Answer to Question #141977 in Quantum Mechanics for Desmond

Question

The energy levels of an atom are given as follows: —2.8 eV, —6.5 eV, —8.6 eV, —13.6 eV.
1. Why are the values negative?
2. What is the ionization energy of the atom?
3 find the wavelength of the photon emitted when the atom falls from —6.5 eV to the -13.6 eV

Answer 1

1. The "-" sign indicates that the energyelectron inside the atom is negative, i.e.the electron is in a bound state.
2. The ionization energy of the atom is the lowest value of the energy (-13.6 eV).
3. $h\nu=h\frac{c}{\lambda}=\Delta E$

$\lambda=\frac{hc}{\Delta E}=\frac{6.63\cdot10^{-34}\cdot3\cdot10^8}{(-6.5-(-13.6))\cdot1.6\cdot10^{-19}}=175\cdot10^{-9}$ m.