Download Physical Chemistry II – Exam 1 SOLUTIONS

Chemistry 362
Spring 2013
Dr. Jean M. Standard
February 8, 2013
Name _____KEY_____________________
Physical Chemistry II – Exam 1 SOLUTIONS
Constant
Value in SI Units
h
! = h /2π
c
e
me
ε0
R (Rydberg const.)
€
6.62607×10–34 J s
1.05457×10–34 J s
2.99793×108 m/s
1.60217×10–19 C
9.10938×10–31 kg
8.85419×10–12 C2J–1m–1
109737.3 cm–1
1.) (17 points) When light with a wavelength of 3000 Å shines on a metal surface, the
kinetic energy of ejected electrons is 2.960×10–19 J. When light with a wavelength of
5400 Å shines on the same metal surface, the kinetic energy of ejected electrons is
1.362×10–21 J. From this information, determine the value of Planck's constant (in
units of J s). [1 Å = 10–10 m]
For the photoelectric effect, KE = hν − φ . From this equation, we can see that a plot with
kinetic energy on the y-axis and frequency on the x-axis yields a straight line with the slope
equal to Planck's constant h. Therefore, one way to determine Planck's constant from the
data given is to calculate the slope,
€
h = slope =
ΔKE
.
Δν
The kinetic energy values are given in the problem, but the frequencies must be calculated
from the wavelengths.
€
c
c
ν1 =
ν2 =
λ1
λ2
=
2.99793 × 10 8 m/s
& 10−10 m )
(3000 Å)( 1 Å +
'
*
=
ν1 = 9.993 × 1014 s–1 .
€
2.99793 × 10 8 m/s
& 10−10 m )
(5400 Å)( 1 Å +
'
*
ν 2 = 5.552 × 1014 s–1 .
€
2
1.) continued
These frequency values may now be substituted, along with values of the kinetic energy,
into the equation for the slope,
ΔKE
Δν
1.362 × 10−21 J – 2.960 × 10−19 J
=
5.552 × 1014 s–1 – 9.993 × 1014 s–1
h = 6.634 × 10−34 J s .
h =
€
3
−ax
2.) (17 points) Consider two operators, Aˆ = d /dx and Bˆ = x . For the function f (x) = x e
where a is a constant, evaluate the following quantities.
a.) Aˆ Bˆ f ( x )
€
€
d
Aˆ Bˆ f ( x ) =
x x e−ax
dx
d
=
x 2 e−ax
dx
= 2x e−ax − a x 2 e−ax
Aˆ Bˆ f ( x ) = (2 − ax ) x e−ax .
(
(
)
)
€
b.) Bˆ Aˆ f ( x )
d
Bˆ Aˆ f ( x ) = x
x e−ax
dx
(
€
)
(
= x e−ax − a x e−ax
)
Bˆ Aˆ f ( x ) = (1 − ax ) x e−ax .
€
c.) Do the operators Aˆ and Bˆ commute?
(a yes or no answer is sufficient)
NO. The operators Aˆ and Bˆ do not commute in this case.
€
€
€
€
,
4
3.) (17 points) An electron in a one-dimensional box of width 6.0 Å undergoes a transition
from an initial state with n=2 to some final state. The wavelength of light absorbed in
this transition was determined to be 56.5 nm. Calculate the quantum number of the
final state. [1 Å = 10–10 m; 1 nm = 10–9 m]
The particle in a box energies are given by
n 2h 2
.
8mL2
En =
For a transition from n=2 to some final state n, the energy difference ΔE is
€
ΔE = E n − E 2
n 2h 2
2 2 h 2€
−
8mL2
8mL2
h2
2
2
ΔE = n − 2
.
8mL2
=
(
)
Solving for the unknown quantum number n yields,
€
8mL2 ΔE
n −2 =
h2
8mL2 ΔE
n2 = 4 +
.
h2
2
2
A photon with an energy corresponding to ΔE would have a frequency given by
λ
E photon = ΔE = hν . Since,
€ for light, λν = c , we can substitute ν = , and calculate the
c
energy difference ΔE ,
€
€
hc
ΔE =
λ
€
€
€
(6.62607 × 10
−34
=
)
−9
56.5 × 10 m
ΔE = 3.5158 × 10−18 J .
€
)(
J s 2.99793 × 10 8 m/s
5
3.) continued
Substituting this energy difference ΔE into the expression above and solving,
8mL2 ΔE
n = 4 +
€ h2
2
= 4 +
(
)(
(6.62607 × 10
−34
2
) (3.5158 × 10 J)
J s)
8 9.10938 × 10−31 kg 6.0 × 10−10 m
2
= 4 + 21
n 2 = 25 ,
or n = 5 .
Therefore, the final state in the transition is to the quantum number n=5.
€
−18
6
An acceptable wavefunction must satisfy the following conditions:
(1) It must be continuous.
(2) It must be single-valued.
(3) It must go to 0 as x →±∞ .
A sketch of the wavefunction shows that it is continuous and single-valued. (Note that
the sketch given
€ is for L=1, but the same functional shape would be observed for any
value of L.)
0.3
0.2
ψ(x)
€
$ x'
4.) (17 points) Consider the function ψ (x) = x &1 − ) on the interval 0 ≤ x ≤ L , where L is
% L(
a constant. The function is equal to zero outside this region, ψ (x) = 0 for x < 0 and
x > L.
€
€
a.) Other than normalization, what are the conditions
wavefunction?
€
€ for an acceptable
Does the wavefunction given above meet these conditions? Explain.
0.1
0.0
0.0
0.2
0.4
0.6
0.8
1.0
x
Since the function equals 0 for for x < 0 and x > L , it also satisfies condition (3), that it
goes to 0 as x →±∞ .
€
The main concern for condition
(1)€is that the function be continuous at the boundaries
€
(x=0
€ and x=L), so that it matches the wavefunction outside this region ψ (x) = 0 for x < 0
and x > L . From the formula for the function given above in the region 0 ≤ x ≤ L , we
see that it equals 0 at the boundaries,
€
€
% 0(
€
ψ (0) = 0⋅ '1 − * = 0
& L)
% L(
ψ (L) = L⋅ '1 − * = 0 .
& L)
Therefore, the function is continuous.
€
7
4.) continued
b.) Verify whether or not the wavefunction given above on the interval 0 ≤ x ≤ L is an
eigenfunction of the one-dimensional kinetic energy operator, Tˆ . If it is an
eigenfunction, give the eigenvalue.
€
The one-dimensional kinetic energy operator is
!2 d 2
ˆ
T = −
.
2m dx 2
$ x'
To verify whether or not ψ€(x) = x &1 − ) is an eigenfunction of the kinetic energy
% L(
ˆ
operator, we operate T on it,
€
!2 d 2 * $
x 'ˆ
+
Tψ ( x ) = −
x
1
−
&
).
2m dx 2 , %
L (/
!2 d 2 $
x2 '
= −
x
−
&
)
2m dx 2 %
L(
= −
!2 d $
2x '
&1 −
)
2m dx %
L(
= −
!2 d $
2'
&0 − )
2m dx %
L(
!2
Tˆψ ( x ) =
.
mL
This result does not correspond to a constant times the original function; therefore, this
function is NOT an eigenfunction of the kinetic energy operator.
€
8
5.) (16 points) The idea of quantization is one of the key elements of quantum theory.
Explain the concept of quantization. Then, using one example system, describe at least
one property of the system that exhibits quantization. Make sure to include in your
discussion an equation that illustrates quantization of the property of interest.
Quantization is the idea that a physical property may take only certain allowed (discrete)
values. Such properties would be predicted by classical theories to be continuous; that is, the
property could take any value.
A variety of systems could be used to demonstrate quantization of a physical property. One
example that we discussed in class was the Bohr theory of the hydrogen atom. The property
that was quantized was the angular momentum, L = mvr = n! , where L is the angular
momentum, m is the mass of the electron, v is the electron velocity, ! is a constant, and n is
an integer (the quantum number). This result also led to quantization of the radius of the
orbit and the electron energy.
€
€
The quantization of the energy of the electron also was demonstrated by experiments
involving atomic line spectra. In that case, quantized energy levels in the atoms led to only
specific wavelengths of light observed in the spectrum. For hydrogen, the observed
$1
1
1'
= R & 2 − 2 ) , where λ is the wavelength, R
wavelengths are related to the formula
λ
n2 (
% n1
is a constant, and n1 and n 2 are integers which label the quantized energy levels of the
transition.
€
€
Also in €
the case€of the atomic line spectrum of hydrogen, the quantized property was the
m e4
, where m is the electron mass, e
8 ε 02 n 2 h 2
is the electron charge, ε 0 is the permittivity of free space, h is a constant, and n is an integer
that labels the quantized energy levels.
energy, which was found to be of the form E n = −
€
€ we discussed that exhibits quantization is the particle in a box. In this case,
Another system
n 2h 2
the energy of the particle is quantized and given by E n =
, where m is the mass, L is
8mL2
the box width, h is a constant, and n is an integer that labels the quantized energy levels.
€
9
6.) (16 points)
a.) For the particle in a one-dimensional box of width L, make a sketch of the second
excited state (n=3) wavefunction and the corresponding probability density. How
many nodes does this wavefunction possess? Where are they located?
2
Graphs of the second excited state wavefunction, ψ 3 ( x ) , and probability density, ψ 3 ( x ) ,
are shown below. (Your sketches should capture the key features of these functions, and
in addition should be properly labeled.)
€
2.0
€
2.0
1.5
ψ32(x)
ψ3(x)
1.0
0.0
0.0
0.2
0.4
0.6
0.8
1.0
-1.0
1.0
0.5
0.0
-2.0
0.0
x/L
0.2
0.4
0.6
0.8
1.0
x/L
From the graphs we see that the second excited state wavefunction possesses two nodes,
at x=L/3 and x =2L/3.
b.) What is the most probable position for finding the particle in the second excited
state?
The most probable position (or positions) corresponds to the location of the maximum in
the probability density. From the graph of probability density for the second excited
2
state, ψ 3 ( x ) , given above, there are three equal maxima and therefore three equivalent
most probable locations: x = L/6, L/2, and 5L/6.
€
c.) Determine by explicit calculation the probability of finding the particle in the left
one third of the box if the particle is in the second excited state.
2
Since the probability is given by ψ ( x ) dx , if we want the probability of finding the
particle in the left one third of the box (between x=0 and L/3), we must evaluate the
integral,
€
L/3
Probability =
∫ ψ * ( x ) ψ ( x ) dx .
0
€
10
6 c.) continued
Substituting the particle in a box wavefunction for the second excited state (n=3),
$ 3π x '
2
ψ 3 (x) =
sin&
) , the probability integral becomes
% L (
L
2
Probability =
L
€
L/3
# 3π x &
( dx .
L '
∫ sin2%$
0
This integral can be evaluated using the list provided. From the list of integrals given, we
€ integral:
find the indefinite
∫ sin2 b x dx
Replacing b with
€
=
x
sin2b x
−
.
2
4b
3π
yields
L
€
x=L / 3
*
$ 6π x ' sin&
)/
2 ,x
% L (/
,
Probability =
−
.
$ 3π '
L ,2
4& ) /
,+
% L ( /. x=0
Finally, evaluating the expression at the limits leads to
€
' $
'2 *$ L
L
L
Probability =
sin(2π )) − &0 −
sin(0))/
,& −
( %
(.
L +% 6
12π
12π
2 *L
=
− 0 − 0 + 0/
,
.
L +6
1
Probability =
.
3
[Note: This result for the probability makes sense because the probability density for the
€
second excited state is divided into three equal humps; therefore, the probability in the
left third of the box has to be 1/3.]