Answer to Question #115301 in Atomic and Nuclear Physics for Vidurjah Perananthan

Question

A photon falls against a hydrogen atom that is in the first excited state. The photon is absorbed, the hydrogen atom is ionized and the emitted electron receives an energy of 5.6 eV.
a) What energy must the incident photon have?
b) What energy would the emitted electron have received if the atom had instead been in its basic state?

Answer 1

a) The energy of a hydrogen atom that is in the first excited state is -3.4 eV. The atom is ionized and the electron received 5.6 eV, which means that the atom made transition from -3.4 eV to 0 eV. Therefore, the energy of the photon was

$$E_p=3.4+5.6=9\text{ eV}.$$

b) In the basic state, the hydrogen atom has -13.6 eV of energy. Therefore, the electron would receive

$$E_e=13.6-9=4.6\text{ eV}.$$