Question

You have a cold gas of atoms, and you observe that if you shine light consisting of photons with energy 10 eV through the gas, some free electrons are observed, implying that a photon of this energy is able to ionize an atom in the gas.

If you find that the emitted electrons from the gas have a kinetic energy of 1 eV, what is the ionization energy of the cold atom? Your answer should be in eV_____

Now you shine light with a continuous energy distribution and you observe absorption lines at the following photon energies: 1eV, 3eV, 4eV, and 7eV. Using the information from the two experiments described, select the energies that correspond to all the energy levels of this atom. Hint: Similar to potential energy, the absolute values of the energy levels are arbitrary to a constant, but the difference in energies between levels is not.

Answer 1

The ionization energy of the atom when ionized with a 10 eV photon and leaving a 1 eV kinetic energy electron is 9 eV. Absorption lines at 1 eV, 3 eV, 4 eV, and 7 eV indicate transitions between energy levels. The energy levels of the cold atom are 9 eV (ionization level), 8 eV, 6 eV, 5 eV, and 2 eV.

Step-by-step explanation:

The question involves finding the ionization energy of an atom, given that a photon with 10 eV energy ionizes the atom, resulting in a free electron with 1 eV of kinetic energy. Since the ionization process uses up energy to remove an electron from the influence of the nucleus leaving some as kinetic energy to the electron, the initial photon energy is the sum of the ionization energy and the kinetic energy of the ejected electron.

Therefore, the ionization energy of the atom is the energy of the photon minus the kinetic energy of the emitted electron, which is 10 eV - 1 eV = 9 eV.

When light with a continuous energy distribution is shone through the gas, we observe absorption lines at 1 eV, 3 eV, 4 eV, and 7 eV. These represent the energy levels of the atoms as photons with these specific energies are absorbed when electrons transition between these energy levels and the already known ionization level.

Since the absolute values are arbitrary apart from the highest energy level (ionization energy), we can establish the energy levels by subtracting the given photon energies from the ionization energy. Therefore, we have energy levels corresponding to 9 eV (ionization energy), 8 eV (9 eV - 1 eV), 6 eV (9 eV - 3 eV), 5 eV (9 eV - 4 eV), and 2 eV (9 eV - 7 eV).