Chemistry 362
Dr. Jean M. Standard
Problem Set 8 Solutions
1.
Using the expression for the energy of a hydrogen-like ion, determine the ionization energy for Li2+.
Express your answer in hartrees, eV, and Joules.
For a hydrogen-like ion such as Li2+, the energy in atomic units (also called hartrees) is given by
En = −
Z2
.
2n 2
The ionization process can be described by the equation
€
Li2+ → Li3+ + e − .
The ionization energy is just the energy that it takes for the ionization process. Thus, the ionization energy
can be defined as
€
IE = En =∞ − En =1 .
€
Substituting the hydrogen-like ion energy expression yields
€
% 32 (
32
− '−
*
2
2
2×∞
& 2 ×1 )
% 9(
= 0 − '− *
& 2)
IE = 4.5 a.u. (or hartrees).
IE = −
Using the conversion factors, 1 hartree = 27.211 eV and 1 hartree = 4.3597 × 10 –18 Joules, we have
€
IE = 122.4 eV
= 1.962 €
× 10 -17 J .
€
IE
2
2.
The ground state wavefunction of the hydrogen-like ion is
% Z (3 / 2 −Z r / a
o Y
ψ100 ( r,θ ,φ ) = 2 ' *
e
00 (θ ,φ ) .
& ao )
Show that this function is an eigenfunction of the Hamiltonian operator of the hydrogen-like ion. Hint: to
evaluate the angular €
portion, use the result that
Lˆ2 Yℓm ℓ (θ ,φ ) = " 2ℓ (ℓ + 1) Yℓm ℓ (θ ,φ ) .
We want to show that Hˆ ψ€100 ( r,θ ,φ ) = E1 ψ100 ( r,θ ,φ ) with the ground state energy given by
E1 = −
€
Z 2e 4 m e
2! 2 ( 4 πε 0 )
2
, (n = 1) .
The Hamiltonian operator for the hydrogen-like ion is
€
Lˆ2 (θ ,φ )
!2 $ ∂ 2
2 ∂ '
Ze 2
Hˆ = −
−
.
& 2 +
)+
2me % ∂ r
r ∂r(
4 πε 0 r
2me r 2
If we operator the angular momentum part of the Hamiltonian operator on the ground state wavefunction, we
get
€
Lˆ2 (θ ,φ )
2me r 2
ψ100 ( r,θ ,φ ) =
Lˆ2 (θ ,φ )
R10 ( r ) Y00 (θ ,φ )
2m e r 2
1
=
R10 ( r ) Lˆ2 (θ ,φ ) Y00 (θ ,φ ) .
2me r 2
Note here that the radial function can be pulled through the angular momentum operator because the radial
function only depends
€ on r and so is unaffected by an operator that depends only on the angular coordinates.
2
Since Lˆ2Yℓm ℓ (θ ,φ ) = " 2ℓ(ℓ + 1) Yℓm ℓ (θ ,φ ) , then for ℓ = 0 , we have Lˆ Y00 (θ ,φ ) = 0 . Therefore,
€
€
Lˆ2 (θ ,φ )
2me r 2
ψ100 ( r,θ ,φ ) = 0 .
€
Working out the rest,
€
/ !2
ˆ
H ψ100 ( r,θ ,φ ) = 1−
10 2me
/ !2
= 1−
10 2me
' ∂2
2 ∂*
Ze 2 2
4 ψ100 ( r,θ ,φ )
) 2 +
,−
r ∂r +
4 πε 0 r 43
( ∂r
.
' ∂2
2 ∂*
Ze 2 2
4 R10 ( r ) Y00 (θ ,φ )
) 2 +
,−
r ∂r +
4 πε 0 r 43
( ∂r
/ !2
= Y00 (θ ,φ ) 1−
10 2m e
€
' ∂2
2 ∂*
Ze 2 2
4 R10 ( r ) .
) 2 +
, −
r ∂r +
4 πε 0 r 43
( ∂r
3
2.
Continued
Notice that since the remaining parts of the Hamiltonian operator are independent of the angular coordinates, it
does not operate on the function Y00 (θ ,φ ) , and we can pull this function through.
Substituting the expression for R10 ( r ) yields the result
€
/ !2
Hˆ ψ100 ( r,θ ,φ ) = Y00 (θ ,φ ) 1−
€
10 2m e
' ∂ 2 2 ∂ * Ze 2 2
4
) 2+
,−
r ∂r + 4 πε 0 r 43
( ∂r
/ !2
' Z *3 / 2
= 2 ) , Y00 (θ ,φ ) 1−
10 2m e
( ao +
' Z *3 / 2
2 ) , e −Zr / a 0
( ao +
' ∂ 2 2 ∂ * Ze 2 2
4 e −Zr / a 0 .
) 2+
,−
r
∂
r
4
πε
r
4
( ∂r
+
0 3
The next step is to evaluate the partial derivatives. We have
€
∂ −Zr / a 0
Z −Zr / a 0
e
= −
e
,
∂r
ao
and
€
∂ 2 −Zr / a 0
Z 2 −Zr / a 0
e
=
e
.
∂r 2
a o2
Substituting,
€
. !2
% Z (3 / 2
ˆ
Hψ100 ( r,θ ,φ ) = 2 ' * Y00 (θ ,φ ) 0−
0/ 2me
& ao )
% Z 2 2 Z ( Ze 2 1
3 e −Zr / a 0 .
' 2−
*−
& a o r a o ) 4 πε 0 r 32
−Zr / a 0
Note that the exponential e
is common to all the terms remaining, so it has been factored out. Finally, we
must use€the definition of the constant a o ,
€
€
ao =
4 πε 0 ! 2
.
me e 2
Substituting,
Hˆ ψ100
€
€
# Z &3 / 2 , ! 2 # Z 2 m 2e 4
2Zme e 2 &
Ze 2 / −Zr / a 0
e
1e
= 2 % ( Y00 .−
−
−
.
%
(
4 πε 0 r 10
.- 2me $ 16 π 2ε 02 ! 4 4 πε 0 ! 2 r '
$ ao '
4
2. Continued
Simplifying,
# Z 2e 4 m
# Z &3 / 2
Ze 2
Ze 2 & −Zr / a 0
e
ˆ
Hψ100 = 2 % ( Y00 %−
+
−
(e
2 2 2
4 πε 0 r
4 πε 0 r '
$ ao '
$ 32π ε 0 !
# Z 2e 4 m &
# Z &3 / 2
e
= 2 % ( Y00 %−
( e −Zr / a 0
2 2 2
a
$ o'
$ 32π ε 0 ! '
= −
Hˆ ψ100 = −
Z 2e 4 me
Y00 (θ ,φ )
32π 2ε 02 ! 2
Z 2e 4 me
2
2( 4 πε 0 ) ! 2
# Z &3 / 2
2 % ( e −Zr / a 0
$ ao '
ψ100 ( r,θ ,φ ) .
Therefore, the wavefunction ψ100 ( r,θ ,φ ) is an eigenfunction of the hydrogen-like ion Hamiltonian operator.
The eigenvalue
€ is therefore the ground state energy. Since
€
E1 = −
we have Hˆ ψ100 ( r,θ ,φ ) = E1 ψ100 ( r,θ ,φ ) .
€
€
Z 2e 4 m e
2! 2 ( 4 πε 0 )
2
,
5
3.
The radial wavefunction for the 2s orbital for the hydrogen-like ion is
" Z %3 / 2 "
Z r % −Z r / 2a o
R20 ( r ) = $
.
' $2 −
' e
ao &
# 2a o & #
Show that this function is normalized.
€
To show that R20 ( r ) is normalized, we must show that
∞
∫ [R
0
€
20
( r )]
2 2
r dr = 1 .
2
Note that the volume element r dr must be used for the radial coordinate and the limits of integration for the
radial coordinate range from 0 to €
∞ . Using the form for R20 ( r ) given above, the normalization integral is
€
∞
€
∫ [R
0
20
( r )]
$ Z '3
r dr = &
)
€% 2a o (
2 2
∫
2
+
Zr . −Zr / a 0 2
r dr .
-2 −
0 e
ao /
,
∞
0
Expanding,
€
∞
∫ [R
0
20
( r )]
$ Z '3
r dr = &
)
% 2a o (
2 2
∫
∞
0
+
4Zr
Z 2 r 2 . −Zr / a 0 2
+
r dr .
-4 −
0e
ao
a o2 /
,
This expression can be broken up into a sum of 3 integrals. This yields
€
∞
∫ [R
0
€
20
( r )]
$ Z '3 +
r dr = &
) ,4
% 2a o ( -
2 2
∫
∞
0
r 2 e −Zr / a 0 dr −
4Z
ao
∫
∞
0
r 3 e −Zr / a 0 dr +
Z2
a o2
∫
∞
0
.
r 4 e −Zr / a 0 dr / .
0
We can evaluate the 3 integrals using definite integrals found in the CRC or some other source. The integrals
all have the form
∫
∞
0
x n e −ax dx =
n!
a n +1
.
The 3 integrals are therefore
€
∫
and
€
∫
∞
0
∞
0
% a (3
r 2e −Zr / a 0 dr = 2 ' o * ,
&Z)
% a (5
r 4 e −Zr / a 0 dr = 24 ' o * .
&Z)
∫
∞
0
% a (4
r 3e −Zr / a 0 dr = 6 ' o *
&Z)
6
3.
Continued
Substituting,
∞
∫ [R
0
20
( r )]
$ Z '3
r dr = &
)
% 2a o (
4
5
+- $ a '3
24Z $ a 0 '
24Z 2 $ a 0 '
,8 & 0 ) −
& ) +
&
)
a0 % Z (
a 02 % Z (
.- % Z (
2 2
3
1$ Z '
= & )
8 % ao (
+- $ a '3
$ a '3
$ a '3 /,8 & 0 ) − 24 & 0 ) + 24 & 0 ) 0
%Z(
% Z ( -1
-. % Z (
3
=
∞
∫ [R
0
20
( r )]
3
1 $ Z ' $ ao '
& ) 8& )
8 % ao ( % Z (
2 2
r dr = 1.
Therefore, the radial function R20 ( r ) is normalized.
€
€
/0
1-
7
4.
Evaluate the average value of r for the hydrogen 1s electron. The ground state wavefunction is given by
ψ100 ( r, θ , φ ) = 3/2
1 !Z$
−Z r/ ao
.
# & e
π " ao %
Your answer may be expressed in terms of the constant a o .
For hydrogen, we have Z=1, so the wavefunction simplifies to
€
ψ100 ( r, θ , φ ) = 1
π
! 1 $3/2 −r/ a
# & e o .
" ao %
To determine the expectation value of r, we have to use the expression
∞ π 2π
r = ∫∫∫
*
ψ100
(r, θ, φ ) r ψ100 (r, θ, φ ) r 2 dr sinθ dθ dφ .
0 0 0
Substituting the form of the wavefunction,
r = = 1
π
! 1 $3 ∞
# & ∫
" ao % 0
π 2π
∫∫
0 0
$3 ∞ π 2 π
1!1
# & ∫
π " ao % 0
e−r/ ao r e−r/ ao r 2 dr sin θ dθ d φ ∫∫
r 3 e−2r/ ao dr sin θ dθ d φ 0 0
3
1 ! 1 $ ∞ 3 −2r/ ao
r = # & ∫ r e
dr π " ao % 0
2π
π
∫ sinθ dθ ∫
0
dφ .
0
The integrals over theta and phi are easily evaluated,
2π
∫
d φ = 2π ,
0
π
∫ sinθ dθ
= 2 .
0
Substituting, we are just left with the radial integral,
3
1 ! 1 $ ∞ 3 −2r/ ao
r = # & ∫ r e
dr π " ao % 0
∞
4
r = 3 ∫ r 3 e−2r/a0 dr .
0
ao
π
2π
∫ sinθ dθ ∫
0
0
dφ
8
4.
Continued
From integral tables, the definite integral can be evaluated using
∫
∞
0
x n e −ax dx =
n!
.
a n +1
Working out the integral,
€
r
=
24a o4
16a o3
3
=
ao .
2
=
r
€
4
3!
⋅
a o3 ( 2 / a o ) 4
9
5.
Determine the most probable location for finding an electron in a 1s orbital in the hydrogen atom. Use
only the radial portion of the wavefunction, which is given by
" Z %3 / 2 −Z r / a
o .
R10 ( r ) = 2 $ '
e
# ao &
2
2
The radial distribution function is r [ Rnℓ ( r )] . To find the most probable location, we need to find where this
€ the maximum of the radial distribution function, we take
function is a maximum. To find
2
d
r 2 [ Rnℓ ( r )] = 0 ,
dr
{
€
}
and solve for r. The radial distribution function for the 1s orbital is
€
2
r [ R10 ( r )]
" Z %3 −2Zr / a
0
= 4r $ ' e
.
# ao &
2
2
Setting Z=1 (for the hydrogen atom) and taking the derivative leads to the expression
€
2
d 2
r [ R10 ( r )]
dr
{
" 1 %3
" 1 %3 " 2 %
= 4 $ ' ( 2r ) e −2r / a 0 + 4 $ ' r 2 $− ' e −2r / a 0 .
# ao &
# ao &
# ao &
}
Collecting terms yields
€
2
d
r 2 [ R10 ( r )]
dr
{
}
=
8 #
r 2 & −2r / a 0
r
−
.
%
(e
ao '
a o3 $
We now have to set the derivative equal to zero and solve for r to get the most probable value. Thus, we have
€
8 #
r 2 & −2r / a 0
= 0.
%r −
(e
3
ao '
ao $
8 −2r / a 0
Dividing both sides of this equation by 3 e
(which is okay as long as r is not infinity) gives
ao
€
r −
€
r2
= 0.
ao
To simplify solving this equation, we can factor out r,
€
€
#
r &
r %1 −
( = 0.
ao '
$
10
5.
Continued
This equation has a root at r=0 (which is a minimum, not a maximum) and at the values of r that satisfies the
equation in parentheses,
r
= 0,
ao
or r = a o .
1−
€
6.
Show that the most probable location for finding an electron in a 2s orbital in the hydrogen atom occurs
at r = 5.236 a o . Use only the radial portion of the wavefunction, which is given by
" Z %3 / 2 "
Z r % −Z r / 2 a o
R20 ( r ) = $
.
' $2 −
' e
ao &
# 2a o & #
€
2
2
The radial distribution function
is r [ Rnℓ ( r )] . To find the most probable location, we need to find where this
€
function is a maximum. To find the maximum of the radial distribution function, we take
2
d
r 2 [ Rnℓ ( r )] = 0 ,
dr
{
€
}
and solve for r. The radial distribution function for the 2s orbital is
€
2
r [ R20 ( r )]
2
2
" Z %3 2 "
Zr % −Zr / a 0
= $
.
' r $2 −
' e
ao &
# 2a o & #
Substituting Z=1 for the hydrogen atom and expanding, we get
€
2
r [ R20 ( r )]
2
" 1 %3 " 2
4r 3
r4 %
= $
+ 2 ' e −r / a 0 .
' $ 4r −
ao
ao &
# 2a o & #
Taking the derivative leads to the expression
€
2
d
r 2 [ R20 ( r )]
dr
{
}
" 1 %3 "
" 1 %3 " 1 %" 2
12r 2
4r 3 % −r / a 0
4r 3
r4 %
= $
+
e
+
+ 2 ' e −r / a 0 .
'
' $8r −
$
' $− '$ 4r −
2
ao
ao
ao &
ao &
# 2a o & #
# 2a o & # a o &#
Collecting terms yields
€
2
d
r 2 [ R20 ( r )]
dr
{
€
}
" 1 %3 "
16r 2 8r 3 r 4 % −r / a 0
= $
+ 2 − 3'e
.
' $8r −
ao
ao ao &
# 2a o & #
11
6.
Continued
We now have to set the derivative equal to zero and solve for r to get the most probable value. Thus, we have
" 1 %3 "
16r 2
8r 3
r4 %
+
− 3 ' e −r / a 0 = 0 .
$
' $8r −
2
ao
ao
ao &
# 2a o & #
3
" 1 % −r / a
0
Dividing both sides of this equation by $
(which is okay as long as r is not infinity) leads to the
' e
€
# 2a o &
equation
€
8r −
16r 2
8r 3
r4
+
−
= 0.
ao
a o2
a o3
3
To simplify solving this equation, we can factor out r and multiply by −a o ,
€
r r 3 − 8a o r 2 + 16a o2 r − 8a o3 = 0 .
(
)
€
This equation has a root at r=0 (which is a minimum, not a maximum) and at the values of r which satisfy the
cubic equation. The roots€can be found using a programmable calculator, or by using a computer. Rather than
solving the cubic equation, to show that r = 5.236a o is a solution, we can substitute r = 5.236a o ,
3
2
− 8a o ( 5.236a o ) + 16a o2 ( 5.236a o ) − 8a o3 = 0
€ 3 − 219.326a 3 + 83.776a 3 − 8a 3 =€0
143.549a
o
o
o
o
or 0 = 0.
( 5.236a o )
Therefore, r = 5.236a o is a solution to the equation, and corresponds to the most probable value of r.
€
€