Question

Consider an electron for a hydrogen atom in an excited state. The maximum wavelength of electromagnetic radiation that can completely remove (ionize) the electron from the H atom is \(1460 \mathrm{nm}\). What is the initial excited state for the electron \((n=?) ?\)

Short answer

The initial excited state for the electron in the hydrogen atom is \(n=2\).

Step-by-step solution

Understand the Rydberg formula

The Rydberg formula relates the wavelength of the emitted or absorbed electromagnetic radiation in the hydrogen atom to the quantum numbers of the initial and final states. It is given by: \[\frac{1}{\lambda} = \mathrm{R}( \frac{1}{n_1^2} - \frac{1}{n_2^2} )\] where \(\lambda\) is the wavelength, \(n_1\) and \(n_2\) are the quantum numbers (principal quantum numbers) of the initial and final states respectively, and R is the Rydberg constant (\(1.097 \times 10^7 m^{-1}\)). In this problem, the electron is ionized, so \(n_1 = \infty\). The maximum wavelength corresponds to the case when the final state is \(n_2 = 1\).

Calculate the energy to ionize the electron

The maximum wavelength corresponds to the minimum energy required to ionize the electron. We can convert the wavelength to energy using the following equation: \[E = \frac{hc}{\lambda}\] where \(E\) is the minimum energy required, \(h\) is the Planck's constant (\(6.63 \times10^{-34} Js\)), \(c\) is the speed of light (\(3.0 \times10^{8} m/s\)), and \(\lambda\) is the maximum wavelength (\(1460 \times10^{-9} m\)). Calculating the energy: \[E = \frac{(6.63 \times 10^{-34} Js)(3.0 \times 10^8 m/s)}{1460 \times 10^{-9} m}\] \[E \approx 1.361 \times 10^{-19} J\]

Convert energy to electron volts (eV)

It is more convenient to work with electron volts (eV) instead of Joules (J) when dealing with atomic energy levels. To convert Joules to electron volts, divide the energy in Joules by the elementary charge (\(1.602 \times 10^{-19} C\)): \[E_{eV} = \frac{1.361 \times 10^{-19} J}{1.602 \times 10^{-19} C}\] \[E_{eV} \approx 0.851 eV\]

Find the initial excited state using the energy formula

For the hydrogen atom, the energy levels for different quantum states are given by: \[E_n = -\frac{13.6}{n^2} eV\] where \(E_n\) is the energy of the atom at the quantum state of principal quantum number \(n\). To find the initial excited state, we can equate the energy difference between the excited state and the ground state (\(n=1\)) to the energy we found in step 3: \[0.851 eV = 13.6 eV - \frac{13.6 eV}{n^2}\]

Solve for the quantum number n

Now, we can solve the above equation for the quantum number \(n\): \[0.851 eV = 12.749 eV - \frac{13.6 eV}{n^2}\] \[\frac{13.6 eV}{n^2} = 11.898 eV\] \[n^2 = \frac{13.6 eV}{11.898 eV}\] \[n^2 \approx 1.143\] Taking the square root of both sides, we obtain: \[n \approx 1.07\] Since n must be an integer and it cannot be 1 (because the question asks for an excited state), we conclude that the initial excited state for the electron is n=2.

Key concepts

Rydberg formula

The Rydberg formula is a fundamental equation in quantum mechanics that describes the wavelengths of light emitted or absorbed by electrons in a hydrogen atom. This formula is crucial for understanding how electrons transition between different energy levels. It is expressed mathematically as: \[ \frac{1}{\lambda} = \mathrm{R} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Here:

• \(\lambda\) is the wavelength of the light.
• \(\mathrm{R}\) is the Rydberg constant, approximately \(1.097 \times 10^7 \; \mathrm{m}^{-1}\).
• \(n_1\) and \(n_2\) are the principal quantum numbers of the electron's initial and final states.

In this equation, when an electron moves to a lower energy level, it emits a photon with a wavelength described by the Rydberg formula. In contrast, absorbing a photon causes the electron to move to a higher energy level. Understanding this formula helps unravel the complex interactions happening within the simple hydrogen atom.

hydrogen atom

The hydrogen atom is the simplest atom and serves as a fundamental model in quantum mechanics. It consists of one proton and one electron. Despite its simplicity, the hydrogen atom plays a crucial role in the development of quantum theory. Each possible energy level of an electron within a hydrogen atom is quantized, which means electrons can only inhabit specific energy levels.Each level is associated with a "principal quantum number," denoted by \(n\). The electron's behavior in these energy levels gives rise to the atom's emission spectra, which are unique patterns of light that can be measured. These spectra consist of discrete lines, each corresponding to an electron transition between two energy levels. This behavior can be predicted using the Rydberg formula, tying together quantum mechanics with observable properties of atoms.

principal quantum number

The principal quantum number, represented by \(n\), is a key integer in quantum mechanics that characterizes the overall size and energy of an electron's orbit in an atom. In the context of a hydrogen atom, \(n\) determines the energy level at which an electron resides. The principal quantum number is always a positive integer (\(n = 1, 2, 3, \ldots\)).

• When \(n = 1\), the electron is in the ground state, the lowest energy level.
• As \(n\) increases, the electron occupies higher energy levels.
• The difference in energy between two levels \((n_1 > n_2)\) can be used to calculate the wavelength of the photon absorbed or emitted using the Rydberg formula.

Understanding \(n\) is essential for determining electron configurations, ionization energies, and spectral lines for hydrogen and similar one-electron ions. The principal quantum number effectively outlines the electron's "address" within the atom.

electron volts

Electron volts (eV) are a unit of energy commonly used in atomic and particle physics. One electron volt is the amount of kinetic energy gained by an electron when it accelerates through an electric potential difference of one volt. To convert energy from Joules to electron volts, the relationship is given by: \[ 1 \; \mathrm{eV} = 1.602 \times 10^{-19} \; \mathrm{J} \]In the study of atomic structure and interactions, using eV is more convenient than Joules because:

• eV simplifies calculations involving atomic and subatomic particles, as these energies typically fall in the eV range.
• It provides a more intuitive sense of scale when discussing energies at a microscopic level.
• Quantum energies like ionization energy, or the energy required to move an electron from one energy level to another, are typically measured in eV.

By expressing energy in electron volts, physicists can easily compare and calculate energies involved in chemical reactions, particle collisions, and more.