Download F1 In the Bohr model, the quantum number n gives the orbital

F1
In the Bohr model, the quantum number n gives the orbital angular momentum of the electron in its circular
orbit in units of h/2π. The angular momentum is, therefore, 5h/2π.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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S570 V1.1
F2
The radius of the Bohr orbit with n = 5 is 52 a0 , where a 0 is the Bohr
radius. In the stationary state, the orbit holds 5 wavelengths.
r
It appears, therefore, as sketched in Figure 3.
p
Figure 34A stationary electron wave
‘orbiting’ a proton. Note that the
circumference of the circle contains a
whole number of wavelengths of the
wave.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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S570 V1.1
F3
z-axis
(a) The coordinates r, θ and φ are defined as shown in Figure 6.
(r, θ, φ)
θ
(b) The three quantum numbers of the electron state are n, l and m.
(c) The quantum numbers related to the angular momentum of the
electron are l and m. The z-component of angular momentum is m˙,
where ˙ is h/2π.
The magnitude of the orbital angular momentum is
l(l + 1) ˙.
φ
r
y-axis
x-axis
Figure 64Alternatively, the point
shown in Figure 5 may be specified
using spherical polar coordinates r,1θ,1φ.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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S570 V1.1
F4
(a) The 3d state has the quantum numbers n = 3, l = 2. The degeneracy of an l-state is 2l + 1, so the 3d state is
5-fold degenerate. (b) The energy of an electron in a hydrogen atom is given by: En = −13.6 eV n2 .
The 3d state therefore has energy −1.511eV and the energy of the 2p state is −3.401eV.
The transition from one state to the other therefore releases about 1.891eV.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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S570 V1.1
R1
The magnitude of the momentum is 9.1 × 10−311kg × 2.0 × 107 1m1s−1 = 1.82 × 10−231kg1m1s−1.
Therefore the magnitude of the orbital angular momentum L is equal to
1.82 × 10−231kg1m1s−1 × 5 × 10−101m = 9.1 × 10−331kg1m2 1s−1 = 9.1 × 10−331J1s
In units of h L =
9.1 × 10 −33
h ≈ 14h
6.6 × 10 −34
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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R2
Energy change = energy before transition − energy after = −51eV − (−151eV) = +101eV
This energy is emitted as a quantum of electromagnetic radiation whose frequency, f, is given by the
Planck–Einstein formula: ∆E = hf.
Therefore the frequency is:
10 × 1.6 × 10 −19 J
f =
= 2. 4 × 1015 Hz
6.6 × 10 −34 s
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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S570 V1.1
R3
The time-independent Schrödinger equation in one dimension may be written:
˙2 d 2 ψ (x)
−
+ U(x) ψ (x) = Etot ψ (x)
2m dx 2
where Etot is the total energy of the electron.
The energy eigenfunction ψ(x) is the spatial part of a stationary state wavefunction.
It is also the probability density function for the electron. That is, the probability of finding the electron in the
small region from x to x + ∆x is given by |1 ψ(x)1|2 1∆x.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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THE OPEN UNIVERSITY
S570 V1.1
T1
The Bohr model gives, for the energy of the electron in the hydrogen atom, the formula:
−13.6 eV
En =
n2
where the electron has quantum number n.
Its angular momentum is n h (2π) ≡ n˙.
Now since 0.544 = 13.6/n2, n = 5 and the electron has an angular momentum of magnitude 5˙.
The state with one unit of angular momentum less has n = 4 and energy E4 = −13.6 eV 42 = −00.8501eV.
The energy released is −00.5441eV − (−00.8501eV) = 0.3061eV.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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S570 V1.1
z-axis
T2
(r, θ, φ)
(a) The values on the positive z-axis correspond to θ = 0.
θ
(b) The values on the negative z-axis correspond to θ = π (i.e. 180°).
Looking at Figure 6, points in the x–y plane correspond to values of
θ = π 2 (i.e. 90°).
(i) Points on the x-axis have values of φ = 0. The point (2,10,10) is two
units away from the origin; so the Cartesian coordinates (2,10,10) have
the (r,1θ,1φ) equivalent of (2,1π/2,10).
(ii) The point (1,11,10) is also in the x–y plane, but with φ = π/4 (i.e. 45°).
Its distance from the origin is given by Pythagoras’ theorem to be 2 .
r
φ
x-axis
Figure 64Alternatively, the point
shown in Figure 5 may be specified
using spherical polar coordinates r,1θ,1φ.
So the Cartesian coordinates (1,11,10) have the (r,1 θ,1φ) equivalent of ( 2 ,1π/2,1π/4,).
[There is an algebraic formula relating x,1y,1z to r,1 θ,1φ but it is not necessary for our purposes to use it.]
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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y-axis
T3
The magnitude of the angular momentum corresponding to quantum number l is
Therefore the magnitude for l = 2 is
6 ˙.
The permitted values of the z-component are in general the (2l + 1) values
−l, −l + 1, −l + 2, …, −1, 0, 1, …, l − 1, l in units of ˙.
The five allowed values for l = 2 are therefore −2˙, −˙, 0, ˙, 2˙.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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S570 V1.1
l(l + 1) ˙.
T4
z-axis
z-axis
The orientation of the angular momentum vector that makes
the smallest angle with the z-axis in the case where
l = 2 is given in Figure 15. The angle θ2,min is given by:
l˙
2˙
cos1θ2,min = 2
θ2,min
6 , i.e. θ2,min = 35.3°
θl,min
l(l+1) ˙
6˙
For l = 100, the angle would be given by:
cos1θ100,min = 100
100 × 101 , i.e. θ100,min = 5.7°
Although the angle becomes smaller for larger l, l(l + 1) ˙
can never be equal to l˙, so it is not possible for the angular
momentum vector to point along either the positive or
negative z-direction.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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(a)
Figure 154See Answer T4.
(b)
The radial probability density is
shown in Figure 16a (see also
Figure 7c). Note the increase in
the size of the maxima of the
radial probability density with r.
This is because of the r2 term in
Pnl (r) ∆r = | Rnl (r)|2 4πr 2 ∆r .
R3,0(r)
radial probability density
T5
0
0
2
(a)
4
6 8 10 12 14
r/10 −10 m
Figure 16a4See Answer T5.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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(c)
2
4
6 8 10 12
r /10 −10 m
Figure 7c4The radial function
Rn0l(r) for the hydrogen atom for the
case with principal quantum
number n equal to 3, and angular
momentum quantum number l
equal to zero. Note that the function
Rn00 0(r) has n − 1 zeros (excluding
the limit r → ∞).
T6
Table 94See Answer T6.
orbital angular momentum quantum number, l
The completed Table 4 is shown as Table 9:
0
1
2
3
principal
4
4s
4p
4d
4f
quantum
3
3s
3p
3d
number,
2
2s
2p
n
1
1s
Note that the places left blank correspond to combinations of
quantum numbers that are not permitted. There is no 3g state.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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4
T7
z
The electron cloud picture is shown in Figure 17.
ml = 0
Figure 174See
Answer T7.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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THE OPEN UNIVERSITY
S570 V1.1
T8
The energy is given by En = −13.6 eV n2 . The ground (lowest) state has quantum number n = 1. Its energy is
therefore E1 = −13.61eV. The first and second excited states have energies of, respectively, E2 = −3.41eV and
E3 = −1.51eV. The energies required are therefore,
(a) −3.41eV − (−13.61eV) = 10.21eV, and
(b) −1.51eV − (−13.61eV) = 12.11eV.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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T9
The solution is given in Table 10.
If you look at the totals in the
final column, you will see that
there are n2 states corresponding
to principal quantum number n.
[We have illustrated special cases
of the arithmetic series
summation:
1 + 3 + 5 + … + (2n − 1) = n 2 ,
since it may be seen from the
penultimate column of Table 10
that, for a given n, the number of
states is the sum of the odd
integers from one to 2n − 1.]
Table 104The quantum numbers of the levels of the hydrogen atom. See Answer T9
Value
of n
Values
of l
Values
of ml
1
2
0
0
1
0
0
3
4
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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−1, 0, 1
0
Number of states Number of states
for each n and l
for each n
1
1
3
0
1
−1, 0, 1
1
3
2
−2, −1, 0, 1, 2
5
0
0
1
1
−1, 0, 1
3
2
−2, −1, 0, 1, 2
5
3
−3, −2, −1, 0, 1, 2, 3
7
S570 V1.1
1
4
9
16
T10
(a) In the Bohr model, there is one state with orbital angular momentum of magnitude 3˙. It is the state with
quantum number n = 3. Its energy is
E = −13.6 eV 32 = −1.5 eV
(b) In the Schrödinger model, the magnitude of the orbital angular momentum is l(l + 1) ˙ with l an integer.
The magnitude is therefore never equal to an integral multiple of ˙. In particular, there are no states with orbital
angular momentum equal to 3˙.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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T11
(a) The only change necessary in the calculation outlined in Subsection 2.1 is the replacement of the force term
appearing in Equation 2, e 2 (4πε 0 r 2 ) , by the gravi-tational equivalent GmM r 2 .
mSv2 GmS mE
Equating
=
r
r2
GmS mE
with
n 2 ˙2 = mS2 v2 r 2 (= L2 ) = mS r 3
r2
n 2 ˙2 = GmS2 mE r
so
n2 =
GmS2 mE r
˙2
(
n 2 = 6.67 × 10 −11 N m 2 kg −2 × (6. 0 × 10 3 )2 kg 2 × 6. 0 × 10 24 kg × (6. 4 + 0.1) × 10 6 m / 6.63 × 10 −34 J s 2π
= 8.41 × 10961m
so n = 2.90 × 1048
FLAP P11.3
Schrödinger’s model of the hydrogen atom
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)
2
(b) The change in radius is ∆r = −2r/n = −4.5 × 10−421m.
(c) v = ˙ nma0 = 7848 m s −1 ,
so that ∆v = v/n = 2.7 × 10−451m1s−1.
FLAP P11.3
Schrödinger’s model of the hydrogen atom
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1