Question

A hydrogen atom is in its fifth excited state, with principal quantum number \(n=6 .\) The atom emits a photon with a wavelength of \(410 \mathrm{nm}\). Determine the maximum possible orbital angular momentum of the electron after emission.

Short answer

Answer: The maximum possible orbital angular momentum of the electron after the photon emission is approximately \(1.49 \times 10^{-34} \mathrm{Js}\).

Step-by-step solution

Calculate the energy of the emitted photon.

The energy of a photon can be calculated using its wavelength and the Planck's constant, with the formula \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), \(c\) is the speed of light (\(2.998 \times 10^{8} \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the photon (\(410 \mathrm{nm}\)). Converting the wavelength to meters, we get \(\lambda = 410 \times 10^{-9} \mathrm{m}\). Now, we can calculate the energy of the photon: \(E = \frac{(6.626 \times 10^{-34} \mathrm{Js})(2.998 \times 10^{8} \mathrm{m/s})}{410 \times 10^{-9} \mathrm{m}} \approx 4.839 \times 10^{-19} \mathrm{J}\).

Calculate the initial energy level of the hydrogen atom.

The energy of a hydrogen atom with a principal quantum number \(n\) is given by the formula \(E_n = -\frac{13.6 \mathrm{eV}}{n^2}\). Since the atom is initially in its fifth excited state (with \(n = 6\)), its initial energy can be calculated as: \(E_6 = -\frac{13.6 \mathrm{eV}}{6^2} \approx -0.377 \mathrm{eV}\). Converting the energy to joules, we get \(E_6 \approx -6.044 \times 10^{-20} \mathrm{J}\).

Calculate the final energy level of the hydrogen atom.

Since the energy of a photon is equal to the difference in energy levels between the initial and final states, we can calculate the final energy level (\(E_f\)) after photon emission as follows: \(E_f = E_6 - E_{photon} \approx -6.044 \times 10^{-20} \mathrm{J} - 4.839 \times 10^{-19} \mathrm{J} \approx -5.443 \times 10^{-19} \mathrm{J}\). Converting the energy back to electron volts, we get \(E_f \approx -3.39 \mathrm{eV}\).

Determine the maximum value of the azimuthal quantum number (l).

To find the final principal quantum number \(n_f\), we can compare the obtained energy to the energy levels of the hydrogen atom: \(E_{n_f} = -\frac{13.6 \mathrm{eV}}{n_f^2}\). Setting \(E_{n_f} = -3.39 \mathrm{eV}\), we can solve for \(n_f\): \(n_f \approx \sqrt{\frac{13.6 \mathrm{eV}}{3.39 \mathrm{eV}}} \approx 2\). The maximum possible value of the azimuthal quantum number (\(l\)) is \(n_f - 1\) for a given energy level. Therefore, the maximum value of \(l\) for the final state is: \(l_{max} = n_f - 1 = 2 - 1 = 1\).

Calculate the maximum possible orbital angular momentum.

The orbital angular momentum (\(L\)) can be calculated using the formula \(L = \sqrt{l(l+1)}\hbar\), where \(\hbar\) is the reduced Planck's constant (\(1.054 \times 10^{-34} \mathrm{Js}\)). Using the maximum value of \(l\), we can calculate the maximum possible orbital angular momentum: \(L = \sqrt{1(1+1)}(1.054 \times 10^{-34} \mathrm{Js}) \approx 1.49 \times 10^{-34} \mathrm{Js}\). The maximum possible orbital angular momentum of the electron after the photon emission is approximately \(1.49 \times 10^{-34} \mathrm{Js}\).

Key concepts

Hydrogen Atom

The hydrogen atom is the simplest and most abundant atom in the universe. It consists of a single proton in its nucleus and one electron orbiting this nucleus. The simplicity of the hydrogen atom makes it an excellent model for studying atomic physics and quantum mechanics.
The behavior of the electron in a hydrogen atom can be described using quantum numbers, which specify certain characteristics of the electron's orbit. The principal quantum number, denoted by \(n\), indicates the energy level of the electron.

• A higher \(n\) means the electron has a higher energy and is further from the nucleus.
• In our example, the hydrogen atom is in its fifth excited state where \(n=6\), indicating it has absorbed energy to reach this high level.

Understanding these energy states and their transitions forms the basis for explaining phenomena such as photon emissions.

Photon Emission

Photon emission occurs when an electron in an excited state of an atom transitions to a lower energy state, releasing energy in the form of a photon. This process is fundamental in understanding how atoms interact with light and other electromagnetic radiation.
In the context of the hydrogen atom, when the electron drops from a higher energy level to a lower one, it releases energy that corresponds to a photon of a specific wavelength or frequency.

• The wavelength of the emitted photon is inversely proportional to its energy; a shorter wavelength denotes higher energy.
• In our exercise, the emission of a photon with a wavelength of 410 nm indicates a transition to a lower energy state.

By analyzing the wavelength of the emission, we can deduce the energy difference between the initial and final states of the electron in the atom.

Orbital Angular Momentum

Orbital angular momentum is a critical concept in quantum mechanics, especially when discussing the motion of electrons around an atomic nucleus. It provides insight into the structure and behavior of atoms.
Quantum mechanics describes the angular momentum of electrons in terms of quantum numbers with the azimuthal quantum number \(l\) being of particular importance here. The allowed values of \(l\) range from 0 to \(n-1\), where \(n\) is the principal quantum number.

• The orbital angular momentum \(L\) is given by the formula \(L = \sqrt{l(l+1)}\hbar\), where \(\hbar\) is the reduced Planck's constant.
• In this exercise, after the photon emission, the maximum \(l\) is 1, resulting in a maximum orbital angular momentum of approximately 1.49 x 10^{-34} Js.

Knowing \(L\) helps us understand the dynamics and transitions of the electron in an atom during processes like photon emission.

Quantum Numbers

Quantum numbers are fundamental in quantum mechanics, specifying the properties of electrons in an atom. They play a crucial role in defining the atomic orbitals and the arrangement of electrons.
Each electron in an atom is described by a set of four quantum numbers:

• **Principal Quantum Number (n):** Indicates the energy level and size of the orbital, with values starting from 1.
• **Azimuthal Quantum Number (l):** Defines the shape of the orbital, with values ranging from 0 to \(n-1\).
• **Magnetic Quantum Number (m_l):** Describes the orientation of the orbital in space, with values from \(-l\) to \(+l\).
• **Spin Quantum Number (m_s):** Specifies the electron's spin direction, either +1/2 or -1/2.

In our exercise scenario, the principal quantum number changes from 6 to 2 after photon emission, and this determines other quantum numbers, ultimately affecting the electron's physics and behavior within the atom.