Question

From the information below, identify element \(\mathrm{X}\). a. The wavelength of the radio waves sent by an FM station broadcasting at \(97.1 \mathrm{MHz}\) is \(30.0\) million \(\left(3.00 \times 10^{7}\right)\) times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state. b. Let \(V\) represent the principal quantum number for the valence shell of element \(X\). If an electron in the hydrogen atom falls from shell \(V\) to the inner shell corresponding to the excited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of \(570 . \mathrm{m} / \mathrm{s}\) c. The number of unpaired electrons for element \(\mathrm{X}\) in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations \(n=2\), \(m_{\ell}=-1\), and \(m_{s}=-\frac{1}{2}\) d. Let \(A\) equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of \(A\) also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element \(\mathrm{X}\).

Short answer

After analyzing all the given information, we can deduce that element X has unpaired electrons in the second shell's \(p\) subshell and one unpaired electron. Therefore, element X is most likely Boron.

Step-by-step solution

Solving for part a#

We know that the wavelength of the radio waves sent by the FM station broadcasting at \(97.1\, \mathrm{MHz}\) is \(30.0 \times 10^7\) times greater than the wavelength of the energy difference of the excited state of the hydrogen atom and the ground state. The frequency \(f\) is given as: \[f = 97.1 \times 10^6\, \mathrm{Hz}\] To get the wavelength \(\lambda_1\), we can use the speed of light, \(c\): \[c = 3.00 \times 10^8\, \mathrm{m/s}\] and the formula, \[c = \lambda_1 \times f\] Solving for wavelength \(\lambda_1\): \[\lambda_1 = \frac{c}{f}\] Now, \[\lambda_2 = 30.0 \times 10^7 \times \lambda_1\] Where \(\lambda_2\) corresponds to the energy difference between the excited state of the hydrogen atom and the ground state.

Solving for part b#

We are given that if an electron falls from shell V to the inner shell of excited state, the wavelength of emitted light is the same as the wavelength of an electron moving at a speed of \(570\, \mathrm{m/s}\), which is \(\lambda_2\). Using the de Broglie wavelength formula, \[\lambda_2 = \frac{h}{mv}\] Where \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(v\) is its speed. Now we have: \[V = \frac{n_1^2 + n_2^2}{n_1^2 - n_2^2}\] Where \(n_1\) and \(n_2\) are principal quantum numbers, and \(V\) represents the principal quantum number for the valence shell of element X.

Solving for part c#

Now we need to find the number of unpaired electrons for element X, which is the same as the maximum number of electrons in an atom that can have the quantum number \(n=2, m_\ell=-1\), and \(m_s=-\frac{1}{2}\). Since there is only one electron that can have these designations, the element X must have one unpaired electron.

Solving for part d#

Lastly, we need to find the charge A of the stable ion that would form when the unknown element 120 forms ionic compounds. This charge A also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element X. Angular momentum quantum number \(l\) is given by \(A=l\): Since \(A=1\), the subshell containing the unpaired electron(s) would have the quantum number representation of: \[n=2, l=1, m_\ell=-1\] #Conclusion#: After analyzing all the given information, we can deduce that element X has unpaired electrons in the second shell's \(p\) subshell and one unpaired electron. Therefore, element X is most likely Boron.

Key concepts

Hydrogen Atom

The hydrogen atom is the simplest atom that consists of only one proton and one electron. Its simplicity makes it an important subject of study in quantum mechanics. The hydrogen atom's energy levels can be described by the Rydberg formula, which calculates the wavelengths of light emitted or absorbed when an electron transitions between energy levels.

In the given exercise, an energy difference exists between an excited state and the ground state of the hydrogen atom. This transition corresponds to the release of a photon, which can be calculated using quantum mechanics principles. Understanding the hydrogen atom's transitions is crucial for solving problems related to its spectral lines and provides the basis for the Bohr model of the atom, which depicts electrons in defined circular orbits around the nucleus.

Quantum Numbers

Quantum numbers are used to describe the properties of electrons in an atom. They are essential for understanding electron configurations and the distribution of electrons around the nucleus. Each electron in an atom is described by a set of four quantum numbers:

• Principal quantum number \(n\): Indicates the energy level or shell of the electron.
• Azimuthal quantum number \(l\): Describes the shape of the electron's orbital.
• Magnetic quantum number \(m_\ell\): Defines the orientation of the orbital in space.
• Spin quantum number \(m_s\): Indicates the direction of the electron's spin.

In the exercise, quantum numbers help identify the possible states an electron may occupy. For example, the principal quantum number is significant in determining the valence shell of element X, while other quantum numbers help describe unpaired electrons and their specific energy states.

De Broglie Wavelength

The de Broglie wavelength is a concept used to describe the wave-like properties of particles. According to Louis de Broglie, every moving particle or object has an associated wavelength, given by the formula:\[ \lambda = \frac{h}{mv} \]where \( \lambda \) is the de Broglie wavelength, \( h \) is the Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

In the context of the exercise, the de Broglie wavelength concept is applied to relate the wavelength of light emitted during the electron transition in the hydrogen atom to the wavelength of an electron moving at a known speed. This connection is vital in quantum mechanics as it links macroscopic physical properties (like speed) to microscopic phenomena (like wavelength).

Angular Momentum Quantum Number

The angular momentum quantum number \( l \) determines the shape of the electron's orbital and is integral to the atom's quantum chemistry. It defines the subshells within a given principal energy level characterized by the principal quantum number \( n \), where \( l \) takes values from 0 to \( n-1 \).

The angular momentum quantum number directly relates to the orbital angular momentum of an electron in its shell, with its name derived from this association. Within the context of the exercise, it helps identify the subshell containing the unpaired electron(s) for element X. The problem indicates that \( l = 1 \), which corresponds to the \( p \) subshell, aligning with the behavior and structure of the identified element, Boron. Understanding these quantum numbers is crucial as they predict chemical bonding and the interaction of atoms with energy.