An electron is excited from the \(n=1\) ground state to the \(n=3\) state in a hydrogen atom. Which of the following statements are true? Correct the false statements to make them true. a. It takes more energy to ionize (completely remove) the electron from \(n=3\) than from the ground state. b. The electron is farther from the nucleus on average in the \(n=3\) state than in the \(n=1\) state. c. The wavelength of light emitted if the electron drops from \(n=3\) to \(n=2\) will be shorter than the wavelength of light emitted if the electron falls from \(n=3\) to \(n=1\). d. The wavelength of light emitted when the electron returns to the ground state from \(n=3\) will be the same as the wavelength of light absorbed to go from \(n=1\) to \(n=3\). e. For \(n=3\), the electron is in the first excited state.

In an atom of hydrogen, the simplest and most widely studied system in quantum mechanics, electrons occupy specific orbits around the nucleus, each corresponding to a particular energy level. The energy levels are defined by the formula: \begin{align*}E_n = -\frac{13.6 eV}{n^2}\tag{1}\begin{align*}where \(E_n\) is the energy of the electron in the \(nth\) energy level and \(n\) represents the principal quantum number, an integer that increases with the electron's distance from the nucleus. The factor of 13.6 eV is the ionization energy of hydrogen, meaning the energy required to completely remove the electron from the ground state. The negative sign indicates that the electrons are bound to the nucleus, and energy is required to escape from this attraction.Understanding these energy levels is fundamental to explaining how electrons transition within the atom. When electrons absorb energy, they can move to a higher energy level (excited state). Conversely, when they lose energy, they fall to a lower energy level, emitting light in the process. The energy levels get closer together as the value of \(n\) increases, which is why less energy is required to ionize the electron from a higher state as compared to the ground state.

Ionization energy refers to the minimum amount of energy needed to remove an electron entirely from an atom or ion. In the hydrogen atom, the ground state refers to the electron being in the lowest energy level, which is \(n=1\). Since the energy levels increase as the quantum number (\(n\)) increases, electrons in higher energy states are more easily ionized because they're already further from the attractive pull of the nucleus and possess a higher (less negative) energy.For a hydrogen atom, the ionization energy from the ground state is 13.6 eV, while from the \(n=3\) state, it is significantly lower. Correcting a common misunderstanding: it takes less energy to ionize an electron from a higher energy level than from a lower one. In essence, the higher the energy level an electron occupies, the less 'tightly bound' it is to the atom, and thus the easier it is to ionize.

When an electron transitions between energy levels in an atom, it either emits or absorbs photons - particles of light. This photon has a specific wavelength that is inversely proportional to the energy involved in the transition. Using the formula: \begin{align*}E = hf = \frac{hc}{\tag{2}}{\bdatatataE = hf = \frac{hc}{\tag{2}}\tag{1}\begin{align*}where \(E\) is the energy difference between the initial and final levels, \(h\) is Planck's constant, \(f\) is the frequency, and \(c\) is the speed of light, we can find the wavelength (\(\lambda\)). A larger energy difference results in a shorter wavelength of the emitted photon. Therefore, an electron dropping from a higher energy level (e.g., \(n=3\) to \(n=1\)) will release more energy and thus emit light of a shorter wavelength than if it were dropping to just the next level down (e.g., \(n=3\) to \(n=2\)).

Quantum numbers are a set of numerical values that describe the unique quantum state of an electron in an atom. Each electron in an atom is described by four quantum numbers:- Principal quantum number (\(n\)) specifies the electron's shell and its energy level.- Orbital quantum number (\(l\)) defines the shape of the electron's orbit.- Magnetic quantum number (\(m_l\)) describes the orientation of the orbit in space.- Spin quantum number (\(m_s\)) indicates the direction of the electron's spin.For the hydrogen atom, since it only has one electron, the principal quantum number (\(n\)) plays the most prominent role in determining the energy level. Corrections from the exercise illustrate misunderstanding in quantum numbers, specifically, the principal quantum number. As \(n\) increases, the electron's energy and distance from the nucleus increase. It's also important to note that \(n=1\) is the ground state, \(n=2\) is the first excited state, and so on.

The frequency of an electronic transition refers to the number of times per second that an electron in an atom oscillates between two energy levels. When an electron jumps between different energy levels, it absorbs or emits a photon with a specific frequency that corresponds to the energy difference between those levels. The relationship between the emitted light's frequency (\(f\)), its wavelength (\(\lambda\)), and the speed of light is given by:\begin{align*}c = f\lambda\tag{3}\begin{align*}Using this equation, we can deduce that the frequency of the emitted photons during an electron's transition from a higher to a lower energy state is higher for transitions involving greater energy differences. Such calculations are consistent with the energy levels of the hydrogen atom and are crucial for understanding phenomena like the emission spectrum of hydrogen, where lines in the spectrum correspond to different electronic transition frequencies.