CHAPTER 16 DRILL
- B Combining the equation E = hf with f = c/ λ gives us
- D The energy of the incident photons is
E = hf = (4.14 × 10−15 eV · s)(7.2 × 10^15 Hz) = 30 eV
Since E > φ, photoelectrons will be produced, with maximum
kinetic energy
Kmax = E − φ = 30 eV − 6 eV = 24 eV
- A If the atom’s ionization energy is 25 eV, then the electron’s ground-state
energy must be −25 eV. Making a transition from the −16 eV energy level to
the ground state will cause the emission of a photon of energy.
∆E = (−16 eV) − (−25eV) = 9 eV
- E The gap between the ground-state and the first excited state is
−10 eV − (−40 eV) = 30 eV
Therefore, the electron must absorb the energy of a 30 eV photon (at
least) to move even to the first excited state. Since the incident
photons have only 15 eV of energy, the electron will be unaffected.
- C The de Broglie wavelength of a particle with momentum p, is λ = h/p. For
this proton, we find that
- A In β− decay, a neutron is transformed into a proton and an electron.
