CHEM 301: Homework assignment #6
Solutions
1. What transition in He+ has the same frequency as the 2p→1s transition
in H? (20%)
He+ is a hydrogen-like ion (has only one electron). Thus, transitions in
both H and He+ can be described by Rydberg’s formula:
1
1
1
2
= RZ
− 2 ,
λ
n21
n2
where λ is the wavelength of the emitted photon, R is Rydberg’s constant,
Z is the atomic number (1 for H, 2 for He), and n1 and n2 are the principal
quantum numbers for the initial and final states, respectively.
Since there is one-to-one correspondence between the frequency ν and the
wavelength λ, we can compare wavelengths instead of frequencies:
!
!
1
1
1
1
1
2
2
=R·2
− 2
=R·1
− 2
.
λ
n21(He)
n2(He)
n21(H)
n2(H)
We can divide both the left- and the right-hand side in the second equation
by R and, using the initial and final states for H that are given in the
problem (n1(H) = 2 (2p) and n1(H) = 1 (1s)), we arrive at:
! 1
1 1
1
4
−
=
−
n21(He)
n22(He)
4 1
or
1
n21(He)
−
1
n22(He)
=
1
1
− .
16 4
We conclude that n1(He) = 4 and n2(He) = 2.
In order to be allowed, transitions must satisfy the selection rule for the
orbital angular momentum quantum number: ∆l = ±1. Thus, the allowed
transitions in He+ with the same energy as the 2p→1s transition in H are:
4p→2s, 4s→2p, and 4d→2p.
2. How many radial and angular nodes do 4s, 4p, and 4d orbitals of hydrogen
have? How many electrons can the 4s, 4p, and 4d subshells contain?
(15%)
The number of angular nodes for an orbital is l, the number of radial
nodes is n − l − 1. Thus, the 4s orbital (n = 4, l = 0) has no angular
nodes and 3 radial nodes. Likewise, for the 4p orbital (n = 4, l = 1) there
is 1 angular node and 2 radial nodes, while for the 4d orbital there are 2
angular nodes and 1 radial node.
1
In an s subshell (l = 0) there is only one orbital (ml = 0). In a p subshell
(l = 1) there are three orbitals (ml = 0, ±1). In a d subshell (l = 2) there
are five orbitals (ml = 0, ±1, ±2). According to Pauli’s exclusion principle,
each orbital can hold two electrons. Thus, there can be 2 electrons in the
4s subshell, 6 electrons in the 4p subshell, and 10 electrons in the 4d
subshell.
3. Calculate the probability of finding a 1s electron in H within the distance
2rB from the nucleus. (30%)
The probability of finding a 1s electron in H within the distance 2rB from
the nucleus (within a sphere of radius 2rB ) can be obtained by integrating
2
the probability density (|ψ (r)| ) over the volume of this sphere.
2
2
Since the 1s orbital is spherically symmetric, |ψ (r)| = |ψ (r)| (the wavefunction only depends on r). Thus, we can calculate the probability of
finding the electron within a very thin spherical shell between radius r
2
and radius r + dr by multiplying |ψ (r)| (that is assumed to be constant
within this sphere) by the volume of this spherical shell, 4πr2 dr. Then,
we need to integrate over all radii from 0 to 2rB :
Z2rB
2
P (r < 2rB ) =
4πr2 |ψ (r)| dr.
0
In Table 13.1 of the 5th edition of the textbook we find the 1s wavefunction
of a hydrogen-like atom:
32
Z
Zr
1
√
2
exp −
ψ (r) =
rB
rB
2 π
| {z } |
{z
}
angular part
radial part
and introduce it into the equation for the probability, taking Z = 1 for H.
This gives:
3
Z2rB
1
2r
2 1
4
P (r < 2rB ) =
exp −
dr.
4πr
4π
rB
rB
0
Taking the constants out of the integration, we get:
P (r < 2rB ) = 4
1
rB
3 Z2rB
2r
r2 exp −
dr.
rB
0
In a table of integrals (for instance, in the “List of integrals of exponential
functions” on Wikipedia) we find that
2
Z
r
2r
2
2
r exp (cr) dr = exp (cr)
− 2 + 3 .
c
c
c
2
Thus, if we denote c = − r2B , we can rewrite the probability of finding a 1s
electron in H within the distance 2rB from the nucleus as:
2
2r
3
r
2 B
1
2r
=
exp (cr)
P (r < 2rB ) = 4
− 2 + 3 rB
c
c
c
0



3 

1
2 · 2rB
2
2
 4r2

=4
exp − 2rB  B − 2 + 3  −
2

rB
r
B
2
2

− rB
− rB
− rB




2


− exp (0) 0 − 0 + 3  =


− r2B
3 3
3
3
3
1
4rB
2rB
2rB
4rB
=4
−
−
− −
=
exp (−4) −
rB
2
4
8
8
1
1
= 4 exp (−4) −2 − 1 −
+
=
4
4
= [1 − 4 exp (−4) · 3.25] = 0.762.
4. Write down the wavefunction for He in the orbital approximation if both
electrons occupy the 1s orbital. Assume that both electrons are at the
same distance from the nucleus (r1 = r2 = r). (15%)
In the lecture, we have written out the wavefunction for the He atom in
the orbital approximation:
Ψ (r1 , σ1 , r2 , σ2 ) =
1
√
2
|{z}
ψ (r1 ) ψ (r2 ) [α (σ1 ) β (σ2 ) − α (σ2 ) β (σ1 )] .
|
{z
}|
{z
}
spatial part
spin part
normalization
constant
This wavefunction consists of a normalization constant, a symmetric spatial part and an antisymmetric spin part, so that the total wavefunction
satisfies the antisymmetricity requirement of Pauli’s principle.
All we have to do now is to set r1 = r2 = r and to introduce the spatial
wavefunction ψ (r) that corresponds to the 1s orbital (we can find it in
Table 13.1 of the 5th edition of the textbook):
32
Z
Zr
1
√
2
exp −
,
ψ (r) =
rB
rB
2 π
| {z } |
{z
}
angular part
radial part
where Z = 2 for He, and rB is the Bohr radius. Then, the wavefunction
for He in the orbital approximation when both electrons occupy the 1s
orbital becomes:
3
2
2r
1 1
√
Ψ (r, σ1 , r, σ2 ) =
4
exp −2
[α (σ1 ) β (σ2 ) − α (σ2 ) β (σ1 )] ,
rB
rB
2 4π
3
or
Ψ (r, σ1 , r, σ2 ) =
1
√
π 2
2
rB
3
4r
exp −
[α (σ1 ) β (σ2 ) − α (σ2 ) β (σ1 )] .
rB
5. Write out the electronic configurations of Co and Sc2+ . (20%)
Co: [Ar] 4s2 3d7
Sc2+ : [Ar] 3d1
(The electronic configuration of Sc is [Ar] 3d1 4s2 . To form a cation, we
first remove the valence p electrons - there are no such electrons for Sc,
then the s electrons, then the d electrons.)
4