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Chemistry 362 Dr. Jean M. Standard
Problem Set 3 Solutions 1. Verify for the particle in a one-dimensional box by explicit integration that the wavefunction
ψ2 x( ) = 2asin 2π x
a⎛
⎝⎜
⎞
⎠⎟ is normalized.
To verify that
€
ψ2 x( ) is normalized, we have to show that
ψ22(x) dx
0
a
∫ = 1 .
Note that the integration range is from x=0 to x=a since the particle in a box wavefunctions vanish outside that range. Substituting, the integral becomes
ψ22(x) dx
0
a
∫ = 2a
sin 2π xa
⎛
⎝⎜
⎞
⎠⎟⋅
2a
sin 2π xa
⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫
= 2a sin2 2π x
a⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫ .
The integral that we need can be found, for example, in the CRC or in most any "useful" table of integrals:
€
sin2 bx dx∫ = x2−
sin2bx4b
.
Setting b = 2πa
and substituting, the normalization integral becomes
ψ22(x)dx
0
a
∫ = 2L
sin2 2π xa
⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫
= 2a
x2−
1
4 2πa
⎛
⎝⎜
⎞
⎠⎟
sin 4π xa
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
0
a
.
2
1. continued
Evaluating the expression at the limits,
ψ22(x) dx
0
a
∫ = 2a
x2−
1
4 2πa
⎛
⎝⎜
⎞
⎠⎟
sin 4π xa
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
0
a
= 2a
a2−
a8π
sin 4π( )⎡
⎣⎢⎤
⎦⎥ − 0 −
a8π
sin 0( )⎡
⎣⎢⎤
⎦⎥
⎧⎨⎩
⎫⎬⎭
= 2a
a2− 0
⎡
⎣⎢⎤
⎦⎥ − 0 − 0[ ]
⎧⎨⎩
⎫⎬⎭
ψ22(x) dx
0
a
∫ = 1 .
Therefore, the wavefunction
€
ψ2 x( ) is normalized.
3
2. Calculate the energy level spacings in Joules between the ground (n=1) and first excited (n=2) levels for
the following cases.
a.) An electron confined to a one-dimensional box of width 5 Å. The energy level spacing
€
ΔE between the ground and first excited state is
€
ΔE = E2 − E1 . Using the particle in a box energy expression, we can obtain an equation for the energy level spacing,
ΔE = E2 −E1 = 22h2
8ma2 − 12h2
8ma2
ΔE = 3h2
8ma2 .
Substituting,
€
ΔE = E2 − E1 = 3 6.62618×10−34 Js( )2
8 9.10953×10−31 kg( ) 5×10−10 m( )2
ΔE = E2 − E1 = 7.230×10−19 J .
b.) A baseball with a mass of 140 g confined to a one-dimensional box of width 1 meter.
Using the same expression from part (a) for the energy level spacing, we have
ΔE = E2 −E1 = 22h2
8ma2 − 12h2
8ma2
ΔE = 3h2
8ma2 .
Substituting,
€
ΔE = E2 − E1 = 3 6.62618×10−34 Js( )2
8 0.1 kg( ) 1 m( )2
ΔE = E2 − E1 = 1.646×10−66 J . Notice how much smaller the energy level spacing is for the macroscopic object. Energy level spacings this small are not measurable, and therefore the energy level spacings are so small as to be effectively continuous for macroscopic objects.
4
3. An electron in a box of width a undergoes a transition from the lowest energy level (n=1) to the first
excited level (n=2). The wavelength of light absorbed in this transition was determined to be 650 nm. Calculate the width of the box. For a transition from n=1 to n=2, the energy difference
€
ΔE is
€
ΔE = E2 − E1 . A photon with an energy corresponding to
€
ΔE would have a frequency given by
€
Ephoton = ΔE = hν . Since, for light,
€
λν = c , we can
substitute
€
ν =λc , and obtain an expression for the energy difference,
€
ΔE = E2 − E1 = hcλ
.
Substituting,
€
ΔE = E2 − E1 = 6.62618×10−34 Js( ) 2.99793×108 ms−1( )
680 nm( ) 10-9 m1 nm
%
& '
(
) *
ΔE = E2 − E1 = 2.9213×10−19 J . Then, we can use the particle in a box energies to obtain an expression for the energy difference,
ΔE = E2 −E1 = 22h2
8ma2 − 12h2
8ma2
ΔE = 3h2
8ma2 .
Solving for a, the width of the box, yields
a = 3h2
8mΔE
⎛
⎝⎜⎜
⎞
⎠⎟⎟
1/2
.
Substituting,
a = 3h2
8mΔE
⎛
⎝⎜⎜
⎞
⎠⎟⎟
1/2
= 3 6.62618×10−34 Js( )
2
8 9.10953×10−31 kg( ) 2.9213×10−19 J( )
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
1/2
a = 7.866×10−10 m or 7.866 Å .
5
4. Determine the most probable location for the particle in the ground state of the one-dimensional particle
in a box. Also determine the most probable location for the first excited state of the particle in a box.
To obtain the most probable value, we must look at the probability density,
€
ψ 2 x( ) . For the ground state of the particle in a box, the probability density is given by
ψ12 x( ) = 2
asin2 π x
a⎛
⎝⎜
⎞
⎠⎟ .
A plot of this probability density is shown below.
Note that the maximum of this function occurs at x = a2
; this is the most probable value. In this case, the
maximum can be found by inspection. For more general situations to get the maximum value, we would have to take the derivative, set it equal to zero, and solve.
For the first excited state of the particle in a box, the probability density is given by
ψ22(x) = 2
asin2 2π x
a⎛
⎝⎜
⎞
⎠⎟ .
A plot of this probability density is given below.
For this function, we see from the plot that there are two maxima, each with identical amplitude. By inspection,
these maxima occur at x = a4
and x = 3a4
; these two values are therefore the most probable values.
6
5. For a particle in a one-dimensional box of width a, determine the probability of finding the particle in the
right third of the box (between 2/3 a and a) if the particle is in the ground state.
Since the probability is given by
€
ψ 2 x( ) dx , if we want the total probability of finding the particle between 2/3 a and a, we must add up the probability for all the points from 2/3 a to a. Since x is a continuous variable, the sum is really an integral from 2/3 a to a,
Probability = ψ2 x( ) dx .2a/3
a
∫
Substituting the particle in a box wavefunction for the ground state, ψ1(x) = 2a sin π x
a⎛
⎝⎜
⎞
⎠⎟ , the probability
integral becomes
Probability = 2a
sin2 π xa
⎛
⎝⎜
⎞
⎠⎟ dx
2a/3
a
∫ .
This integral can be evaluated using tables. From the CRC Handbook or any other table of integrals, we find the indefinite integral:
€
sin2 bx dx∫ = x2
− sin 2bx4b
.
Replacing b with πa
yields
Probability = 2a x
2 −
sin 2π xa
⎛
⎝⎜
⎞
⎠⎟
4 πa⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
2a/3
a
.
Finally, evaluating the expression at the limits leads to
Probability = 13 + 1
2πsin 4π
3⎛
⎝⎜
⎞
⎠⎟ .
Evaluating the numerical value of the sine function in the expression above, the probability is
€
Probability = 0.3333 − 0.1378= 0.1955 .
7
6. The 1,3,5-hexatriene molecule is a conjugated molecule with 6 pi electrons. Consider the pi electrons free
to move back and forth along the molecule through the delocalized pi system. Using the particle in a box approximation, treat the carbon chain as a linear one-dimensional "box". Allow each energy level in the box to hold 2 pi electrons. Treating 1,3,5-hexatriene as a linear chain, we would have the structure
The pi electrons are free to move through the conjugated system; hence, the length of the molecule would represent the "box." We can add up the lengths of the single and double bonds in the molecule to give the width of the box, a. For 1,3,5-hexatriene, this gives
a = 3rC=C + 2rC−C , where r represents the bond length. The energy of the pi electrons in the molecule would be represented as
En = n2h2
8ma2 , n =1, 2, 3, …,
where a is the box width defined above and m is the electron mass. a.) Calculate the energy of the highest filled level, using 1.54 Å as the carbon-carbon single bond length,
and 1.35 Å as the carbon-carbon double bond length. Using
€
rC =C = 1.35 Å and
€
rC −C = 1.54 Å, the width of the box can be calculated from the equation above.
a = 3(1.35 Å) + 2(1.54 Å) = 7.13 Å (or 7.13×10-10 m). Using the formula above with a = 7.13×10–10 m for the energy leads to the energy diagram shown below.
C1 C2 C3 C4 C5 C6
n=1
n=2
n=3
n=4
n=5
8
6 a.) continued 1,3,5-hexatriene has 6 pi electrons. Placing two electrons in each energy level fills the levels up to the n=3 level, as shown in the figure above. So, n=3 is the highest filled level. Computing the energy of the highest filled level,
E3 = 32h2
8ma2 = 32 6.62618×10−34 Js( )
2
8 9.10953×10−31 kg( ) 7.13×10−10 m( )2
= 32 1.18511×10−19 J( )E3 = 1.0666×10−18 J .
b.) Determine the energy of the lowest unfilled level. The lowest unfilled level corresponds to n=4. Calculating the energy of this state gives
€
E4 = 42h2
8ma2
E4 = 42 1.18511×10−19 J( )E4 = 1.8962×10−18 J
c.) Calculate the wavelength for an electronic transition from the highest filled level to the lowest unfilled level, using your answers from parts a) and b). Compare your result to the experimental ultraviolet absorption maximum of 268 nm. For an electronic transition from n=3 to n=4, the energy difference
€
ΔE is
€
ΔE = E4 − E3 . A photon with an energy corresponding to
€
ΔE would have a frequency given by
€
Ephoton = ΔE = hν . Since, for light,
€
λν = c
, we can substitute
€
ν =λc , and solve for the wavelength to give
€
λ = hcΔE
= hcE4 − E3
.
Inserting numerical values leads to
€
λ = 6.62618×10−34 Js( ) 2.99793×108 ms−1( )
1.8962×10−18 J − 1.0666×10−18 J( )
€
λ = 2.395×10−7 m or
€
λ = 239.5nm. This result is fairly close to the experimental wavelength,
€
λexp = 268nm with about an 11% error. This is surprisingly good agreement given the crudeness of the model.
9
7. Verify, by explicit integration, that
€
ψ2 and
€
ψ3 for the particle in a one-dimensional box are orthogonal.
For a particle in a 1D box, the wavefunctions are given by
ψn (x) = 2a
sin nπ xa
⎛
⎝⎜
⎞
⎠⎟ .
To show that
€
ψ2 and
€
ψ3 are orthogonal, the integral of the product of the two functions (including the complex conjugate of one of the functions if they are not real functions) must be zero. Evaluating the integral (in this case, both functions are real), we have
ψ2*(x)ψ3(x) dx
0
a
∫ = ψ2 (x) ψ3(x) dx0
a
∫ = 2a
sin 2π xL
⎛
⎝⎜
⎞
⎠⎟ sin 3π x
L⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫ .
Note that the integration range is from x=0 to x=a since the particle in a box wavefunctions vanish outside that range. The indefinite integral that we need can be found, for example, in the CRC or in most any table of integrals:
€
sinmx sinnx dx∫ = sin m − n( )x
2 m − n( ) −
sin m + n( )x2 m + n( )
.
Setting m =2πa
and n = 3πa
, we have m− n = −πa
and m+ n = 5πa
. Substituting,
ψ2 (x) ψ3(x) dx0
a
∫ = 2a − a
2πsin −
π xL
⎛
⎝⎜
⎞
⎠⎟ −
a10π
sin 5π xL
⎛
⎝⎜
⎞
⎠⎟
⎡
⎣⎢
⎤
⎦⎥
0
a
.
Using the identity
€
sin −x( ) = sin x , and evaluating the integral at the limits yields
ψ2 (x) ψ3(x) dx0
a
∫ = 2a a
2πsinπ − a
10πsin5π
⎡
⎣⎢⎤
⎦⎥ − 2
a a
2πsin0 − a
10πsin0
⎡
⎣⎢⎤
⎦⎥ .
Since
€
sinπ = sin 5π = sin 0 = 0 , the integral simplifies to
ψ2(x) ψ3(x) dx = 00
a
∫ .
Therefore, the functions
€
ψ2 x( ) and
€
ψ3 x( ) are orthogonal.
10
8. Determine the average value of the position x for the ground state of the one-dimensional particle in a
box. Compare your result with the most probable location. The average value of the position is given by
x = ψ1*(x) x̂ ψ1(x) dx
0
a
∫ = ψ1*(x) x ψ1(x) dx
0
a
∫ ,
where the definition of the position operator,
€
ˆ x = x , has been used. In addition, the limits of the integral are x=0 to x=a because the wavefunction vanishes outside this range. The ground state wavefunction for the particle in a box is given by
ψ1(x) = 2a sin π x
a⎛
⎝⎜
⎞
⎠⎟ .
Upon substitution, the expression for the average value of the position becomes
x = ψ1*(x) x ψ1(x) dx
0
a
∫
= 2a sin π x
a⎛
⎝⎜
⎞
⎠⎟ ⋅ x ⋅ 2
a sin π x
a⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫
= 2a x sin2 π x
a⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫ .
From the CRC Handbook (or handout of integrals),
xsin2 bx dx = x2
4 ∫ − xsin2bx
4b − cos2bx
8b2 .
Replacing b by πa
, the average value becomes
x = 2a
x2
4 −
xsin 2π xa
⎛
⎝⎜
⎞
⎠⎟
4 πa⎛
⎝⎜
⎞
⎠⎟
− cos 2π x
a⎛
⎝⎜
⎞
⎠⎟
8 πa⎛
⎝⎜
⎞
⎠⎟
2
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
0
a
.
11
8. continued
Evaluating the expression at the limits yields the average value of the position for the ground state of the particle in a box:
x = 2a
a2
4 − a
2
4πsin 2π( ) − a
2
8π 2 cos 2π( ) ⎡
⎣⎢
⎤
⎦⎥ − 2
a 0 − 0 ⋅sin 0( ) − a
2
8π 2 cos 0( ) ⎡
⎣⎢
⎤
⎦⎥
= 2a
a2
4 − 0 − a
2
8π 2 ⎡
⎣⎢
⎤
⎦⎥ − 2
a 0 − 0 − a
2
8π 2 ⎡
⎣⎢
⎤
⎦⎥
x = a2
.
For the ground state of the particle in a box, the average value and most probable value of x are identical (the most probable value of a/2 was found in problem 4). This will not always be the case, as we will see in next problem.
12
9. Repeat problem 8 for the first excited state of the particle in a box. Does your result agree this time with
the most probable location for this state?
For the first excited state of the particle in a box, the wavefunction is
ψ2(x) = 2a sin 2π x
a⎛
⎝⎜
⎞
⎠⎟ .
The average value of x for this state is
x = ψ2*(x) x ψ2(x) dx
0
a
∫
= 2a sin 2π x
a⎛
⎝⎜
⎞
⎠⎟ ⋅ x ⋅ 2
a sin 2π x
a⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫
x = 2a xsin2 2π x
a⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫ .
We can use the same integral as we did in the previous problem; in this case, we replace b by 2πa
:
x = 2a
x2
4 −
xsin 4π xa
⎛
⎝⎜
⎞
⎠⎟
4 2πa
⎛
⎝⎜
⎞
⎠⎟
− cos 4π x
a⎛
⎝⎜
⎞
⎠⎟
8 2πa
⎛
⎝⎜
⎞
⎠⎟
2
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
0
a
.
Evaluating this expression at the limits gives the average value of the position for the first excited state,
x = 2a
a2
4 − a
2
8πsin 4π( ) − a2
32π 2 cos 4π( ) ⎡
⎣⎢
⎤
⎦⎥ − 2
a 0 − 0 ⋅sin 0( ) − a2
32π 2 cos 0( ) ⎡
⎣⎢
⎤
⎦⎥
= 2a
a2
4 − 0 − a2
32π 2 ⎡
⎣⎢
⎤
⎦⎥ − 2
a 0 − 0 − a2
32π 2 ⎡
⎣⎢
⎤
⎦⎥
x = a2 .
This is exactly the same result that we got for the ground state. This will not always happen – here we got the same result because of the symmetry of the potential energy and as a result the symmetry of the wavefunction about x=a/2.
From problem 4, we found that the most probable values for the first excited state occur at x = a4
and x = 3a4
.
Note that for the first excited state of the particle in a box, the average value and most probable values of x are not the same.
13
10. Determine the average value of the momentum
€
px for the ground state of the one-dimensional particle in a box.
The average value of the momentum is given by
px = ψ1*(x) p̂x ψ1(x) dx
0
a
∫ = − i! ψ1*(x) d
dx ψ1(x) dx
0
a
∫ ,
where the definition of the momentum operator,
€
ˆ p x = −i! ddx
, has been used. In addition, the limits of the
integral again are x=0 to x=a because the wavefunction vanishes outside this range. The ground state wavefunction for the particle in a box is given by
ψ1(x) = 2a sin π x
a⎛
⎝⎜
⎞
⎠⎟ .
Upon substitution, the expression for the average value of the momentum becomes
px = − i! ψ1*(x) d
dxψ1(x) dx
0
a
∫
= −i! 2a
sin π xa
⎛
⎝⎜
⎞
⎠⎟ ⋅
ddx
2a
sin π xa
⎛
⎝⎜
⎞
⎠⎟
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪ dx
0
a
∫
px = − 2i!a
sin π xa
⎛
⎝⎜
⎞
⎠⎟ ⋅
ddx
sin π xa
⎛
⎝⎜
⎞
⎠⎟
⎧⎨⎩
⎫⎬⎭
dx0
a
∫ .
Next, the derivative must be evaluated,
ddx sin π x
a⎛
⎝⎜
⎞
⎠⎟
⎧⎨⎩
⎫⎬⎭ = π
a cos π x
a⎛
⎝⎜
⎞
⎠⎟ .
Substituting, the average value of the momentum is
px = − 2i!πa2 sin π x
a⎛
⎝⎜
⎞
⎠⎟ ⋅ cos π x
a⎛
⎝⎜
⎞
⎠⎟ dx
0
a
∫ .
From the CRC Handbook (or handout of integrals),
€
sin∫ bx cosbx dx = sin2bx2b
.
14
10. continued
Replacing b by πa
, the average value of momentum becomes
px = − 2i!πa2
sin2 π xa
⎛
⎝⎜
⎞
⎠⎟
2 πa⎛
⎝⎜
⎞
⎠⎟
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
0
a
.
Evaluating this expression at the limits gives the average value of momentum for the ground state,
px = − 2i!πa2 a
2πsin2 π( )
⎡
⎣⎢⎤
⎦⎥ + 2i!π
a2 a2π
sin2 0( ) ⎡
⎣⎢⎤
⎦⎥
= − 2i!πa2 0[ ] + 2i!π
a2 0[ ]
px = 0 .
For the ground state of the particle in a box, the average value of momentum is 0. This is actually true for any state of the particle in a box, and reflects the idea that motion in the positive x-direction, corresponding to positive values of momentum, and motion in the negative x-direction, corresponding to negative values of momentum, are equally probable. Thus, the positive and negative values of momentum cancel, yielding an average value of 0.