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

To calculate the electron affinity (EA) of the \( H_2^+ \) ion given the dissociation energy, we can follow these steps:

### Step 1: Understand the relationship between dissociation energy and electron affinity
The dissociation energy \( D \) of a molecule is the energy required to break it into its constituent atoms. For the \( H_2^+ \) ion, the electron affinity can be related to the dissociation energy of the neutral \( H_2 \) molecule and the dissociation energy of the ion \( H_2^+ \).

### Step 2: Use the given dissociation energy
We are given that the dissociation energy of \( H_2 \) is \( 4.500 \, \text{eV} \).

### Step 3: Calculate the electron affinity
The electron affinity \( EA \) can be calculated using the formula:
\[ EA = D(H_2) - D(H_2^+) \]
Where:
- \( D(H_2) \) is the dissociation energy of \( H_2 \) (which is \( 4.500 \, \text{eV} \)).
- \( D(H_2^+) \) is the dissociation energy of the ion \( H_2^+ \) which is given as \( 2.650 \, \text{eV} \).

### Step 4: Substitute the values into the equation
Substituting the values we have:
\[ EA = 4.500 \, \text{eV} - 2.650 \, \text{eV} \]

### Step 5: Perform the calculation
Calculating the above expression:
\[ EA = 4.500 - 2.650 = 1.850 \, \text{eV} \]

### Final Answer
Thus, the electron affinity \( EA \) of the \( H_2^+ \) ion is approximately \( 1.850 \, \text{eV} \).

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