In atomic units, what is the Hamiltonian for the Be atom? H = -h^2/2m_c summatio

Question

In atomic units, what is the Hamiltonian for the Be atom? H = -h^2/2m_c summatiom_i=1^j = 4 -1/4 pi epsilon_0 Sigma_i = 1 ^ = 4 Ze^2/1_1 + 1/4 pi epsilon_0 summation _i = 1, j = 1, j = 1^j = 4, j =4 e^2/I_j? H = -h^2/2m_c summatiom_i=1^j = 4 -1/4 pi epsilon_0 Sigma_i = 1 ^ = 4 4 e^2/1_1 + 1/4 pi epsilon_0 summation _i = 1, j = 1, j = 1^j = 4, j =4 e^2/I_j ? H = -h^2/2m_c summatiom_i=1^j = 4 -1/4 pi epsilon_0 Sigma_i = 1 ^ = 4 Ze^2/1_1 + 1/4 pi epsilon_0 summation _i = 1, j = 1, j = 1^j = 4, j =4 e^2/I_j ? H = -h^2/2m_c summatiom_i=1^j = 4 -1/4 pi epsilon_0 Sigma_i = 1 ^ = 4 Ze^2/1_1 + 1/4 pi epsilon_0 summation _i = 1, j = 1, j = 1^j = 4, j =4 e^2/I_j

Explanation / Answer

Ans= b

Atomic number of Be = 4

Hamiltonian operator do not involve Z in the final expression because it is atomic number.

For Be, it should be replaced by 4.

Hence, a,d are wrong.

Total no of electrons in Be = 4

But, c involves total electrons as 6.

Hence, c is wrong.

Therefore,

Final answer = b

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