number 2 ( a - c ) For each of the integrals below, indicate whether they evalua

Question

number 2 ( a - c )

For each of the integrals below, indicate whether they evaluate to 0, 1, or non-zero value other than 1. Note that a single integral over dtau denotes integration over all variables. Show your work or briefly justify each result. Hint: you should not have to do much math. dtau Psi2s, Psi1s is the normalized 1s orbital of the hydrogen atom, and Psi2s is the normalised 2s orbital of the hydrogen atom. dtau Psi1s,Li, where Psi1s,H is the normalized 1s orbital of the hydrogen atom, and Psi1s,Li is the normalized 1s orbital of the Li2+ ion. dtau Y2,1, where Psi+=Y2,1 and Yl,mt are the spherical harmonics. dtau Psi-, where Psi+ = Y2,1 + Y2,-1 and Psi- = Y2,1-Y2,-1. dtau (1,2)Psi1(1,2), where Psi0(1,2) is the two-electron wavefunction for the 1s2 ground state configuration of the helium atom, and Psi1 (1,2) is the two-electron wavefunction for the excited 1s1 2s1 triplet configuration of the helium atom: Psi0(1,2) = 1/ 21s(1)1s(2)[alpha(1)beta(2)-beta(1)alpha(2)], Psi1(1,2)= 1/ 2[1s(1)2s(2)-2s(1)1s(2)]alpha(1)alpha(2). The 1s and 2s are the hydrogenic orbitals of the He+ ion, and the alpha and beta are the spin-up and spin-down eigenfunctions, respectively. The first ionization energy, I1, is the energy needed to remove an electron (to infinity) the ground state of a neutral atom in the gas phase. For lithium, the experimental Val- for I1 is 5.39 electron volts(eV). Estimate the effective nuclear charge experienced by the Li 2s electron. Using the effective nuslear charge in part a, determine the most probable distance the Li 2s electron in units of the Bohr radius a0. How does this compare with the most probable distance of the 2s electron in the Li2+ ion? Give a brief physical explanation for the difference in the most probable 2s distance in Li and Li2+. For the second row elements, i.e., the row in the periodic table starting with Li, the value of I1 increases as one goes from left to right, ending with a value of 21.56 eV for He. Explain the Physical reason for this trend.

Explanation / Answer

as the IE1 is 5.39 eV

lamda = 1240/5.39= 230.05 nm

so we also see that the 2s orbital is asked so n1=2 and n2 = infinity

so 1/lamda = RZ^2 [ 1/4-0]

= we use the value of lamda here

1/230.05 = 1.1*10^6 * z^2 [.25]

= z= [ 4 ] approx

the effective nuclear charge is 4

b]R[ most probable distance ] = ao * [n^2/z] = ao* 1

= ao

for lithium the value is = ao * 4/3 = 1.33a0

c]this is because the number of electrons keeps on increasing as we move towards right and this effects in the sigma effect of the electrons which does not allow the electron to be removed from the shell of the elements ..

the Zeffective doesnot change that much but the sigma effect dominates.

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