At a temperature of `0 K`, the total energy of a gaseous diatomic molecule AB is approximately given by:
`E=E_(o)+E_(vib)`
where `E_(o)` is the electronic energy of the ground state, and `E_(vib)` is the vibrational energy.
Allowed values of the vibrational energies are given by the expression:
`E_(vid)=(v-1/2)epsilon " " v=0, 1, 2, ....... " " epsilon=h/(2pi)sqrt(k/mu) " " mu(AB)=(m_(A)m_(B))/(m_(A)+m_(B))`
where h is the planck's constant, is the vibration quantum number, k is the force constant, and is the reduced mass of the molecule. At `0K`, it may be safely assumed that is zero, and `E_(o)` and `k` are independent of isotopic substitution in the molecule.
Deuterium, D, is an isotope of hydrogen atom with mass number `2`. For the `H_(2)` molecule, k is `575.11 N m^(-1)`, and the isotopic molar masses of H and D are `1.0078` and `2.0141 g mol^(-1)`, respectively.
At a temperature of `0K : epsilon_(H_(2))=1.1546 epsilon_(HD)` and `epsilon_(D_(2))=0.8167 epsilon_(HD)`
`ul("Calculate")` the electron affinity, EA, of `H_(2)^(+)` ion in eV if its dissociation energy is `2.650 eV`. If you have been unable to calculate the value for the dissociation energy of `H_(2)` then use `4.500 eV` for the calculate.

Answer

`IP(H)=DeltaE_(n rarr oo)=-13.5984/oo^(2)-13.5984/1^(2)=13.598 eV` (ionozation potential)
{:(H_(2)^(+)+e rarrH_(2),,,EA(H_(2)^(+))=-IP(H_(2))),(H_(2)^(+)rarrH^(+)+H,,,DE(H_(2)^(+))=2.650 eV),(H rarr H+ + e,,,IP(H)=13.598 eV),(H_(2) rarr H +H,,,DE(H_(2))=4.478 eV):}`
`EA(H_(2)^(+))=DE(H_(2)^(+))-IP(H)-DE(H_(2))=2.650-13.598-4.478=-15.426 eV`
Electron affinity `H_(2)^(+)=-15.426 eV`
`=-15.426 eV`