Question

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Short answer

In the given problem, an excited hydrogen atom emits light with a frequency of \(6.90 \times 10^{14} \, s^{-1}\), and we need to find the principal quantum level for the final state. By converting the frequency to wavelength (\(\lambda \approx 4.35 \times 10^{-7}\, m\)) and using the Rydberg formula, we can determine the final principal quantum number. The calculations result in the final principal quantum number \(n_2\) to be approximately \(5.15\). As principal quantum numbers must be integers, we round it to the nearest integer, so the final principal quantum level is \(n_2 = 5\).

Step-by-step solution

Recall the Rydberg formula for hydrogen

The Rydberg formula for hydrogen 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 of the emitted light, \(R_H\) is the Rydberg constant for hydrogen (approximately \(1.097 \times 10^7 \, m^{-1}\)), \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final states respectively, with \(n_1 < n_2\).

Convert the frequency to wavelength

We are given the frequency of the emitted light, but we need the wavelength to use the Rydberg formula. We can convert the frequency to wavelength using the following formula: \[ v = \frac{c}{\lambda} \] where \(v\) is the frequency of the light, and \(c\) is the speed of light (approximately \(3.0\times 10^8 \, m/s\)). Now we can calculate the wavelength of the light: \[ \lambda = \frac{c}{v} = \frac{3.0 \times 10^8 \, m/s}{6.9 \times 10^{14} \, s^{-1}} \approx 4.35 \times 10^{-7}\, m \]

Apply the Rydberg formula

Now we can substitute the calculated wavelength and principal quantum number of the initial state (n=5) into the Rydberg formula and solve for the final principal quantum number (\(n_2\)): \[ \frac{1}{4.35 \times 10^{-7}\, m} = 1.097 \times 10^7 \, m^{-1} \left(\frac{1}{5^2} - \frac{1}{n_2^2}\right) \]

Solve for the final principal quantum number

We will now isolate the term \(\frac{1}{n_2^2}\) and then solve for \(n_2\): \[ \frac{1}{n_2^2} = \frac{1}{5^2} - \frac{1}{ 4.35 \times 10^{-7}\, m \times 1.097 \times 10^7 \, m^{-1}} \approx 0.038 \] Taking the reciprocal and then the square root of both sides, we get: \[ n_2 \approx \frac{1}{\sqrt{0.038}} \approx 5.15 \] Finally, as the principal quantum number must be an integer, we round it to the nearest integer: \[ n_2 = 5 \] So the final principal quantum level for this electronic transition is 5.

Key concepts

Rydberg Formula

The Rydberg formula is a fundamental tool in quantum mechanics, especially for studying atomic spectra. It describes the wavelengths of electromagnetic waves emitted or absorbed during electronic transitions of hydrogen.
For hydrogen, the formula is:

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

Here, \( \lambda \) represents the wavelength, and \( R_H \) is the Rydberg constant (\( 1.097 \times 10^7 \, m^{-1} \)).
\( n_1 \) and \( n_2 \) are principal quantum numbers of the energy levels involved, with \( n_1 < n_2 \).
Electrons absorb or release energy as they move between these levels, and the Rydberg formula helps calculate the specific wavelengths associated with such transitions, crucial for understanding atomic spectra.

Principal Quantum Number

The principal quantum number, denoted by \( n \), is an essential component in quantum mechanics that classifies the energy levels of an electron within an atom.
It is a positive integer and indicates the main shell or level in which an electron resides.

• Higher \( n \) values correspond to electrons that are further from the nucleus.
• Each increment in \( n \) denotes a new electron shell.
• These shells become progressively larger and have higher energy states.
  
• For example, \( n = 1 \) is the closest shell to the nucleus, and as \( n \) increases, the electron’s potential energy likewise increases.

Principal quantum numbers play a crucial role in determining the size and energy of the electron's orbitals, and hence, the atom's chemical properties.

Hydrogen Atom

The hydrogen atom, the simplest atom in the periodic table, consists of one electron orbiting a single proton.
Despite its simplicity, it is crucial in the field of quantum mechanics due to its manageable analytical properties.

• Electronic transitions within hydrogen give rise to the distinct lines in its atomic spectra.
• The Balmer series, a notable wavelength series for hydrogen, arises from transitions ending in the \( n = 2 \) level.
  
• It serves as a model for understanding other elements, despite their more complex interactions.

With its single electron and proton, the hydrogen atom remains a foundational model for studying atomic behavior, energy states, and transitions, providing insights into more complex atomic interactions.

Quantum Mechanics

Quantum mechanics is the fundamental framework utilized to describe the physical properties of nature at the scale of atoms and subatomic particles.
It differs significantly from classical physics, introducing concepts such as quantization of energy, wave-particle duality, and inherent probabilities.

• It explains the behavior of electrons in atoms, illustrating how they exist in discrete energy levels.
  
• The Schrödinger equation, a central element of quantum mechanics, helps predict how quantum states evolve over time.
• Quantum mechanics underpins the technology behind lasers, semiconductors, and MRI machines.

This field has not only transformed our understanding of atoms but also led to groundbreaking technological advances, making it fundamental to modern physics and various applied technologies.