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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. Follow the convention that the ground state energy (the lowest energy state) is the negative of the ionization energy and base your other answers around this. O eV -1 eV ✓-2 eV -3 eV -4 eV ✓-5 eV ✓-6 eV -7 eV -8 eV ✓-9 eV -10 eV -11 eV | -12 eV Now you turn off the light and run a beam of electrons through the gas. The kinetic energy of the electrons is 3.5 eV. Collisions of the electrons in the beam with the atoms in the gas exite the atoms to states above the ground state. Select all the possible energies of the emitted photons from this gas. 1 eV 2 eV 3 eV 4 eV 5 eV 6 eV 7 eV 8 eV 9 eV 10 eV

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Answer:

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Step-by-step explanation:

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User Qi Tang
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The energy level diagram depicts an atom with a ground state at -5 eV, three excited states at -4 eV, -2 eV, and -1 eV, and an ionization energy of 7 eV, illustrating photon absorption and ionization processes.

The image shows an energy level diagram for an atom. The lowest energy level is the ground state, which is labeled as -5 eV. The other energy levels are labeled with their energies above the ground state. The absorption lines at 1 eV, 3 eV, 4 eV, and 7 eV correspond to the energy differences between the ground state and the excited states.

The hint states that the absolute values of the energy levels are arbitrary, but the difference in energies between levels is not. This means that we can shift the entire energy level diagram up or down by a constant amount, but the relative spacing of the levels will remain the same.

Based on the information in the image and the hint, we can identify the following energy levels for the atom:

Ground state: -5 eV

Excited state 1: -4 eV (1 eV above ground state)

Excited state 2: -2 eV (3 eV above ground state)

Excited state 3: -1 eV (4 eV above ground state)

Ionization energy: 0 eV (7 eV above ground state)

The ionization energy is the energy required to remove an electron from the atom completely. In this case, the ionization energy is 0 eV, which means that the electron is very loosely bound to the atom.

The energy level diagram shows that the atom has three excited states below the ionization energy. These excited states are relatively close to the ground state, which is why the atom can easily absorb photons with energies of 1 eV, 3 eV, and 4 eV.

The diagram also shows that the ionization energy is 7 eV above the ground state. This means that it takes a photon with an energy of 7 eV or more to ionize the atom.

Complete question:

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. Follow the convention that the ground state energy (the lowest energy state) is the negative of the ionization energy and base your other answers around this.

Now you shine light with a continuous energy distribution and you observe absorption-example-1
User Eaman
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