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Question
Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.
Solution
Total energy (for Hydrogen and H2 like Atoms): The total energy of the electron in the mth stationary states of the hydrogen. An atom of the hydrogenlike atom of atomic number Z is given by
`E = - ((me^4)/(8ε_0^2h^2)) Z^2/n^2 = - ((me^4)/(8ε_0^2ch^3)) ch Z^2/n^2`
= `- R ch Z^2/n^2`
= `- 13.6 Z^2/n^2 eV`
For a He-nucleus Z = 2, and for ground state n = 1.
Thus, the ground state energy of a He-atom.
`E_n = - 13.6 Z^2/n^2 eV`
= `- 13.6 2^2/1^2 eV`
= – 54.4 eV
Thus, the ground state will have two electrons each of energy E and the total ground state energy would be – (4 × 13.6) eV = – 54.4 eV
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