Question

You have a cold gas of gold atoms, and you observe that if you shine light consisting of photons with energy 10.2 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. You find that the emitted electrons have a kinetic energy of 1.0 eV. What is the ionization energy of the gold atom? What is its ground state energy K + U?

you shine light with continuous energy distribution, and you observe absorption lines at the following photon energies: 1.1 eV, 2.7 eV, 4.6 eV, 5.1eV, 6.7eV, and 7.2eV.

Q/ Using the information from the two experiments describes above, draw a diagram of the energy levelsof one of the atoms in this gas. Draw the diagram expermintly to scale. Label the energy levels with their Values and label the transitions observed from the absorption line Data.

Answer 1

The ionization energy measure the minimum energy required to remove an electron from its ground state:

= 10.2 - 1.0 = 9.2 eV

(the electron has absorbed this energy to become free, the excess energy is the kinetic energy of the electron).

The energy of the ground state has the same value, but opposite sign: - 9.2 eV.

The electron from 6s1 is subject of the first ionization.

For the second part: I can't upload a picture.

Draw it your self.

Draw a vertical line (call it Energy) and put a zero mark at its high end. The values 1.1, 2.7, etc are ionization energies. You have to mark energy level in atom, so take the same values with negative sign and put them on your line (to scale). Draw a horizonthal line for each energy level. Label the energy levels with their values (-1.1 eV, -2.7 eV, etc). Draw transitions (vertical lines) from these energy levels to zero level. Label each transition (e.g., I +1.21 eV, II + 2.7 eV, etc.)