2. (a) Verify the following for each of the six states 1s,2s, 2p,3s,3p,3d of the hydrogen atom. (5²)= (5) 3n²l(+1) 2Z n² [5n²+1-3l(l + 1)] 2Z² Z n² Z n³ ( l + 1²¹ )
2. (a) Verify the following for each of the six states 1s,2s, 2p,3s,3p,3d of the hydrogen atom. (5²)= (5) 3n²l(+1) 2Z n² [5n²+1-3l(l + 1)] 2Z² Z n² Z n³ ( l + 1²¹ )
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Question
Please answer the red boxed equation in #2 for the state 3s only.
![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) = --
a =
μе²
Atomic unit of energy
e²
24
Eatom = =
Atomic unit of length the Bohr radius
ħ
5.292 × 10-⁹ cm
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 8 + 2/3 ²²) 0-²/²
7 5
3√3
e-/3
(5²):
1
8६
27√₁ (1-11) 6-1/2
e-/3
4
=$²e-$/3
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
=
1
Z
(7)=7²
=
n³ ( l + ²/² )
(b) How does the average value of potential
energy compare with the total energy?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3a11ce3d-21c0-472b-9212-b4fe5760ed67%2F5f04233a-06df-4a91-a9fa-fae3f30ae29c%2F5756xuh_processed.jpeg&w=3840&q=75)
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) = --
a =
μе²
Atomic unit of energy
e²
24
Eatom = =
Atomic unit of length the Bohr radius
ħ
5.292 × 10-⁹ cm
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 8 + 2/3 ²²) 0-²/²
7 5
3√3
e-/3
(5²):
1
8६
27√₁ (1-11) 6-1/2
e-/3
4
=$²e-$/3
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
=
1
Z
(7)=7²
=
n³ ( l + ²/² )
(b) How does the average value of potential
energy compare with the total energy?
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