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`
Considering an energy cycle, `ul("determine")` the bond energy `E_(D)( eV)` of `H_(2)^(+)` to the first decimal place. If you were unable to determine the values for `E_(B)` and `E_(C)`, then use `15.0 eV` and `5.0 eV` for `E_(B)` and `E_(C)`, respectively.

Answer

(a) The spectral peak at `5.8 eV` in Fig `2` corresponds to the electron with the highest kinetic energy, which is generated by the reaction
`H_(2)(V=0) rarr H_(2)^(+)(V_(ion)=0)+e`
Accordingly,
`DeltaE_(A1)=21.2 eV-5.8 eV=15.4 eV`
(b) One can estimate from Fig. `2` that the energy difference `DeltaE_(A2)` between `H_(2)^(+)(V_(ion)=0)` and `H_(2)^(+)(V_(ion)=3)` is approximately `0.8 eV`.
The answer are as follows: `DeltaE_(A1)=15.4 eV`
`DeltaE_(A2)=0.8 eV`