EE372_Solutions4

A we first ignore the repulsion between the electrons

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a) We first ignore the repulsion between the electrons; each electron only feels the attraction to the nucleus. The solutions to Schrodinger’s equation are then the same as for hydrogen but with Z = 2. What is the total energy of both electrons in the ground state (1s 2 )? b) Now add in the Coulomb energy of the interaction between the electrons. Since the electrons repel each other, the energy is positive. But to calculate the energy, we need the distance between the electrons. Assume that the electrons stay as far apart as possible within the orbital because of the repulsion. Calculate the mean distance from the nucleus of a 1s orbital for helium using the formula from problem 2. Assume the electrons are at this distance but on opposite sides of the nucleus and calculate the Coulomb energy ( e 2 / 4 π 0 R ) and add this energy to the result from part a to get the total energy of helium. How does your answer compare to the measured value - 79 . 02 eV? Answer: a) The energy of each electron is - 13 . 6 eV(2 2 / 1 2 ) = - 54 . 4 eV. So the total energy of both electrons is - 109 eV. b) The mean distance for the 1s orbital in He (assuming hydrogen-like wavefunctions) is (using the formula from the previous problem) 0 . 75 a 0 . If the electrons stay on opposite sides

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of the numcleus then their typical separation is R = 1 . 5 a 0 = 7 . 9 × 10 - 11 m. The energy from to the Coulomb repulsion is 18.1 eV. Adding this energy to the energy of part a gives - 91 eV which is about 15% lower than the measured value.
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