_H \left(\frac{1}{3^2} - \frac{1}{5^2}\right). \]Now, solve for the wavelength \(\lambda\): \[\lambda = \frac{1}{R_H \left(\frac{1}{3^2} - \frac{1}{5^2}\right)} \approx 6.55 \times 10^{-7}\,\text{m}.\]This wavelength corresponds to visible light in the electromagnetic spectrum.

Summary of Results

a. Transition \(n=4 \rightarrow n=3\): The emitted light has a wavelength of approximately \(1.97 \times 10^{-6}\,m\), corresponding to the infrared region of the spectrum. b. Transition \(n=5 \rightarrow n=4\): The emitted light has a wavelength of approximately \(4.01 \times 10^{-7}\,m\), corresponding to the visible light region of the spectrum. c. Transition \(n=5 \rightarrow n=3\): The emitted light has a wavelength of approximately \(6.55 \times 10^{-7}\,m\), corresponding to the visible light region of the spectrum.

Key concepts

Rydberg Formula

The Rydberg formula is essential in the realm of atomic physics, specifically in understanding the transition of electrons between energy levels in an atom. It provides us with a mathematical expression to calculate the wavelengths of spectral lines in many chemical elements, swiftly becoming a cornerstone in the study of atomic spectra. The equation is stated as:
\[\begin{equation}\frac{1}{\text{λ}} = R \left(\frac{1}{{n_1}^2} - \frac{1}{{n_2}^2}\right),\end{equation}\]where \(λ\) is the wavelength of the electromagnetic radiation, \(R\) is the Rydberg constant specific to the element (for hydrogen, \(R_H = 1.097373 \times 10^7\,\text{m}^{-1}\)), \(n_1\) is the lower energy level, and \(n_2\) is the higher energy level which the electron transitions from.
The Rydberg formula is instrumental for predicting the wavelengths of light emitted during these electronic transitions, and students should appreciate its importance in linking the quantitative aspects of electron energies to the qualitative color outcomes observable in the lab or sky. By plugging in the appropriate values as demonstrated in the provided steps, one can determine the type of electromagnetic radiation emitted. This tool, thus, bridges the gap between abstract quantum phenomena and tangible, measurable entities.

Electromagnetic Radiation

Electromagnetic radiation is a form of energy that is all around us and takes many forms, such as radio waves, microwaves, X-rays and gamma rays. Sunlight is also a form of electromagnetic energy, but visible light is only a small portion of the spectrum, which contains a broad range of energies and wavelengths. When an electron in a hydrogen atom transitions between two energy levels, it either absorbs or emits a photon with an energy that corresponds to the difference between these levels.

• Infrared radiation, like that produced in the transition from \(n=4\) to \(n=3\), is often associated with heat and is used in various technologies, including remote controls.
• Visible light is the portion of the spectrum that is detectable by the human eye. In the exercise, transitions from \(n=5\) to \(n=4\) and \(n=5\) to \(n=3\) yielded visible light.
• Ultraviolet light, which is not produced in the given transitions, but would occur if the transitions were from higher energy levels, can cause skin tanning.

Understanding the nature of electromagnetic radiation not only furthers comprehension in physics but also enriches our understanding of the technological and natural world around us.

Wavelength Calculation

Wavelength calculation plays a vital role in identifying the specific type of electromagnetic radiation. It's about translating the abstract science of electron transitions into concrete numbers that correspond to colors and types of radiation. To calculate the wavelength from the Rydberg formula, one must rearrange the equation to solve for \(λ\), the wavelength. This process involves a bit of algebra but it's straightforward:
\[\begin{equation}λ = \frac{1}{R \left(\frac{1}{{n_1}^2} - \frac{1}{{n_2}^2}\right)}.\end{equation}\]Effective wavelength calculation affords students the opportunity to not merely predict theoretical outcomes, but to connect them with visible and measurable results. For instance, with the right calculation, one can determine that certain transitions correspond to the visible spectrum and predict the color that transition will produce, such as the reddish hue of the hydrogen alpha line in astronomy.
In summary, by methodically applying the Rydberg formula to each specified electron transition, and through careful calculation, students can predict the type of electromagnetic radiation and the corresponding part of the electromagnetic spectrum where these wavelengths fall—whether it be in the infrared, visible, or ultraviolet ranges.