Question

The first ionization potential of ground-state He is \(24.6 \mathrm{eV} .\) The wavelength of light associated with the \(1 s 2 p^{1} \mathrm{P}\) term is \(58.44 \mathrm{nm} .\) What is the ionization energy of the He atom in this excited state?

Short answer

The ionization energy of the He atom in the excited state is approximately 20.469 eV.

Step-by-step solution

Calculate the energy associated with the given wavelength

We use the formula: Energy = \(h \times c / \lambda \) Where h is Planck's constant (\(6.626 \times 10^{-34} kg\cdot m^2/s\)), c is the speed of light (\(3.0 \times 10^8 m/s\)), and λ is the wavelength given in the problem (58.44 nm or \(58.44 \times 10^{-9}m\)). Energy = \(\frac{(6.626 \times 10^{-34}) \times (3.0 \times 10^8)}{58.44 \times 10^{-9}}\) Now, convert this energy to eV by using the fact that 1 Joule = \(6.242\times10^{18}\) eV.

Convert the energy to eV

Energy (eV) = \(\frac{Energy (J)}{1.6 \times 10^{-19}}\) Energy (eV) = \(\frac{(6.626 \times 10^{-34}) \times (3.0 \times 10^8)}{(58.44 \times 10^{-9})(1.6 \times 10^{-19})}\) Energy (eV) = 4.131 eV (rounded to three decimal places)

Determine the ionization energy of the excited state

Now that we have the energy associated with the given wavelength, we will subtract this energy from the ionization energy of the ground state: Ionization energy of the excited state = Ionization energy of the ground state - Energy Ionization energy of the excited state = 24.6 eV - 4.131 eV Ionization energy of the excited state = 20.469 eV (rounded to three decimal places) So, the ionization energy of the He atom in this excited state is approximately 20.469 eV.

Key concepts

Wavelength

The wavelength of light is a fundamental concept in physics and chemistry, especially when dealing with atomic and molecular phenomena. Wavelength, usually denoted by the Greek letter \( \lambda \), is the spatial period of a wave—the distance over which the wave's shape repeats. It determines the color of visible light and is related to the energy of photons.

When light interacts with atoms, as in the ionization energy problem, the wavelength of the light can be used to calculate the energy absorbed or emitted when electrons transit between energy levels. The shorter the wavelength, the higher the energy of the light. For example, ultraviolet light has shorter wavelengths and, therefore, more energy than visible light.

In the He atom exercise provided, the wavelength of light associated with the transition from the \(1s\) to the \(2p^{1}P\) state is \(58.44 \mathrm{nm}\), which is within the ultraviolet range. This detail is crucial for determining the energy involved in this specific transition and, consequently, the ionization energy in the excited state.

Planck's Constant

Planck's constant is a fundamental quantity in quantum mechanics, playing a pivotal role in the relationship between energy and the frequency of radiation. Denoted by \(h\), Planck's constant has a value of approximately \(6.626 \times 10^{-34} \mathrm{J\cdot s}\) and is a part of the equation that links the energy \(E\) of a photon to its frequency \(u\):

\[E = h \times u\]
Because frequency and wavelength are inversely related, Planck's constant also connects energy with wavelength through the formula:

\[E = \frac{h \times c}{\lambda}\]
This equation is central to our understanding of how light behaves as both a wave and a particle, known as wave-particle duality. In the context of the exercise, Planck's constant allows us to calculate the exact amount of energy a photon must have to correspond to the specific wavelength of \(58.44 \mathrm{nm}\), which is used to determine the excited state's ionization energy of a He atom.

Energy Conversion

In physics problems, energy conversion is the process of changing one form of energy into another. In the context of the ionization energy exercise, the relevant conversion is from the energy of a photon calculated in joules (using Planck's constant and the speed of light) to electron volts (eV), a unit commonly used in atomic and nuclear physics to express the energy of electrons and photons.

The conversion factor between joules and electron volts is \(1 \mathrm{eV} = 1.602 \times 10^{-19} \mathrm{J}\). Multiplying or dividing by this conversion factor allows for the expression of energy in the preferred unit. For instance, when we calculate the photon's energy with the wavelength of \(58.44 \mathrm{nm}\), the result is initially in joules, which is then converted to electron volts for easier comparison with typical values for atomic ionization energies.

The ability to convert energy units is an essential skill for students as it enables them to work with the values that are most appropriate for the phenomena being studied, such as the ionization energy of atoms in eV as opposed to joules, which are more suitable for larger-scale energy calculations.