Question

If electrons of energy 12.8 ev are incident on a gas of hydrogen atoms in their ground state, what are the energies of the photons that are emitted by the excited gas?

Answer 1

When 12.8 eV energy electrons excite hydrogen atoms, the atoms are elevated to an energy level just below ionization. To determine the specific energy state, the energy formula for hydrogen atoms is used, then the energies of emitted photons are found by calculating the energy difference between the excited and final states.

Step-by-step explanation:

If electrons with an energy of 12.8 eV are incident on hydrogen atoms in their ground state, we first need to determine the energy level to which the hydrogen atoms are excited. The ground state energy of hydrogen is -13.6 eV. If 12.8 eV is provided, the electron reaches an energy level just below 0 eV (ionization energy), since 12.8 eV is not sufficient to completely ionize the atom (13.6 eV needed).

To find out which energy level the electron will jump to, we use the formula for the total energy of an electron in a hydrogen atom, En = (- 13.6 eV/n²), and solve for n.

Once the electron is excited, it will eventually drop back to a lower energy state, emitting a photon in the process.

The energy of the emitted photon can be calculated using the difference in energy levels, ΔE = E1 - Ef.

Answer 2

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.