Question

In the Brackett series, \(\mathbf{n}_{\mathrm{lo}}=4\) (a) Calculate the wavelength in nanometers of a transition from \(\mathrm{n}=6\) to \(\mathrm{n}=4\) (b) In what spectral region are these lines formed?

Short answer

Answer: The wavelength of the emitted photon during the transition from n=6 to n=4 is 1305 nm, and it belongs to the Near-Infrared (NIR) spectral region.

Step-by-step solution

Write down the Rydberg formula for hydrogen transitions

The Rydberg formula for hydrogen transitions is given as: \(\frac{1}{\lambda} = R_H\left(\frac{1}{n_{lo}^2} - \frac{1}{n_{hi}^2}\right)\) where \(\lambda\) is the wavelength of the emitted photon, \(R_H\) is the Rydberg constant for hydrogen (approximately \(1.097\times10^7 m^{-1}\)), \(n_{lo}\) and \(n_{hi}\) are the principal quantum numbers of the lower and higher energy states, respectively.

Substitute the given values into the Rydberg formula

We are given \(n_{lo} = 4\) and \(n_{hi} = 6\). Substitute these values into the Rydberg formula: \(\frac{1}{\lambda} = R_H\left(\frac{1}{4^2} - \frac{1}{6^2}\right)\)

Calculate the wavelength

Upon substituting the Rydberg constant \(R_H\) and solving for \(\lambda\), we get: \(\frac{1}{\lambda} = 1.097\times10^7\left(\frac{1}{16} - \frac{1}{36} \right)\) \(\frac{1}{\lambda} = 1.097\times10^7\left(0.0208 - 0.0278\right)\) \(\frac{1}{\lambda} = 7.66\times10^5\) \(\lambda = \frac{1}{7.66\times10^5}\) \(\lambda = 1305\times10^{-9} m\) To convert the wavelength into nanometers, the value becomes \(\lambda = 1305\) nm. (a) The wavelength of the emitted photon during the transition from n=6 to n=4 is 1305 nm.

Determine the spectral region

Now that we have the wavelength, we can determine the spectral region where this radiation belongs. Here is a classification of spectral regions by wavelength: - Ultraviolet (UV): \(< 400\) nm - Visible (VIS): \(400 - 700\) nm - Near-Infrared (NIR): \(700 - 2500\) nm - Mid-Infrared (MIR): \(2500 - 50000\) nm Since the wavelength obtained is 1305 nm, it belongs to the Near-Infrared (NIR) region. (b) The spectral region where the emitted photon belongs is the Near-Infrared (NIR) region.

Key concepts

Rydberg Formula

Exploring the mysterious spectra of elements is a fascinating facet of physics, particularly in understanding the transitions between energy levels of electrons in atoms. At the heart of this lies the Rydberg formula, an essential tool for calculating the wavelengths of emitted or absorbed light during these electronic transitions.

The formula, presented as \(\frac{1}{\lambda} = R_H\left(\frac{1}{n_{lo}^2} - \frac{1}{n_{hi}^2}\right)\), is defined by the Rydberg constant \( R_H \) for hydrogen, and it incorporates the principal quantum numbers \( n_{lo} \) and \( n_{hi} \) of the lower and higher energy states, respectively. Here, \( \lambda \) represents the wavelength of the emitted photon when an electron transitions between two specified energy levels.

Spectral Region

When we talk about where a certain wavelength of light 'lives' in the grand theater of electromagnetic spectrum, we're discussing its spectral region. This spectrum is essentially a vast collection of all possible wavelengths of electromagnetic radiation, from radio waves to gamma rays. Each of these wavelengths can be categorized into groups like ultraviolet (UV), visible (VIS), and infrared (IR).

For example, after we use the Rydberg formula to calculate a wavelength, as with the Brackett series problem, determining its spectral region reveals the type of light emitted. The Near-Infrared (NIR) region, where wavelengths range from about 700 to 2500 nm, is critical in telecommunications and astronomy. Given the computed wavelength of 1305 nm falls within this range, we can conclude that the emission is in the NIR spectral region.

Principal Quantum Number

Diving deeper into the arena of quantum mechanics, the principal quantum number, denoted as \( n \), emerges as a fundamental concept that determines the energy levels of an electron in an atom. Think of it as the 'address' that indicates the shell where an electron resides, where the shell number increases with the electron's energy.

The principal quantum number starts at 1 and can take any positive integer value. As \( n \) increases, the energy difference between adjacent levels decreases, which corresponds to emission or absorption of photons with longer wavelengths. This becomes clear while solving problems with the Rydberg formula, where \( n_{hi} \) and \( n_{lo} \) play decisive roles in determining the spectral characteristics of the emitted photon.