The orbital quantum number for the electron in a hydrogen atom is ℓ = 3. What is the smallest possible value (algebraically) for the total energy of this electron? Give your answer in electron volts.

The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 2 as shown in the figure below. Consider the photon of longest wavelength corresponding to a transition shown in the figure. E (eV) -0.378 -0.544 2 -0.850 4 T -1.512 Balmer series ENERGY (a) Determine its energy. 1.889 eV (b) Determine its wavelength. 657.7 nm 0.00 (d) Find its wavelength. nm -3.401 Consider the spectral line of shortest wavelength corresponding to a transition shown in the figure. (c) Find its photon energy. ev (e) What is the shortest possible wavelength in the Balmer series? nm

Chapter 39, Problem 044 A hydrogen atom in a state having a binding energy (the energy required to remove an electron) of -1.51 eV makes a transition to a state with an excitation energy (the difference between the energy of the state and that of the ground state) of 10.200 eV. (a) What is the energy of the photon emitted as a result of the transition? What are the (b) higher quantum number and (c) lower quantum number of the transition producing this emission? Use -13.60 eV as the binding energy of an electron in the ground state. (a) Number Units (b) Number Units (c) Number Units

Suppose you recently discovered a hydrogen like element that has only one electron orbiting around a nucleus containing a proton and a neutron. You found the ground state energy of the electron to be -16 eV. What will be the energy of this electron when it is on the excited state shown in the sketch? Note that all other possible intermediate states are shown by dashed lines. Electron is here Ground state 1.0 eV 16 eV - 1.0 eV -4.0 eV 4.0 eV