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) A molecule H_(2) in the ground state dissociates into its atoms after absorbing a photon of wavelength 77.0 nm. ul("Determine") all possibilities for the electronic states of hydrogen atoms produced. For each case calculate the total kinetic energy, KE, in eV of the disociated hydrogen atoms.

Answer»


ANSWER :`{:(H_(2)+HV,rarr,H,+,H),(,n=,1,,1),(,,1,,2),(,,2,,1),(,,2,,2),(,,.,,.),(,,.,,.),(,,.,,.):}`
The energy of `H2` molecule in its ground STATE is `-31.675 eV`.
`lambda=77.0 nm`
`E_("photon")=(hc)/lambda=(6.6261.10^(-34)xx3.00*10^(-8))/(77.0*10^(-19))=2.58 10^(-18) J`
`E_("photon")=(6.6261*10^(-34))/(1.602*10^(-19))=16.1 eV`
`DeltaE=E_(n_(1))+E_(n_(2))-E_(n_(H_(2)))=-R_(H)/n_(1)^(2)-R_(H)/n_(2)^(2)-(31.675) lt 16.1 eV`
`n_(1)=1"" n_(2)=1`
`DeltaE=-13.5984/1^(2)-13.5984/1^(2)+31.675=4.478 eV`
`KE=16.1-4.478=11.6 eV`
`n_(1)=1, n_(2)=2` or `n_(1)=2, n_(2)=1`
`DeltaE=-ul(13.5984)-ul(13.5984)+31.675=14.677 eV`
`KE=16.1-14.677=1.4 eV`
`n_(1)=2, n_(2)=2`
`DeltaE=-13.5984/2^(2)-13.5984/2^(2)+31.675=24.880 eV gt 16.1 eV`
`K.E.=16.1-14.677=1.4 eV`
Thus, the possibilities are:
`{:(H_(2)+,hv rarr,H,+,H),(,n=,1,,1),(,,1,,2),(,,2,,1):}`


Discussion

No Comment Found