When an atom X absorbs radiation with a photon energy than the ionization energy of the atom, the atom is ionized to generate an ion X^(+) and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is, Photon energy (h)=ionization energy (IE) of X+ kinetic energy of photoelectron. When a molecule, for example, H_(2), absorbs short-wavelength light, the photoelectron is ejected and an H_(2), ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure-2 shows a typical photoelectron spectrum when H_(2) in the lowest vibrational level is irradiated by monochromatic light of 21.2 eV. No photoelectrons are detected above 6.0 eV. (eV is a unit of energy and 1.0 eV is equal to 1.6xx10^(-19) J) Figure 1 Schematic diagram of photoelectron spectroscopy. Figure 2 Photoelectron spectrum of H_(2). The energy of the incident light is 21.2 eV {:a) ul("Determine") the energy difference E_(A1) (eV) between H_(2)(v=0) and H^(+)(V_(ion)=0) to the first decimal place. v and v_(ion) denote the vibrational quantum numbers of H_(2) and H^(+), respectively. {:b) ul("Determine") the energy difference E_(A2) (eV) between H^(+) (v_(ion)=0) and H^(+)(v_(ion)=3) to the first decimal place. The electronic energy levels E_(n)^(H) of a hydrogen atom are given by the equation E_(n)^(H)=-(Ry)/n^(2) "" (n=1, 2, 3 ....) Here n is a principal quantum number, and Ry is a constant with dimensions of energy. The energy from n=1 to n=2 of the hydrogen atom is 10.2 eV.