1.

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»


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`


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