1 2 ● 3 ● ● ● ● VE,,m(r, 0, 0) = fe,e(r)Y (0,0) R(r) = rfe,e(r) Radial Schrödinger equation ħ² d²R ħ²l(l+1)] 2μr² 2μ dr² n + Electron Dimensionless quantity + U(r) = -- Atomic unit of energy e² 24 Eatom = = Atomic unit of length the Bohr radius ħ a = ≈ 5.292 x 10-⁹ cm μе² a ħ² 0 0 1 0 1 2 не r } = a, &= Ck+1 = Ze² E Eatom R(E)=9e-a5W (5) q = (l+1) r e W({) = ΣCKŠk 2 k=0 2qc₁ + (2Z-2aq)c₁ = 0 2ak + 2aq - 2Z ≈ 27.21 eV 0 1 0 2 1 0 (k + 1)(k + 2q) Z α = - n R = ER kmax The radial normalization condition: [ro 0 ƒ²(r)r²dr = 1 Ck Label 1s 2s 2p 3s Un,l,m (r, 0,0) = fn,e(r)Ym (0,4) 3p 3d For convenience, let us modify the radial normalization condition as Label 1s 2s 2p 3s 3p 3d [ro 0 1. Verify the following for the case Z = 1, the hydrogen atom. f²()}² d = 1. fne(r) 2e- (1-7/8)e-²/2 1 ={e-5/2 2√6 2 2 33/7 ( 1 - 3/4 + 1/ / ²²) 0-²/² 2 7 5 e-/3 3√3 8६ 27√6 (5²): 1 = 1- - 1²/2² ) e - ²1 =$²e-$/3 4 81√30 2. (a) Verify the following for each of the six states 1s,2s, 2p,3s,3p,3d of the hydrogen atom. (5) 3n²-l(l + 1) 2Z n² [5n²+1-3l(l + 1)] 2Z² ܀ = e-/3 Z n² Z² n³ ( l + 2/² ) (b) How does the average value of potential energy compare with the total energy?
1 2 ● 3 ● ● ● ● VE,,m(r, 0, 0) = fe,e(r)Y (0,0) R(r) = rfe,e(r) Radial Schrödinger equation ħ² d²R ħ²l(l+1)] 2μr² 2μ dr² n + Electron Dimensionless quantity + U(r) = -- Atomic unit of energy e² 24 Eatom = = Atomic unit of length the Bohr radius ħ a = ≈ 5.292 x 10-⁹ cm μе² a ħ² 0 0 1 0 1 2 не r } = a, &= Ck+1 = Ze² E Eatom R(E)=9e-a5W (5) q = (l+1) r e W({) = ΣCKŠk 2 k=0 2qc₁ + (2Z-2aq)c₁ = 0 2ak + 2aq - 2Z ≈ 27.21 eV 0 1 0 2 1 0 (k + 1)(k + 2q) Z α = - n R = ER kmax The radial normalization condition: [ro 0 ƒ²(r)r²dr = 1 Ck Label 1s 2s 2p 3s Un,l,m (r, 0,0) = fn,e(r)Ym (0,4) 3p 3d For convenience, let us modify the radial normalization condition as Label 1s 2s 2p 3s 3p 3d [ro 0 1. Verify the following for the case Z = 1, the hydrogen atom. f²()}² d = 1. fne(r) 2e- (1-7/8)e-²/2 1 ={e-5/2 2√6 2 2 33/7 ( 1 - 3/4 + 1/ / ²²) 0-²/² 2 7 5 e-/3 3√3 8६ 27√6 (5²): 1 = 1- - 1²/2² ) e - ²1 =$²e-$/3 4 81√30 2. (a) Verify the following for each of the six states 1s,2s, 2p,3s,3p,3d of the hydrogen atom. (5) 3n²-l(l + 1) 2Z n² [5n²+1-3l(l + 1)] 2Z² ܀ = e-/3 Z n² Z² n³ ( l + 2/² ) (b) How does the average value of potential energy compare with the total energy?
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Your Question:
Please answer #1 for 3s only (pointed by the arrow).
![1
2
●
3
●
●
●
●
VE,,m(r, 0, 0) = fe,e(r)Y (0,0)
R(r) = rfe,e(r)
Radial Schrödinger equation
ħ² d²R
ħ²l(l+1)]
2μr²
2μ dr²
n
+
Electron
Dimensionless quantity
+
U(r) = --
Atomic unit of energy
e²
24
Eatom = =
Atomic unit of length the Bohr radius
ħ
a = ≈ 5.292 x 10-⁹ cm
μе²
a ħ²
0
0
1
0
1
2
не
r
} = a, &=
Ck+1 =
Ze²
E
Eatom
R(E)=9e-a5W (5)
q = (l+1)
r
e
W({) = ΣCKŠk
2
k=0
2qc₁ + (2Z-2aq)c₁ = 0
2ak + 2aq - 2Z
≈ 27.21 eV
0
1
0
2
1
0
(k + 1)(k + 2q)
Z
α = -
n
R = ER
kmax
The radial normalization condition:
[ro
0
ƒ²(r)r²dr = 1
Ck
Label
1s
2s
2p
3s
Un,l,m (r, 0,0) = fn,e(r)Ym (0,4)
3p
3d
For convenience, let us modify the radial
normalization condition as
Label
1s
2s
2p
3s
3p
3d
[ro
0
1. Verify the following for the case Z = 1,
the hydrogen atom.
f²()}² d = 1.
fne(r)
2e-
(1-7/8)e-²/2
1
={e-5/2
2√6
2
2
33/7 ( 1 - 3/4 + 1/ / ²²) 0-²/²
2 7 5
e-/3
3√3
8६
27√6
(5²):
1
=
1-
- 1²/2² ) e - ²1
=$²e-$/3
4
81√30
2.
(a) Verify the following for each of the six
states 1s,2s, 2p,3s,3p,3d of the hydrogen
atom.
(5)
3n²-l(l + 1)
2Z
n² [5n²+1-3l(l + 1)]
2Z²
܀
=
e-/3
Z
n²
Z²
n³ ( l + 2/² )
(b) How does the average value of potential
energy compare with the total energy?](https://content.bartleby.com/qna-images/question/3a11ce3d-21c0-472b-9212-b4fe5760ed67/ebb715e1-4f40-4daf-87c4-764ae7eb3488/1bg63ec_processed.jpeg)
Transcribed Image Text:1
2
●
3
●
●
●
●
VE,,m(r, 0, 0) = fe,e(r)Y (0,0)
R(r) = rfe,e(r)
Radial Schrödinger equation
ħ² d²R
ħ²l(l+1)]
2μr²
2μ dr²
n
+
Electron
Dimensionless quantity
+
U(r) = --
Atomic unit of energy
e²
24
Eatom = =
Atomic unit of length the Bohr radius
ħ
a = ≈ 5.292 x 10-⁹ cm
μе²
a ħ²
0
0
1
0
1
2
не
r
} = a, &=
Ck+1 =
Ze²
E
Eatom
R(E)=9e-a5W (5)
q = (l+1)
r
e
W({) = ΣCKŠk
2
k=0
2qc₁ + (2Z-2aq)c₁ = 0
2ak + 2aq - 2Z
≈ 27.21 eV
0
1
0
2
1
0
(k + 1)(k + 2q)
Z
α = -
n
R = ER
kmax
The radial normalization condition:
[ro
0
ƒ²(r)r²dr = 1
Ck
Label
1s
2s
2p
3s
Un,l,m (r, 0,0) = fn,e(r)Ym (0,4)
3p
3d
For convenience, let us modify the radial
normalization condition as
Label
1s
2s
2p
3s
3p
3d
[ro
0
1. Verify the following for the case Z = 1,
the hydrogen atom.
f²()}² d = 1.
fne(r)
2e-
(1-7/8)e-²/2
1
={e-5/2
2√6
2
2
33/7 ( 1 - 3/4 + 1/ / ²²) 0-²/²
2 7 5
e-/3
3√3
8६
27√6
(5²):
1
=
1-
- 1²/2² ) e - ²1
=$²e-$/3
4
81√30
2.
(a) Verify the following for each of the six
states 1s,2s, 2p,3s,3p,3d of the hydrogen
atom.
(5)
3n²-l(l + 1)
2Z
n² [5n²+1-3l(l + 1)]
2Z²
܀
=
e-/3
Z
n²
Z²
n³ ( l + 2/² )
(b) How does the average value of potential
energy compare with the total energy?
