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Determine the maximum of the radial distribution function for the ground state of hydrogen atom. Compare this value with the corresponding radius in the Bohr ...
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He++^ + H → He+^ + H+
Calculate the electronic energy change (in eV). Assume all species in their ground states.
ψ(r, θ, φ) = const
1 − r sin^2 (θ/2)
e−r/^2
is a solution of the Schr¨odinger equation for the hydrogen atom and find the corresponding eigenvalue (in atomic units).
∆ν = 532. 65
8 π 3
|ψ(0)|^2 +
3 cos^2 θ − 1 r^3
MHz
Calculate ∆ν for the 1s and for the 2p 0 states. The result is in MHz when the bracketed terms are expressed in atomic units. (Hint: In the expectation value, do the integral over angles first.)
〈r−^1 〉 =
0
ψ1s(r) r−^1 ψ1s(r) 4πr^2 dr = 1
a 0
0
ψ1s(r)
r
ψ1s(r) 4πr^2 dr = −Z^2
Average kinetic energy:
0
ψ1s(r)
ψ1s(r) 4πr^2 dr = Z^2 / 2
More simply, since total energy E1s = −Z^2 /2, 〈T 〉 = E1s − 〈V 〉. Note that 〈V 〉 = − 2 〈T 〉, consistent with the virial theorem.
const =
π
Noting that sin^2 (θ/2) = (1 − cos θ)/2, the function is found to be an s-p hybrid orbital:
ψ =
(ψ 2 s + ψ 2 pz )
0
|ψ 1 s(r)|^2 4 πr^2 dr = 0. 9
or easier (^) ∫ (^) ∞
R
|ψ 1 s(r)|^2 4 πr^2 dr = 0. 1
We find, using integral table,
R
r^2 e−^2 r^ dr = e^2 R^ (1 + 2R + 2R^2 ) = 0. 1
Solving numerically, R = 2. 6612 a 0 = 1.41 ˚A.
E(α) =
0 e
−αr (− 1 2 ∇
(^2) − Z/r)^ e−αr 4 πr (^2) dr ∫ (^) ∞ 0 e
− 2 αr 4 πr (^2) dr =
α^2 − Zα
E′(α) = 0 for minimum, giving α = Z. Thus ψ(r) = e−Zr^ and E = −Z^2 /2, which in this exceptional case equal the exact eigenfunction and eigenvalue.