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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. |
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Answer» Solution :In a helium atom `(._2He^(4))`, there are two protons and two neutrons in the nucleus. Two electrons are revolving around the nucleus in the first orbit. We are GIVEN that there is no repulsive force between the electrons. for He nuclues, CHANGE, `Z = + 2e`.As two electrons of change `(-2e)` revolve around the nuclues, therefore the formula for energy in nth orbit WOULD be `E_(n) = - (me^(4)(Z^(2)))/(8 in_(0)^(2)n^(2)h^(2)) = - (4 me^(4))/(8 in_(0)^(2)n^(2)h^(2)) = 4E` The ground state will have two electrons, each of energy `E = - (me^(4))/(8 in_0^2 (1)^2h^2)=-13.6 eV` `:.` The total ground state energy of He atom =`4E=4(-13.6)eV=-54.4eV` |
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