1.

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 ul("Calculate") the speed u (m s^(-1)) of the hydrogen atoms generated in the above reaction. H_(2) is assumed to be at rest. If you were unable to determine the value for E_(C), then use 5.0 eV for E_(C)

Answer»


ANSWER :The excess ENERGY is `16.7 eV (=21.2 eV-4.5 eV)`. Because two hydrogen atoms are GENERATED upon dissociation, half of this excess is released as translation energy of the hydrogen atoms.
`1/2 m u^(2)=8.35 eV=1.34xx10^(-18) J`


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