Download book problems c 5.

BOOK PROBLEMS C 5.
1. Only 1 quantum #, n =1, 2, 3, 4 … ∞ , is needed to describe a 1-dimensional standing wave.
n = # of half-wavelengths confined between 2 fixed end points, and there are n-1 nodes.
In this 1-d standing wave there are
4 half wave-lengths and 3 nodes.
n=4
1.5
Fixed
End 1
Point.
0.5
Fixed End Point
½ wave-length
0
0
90
180
270
360
450
540
630
720
-0.5
-1
node
-1.5
0
180
360
540
720
y = sin x. n = 4
y= sin2x, n = 8
y = sin (x/2), n = 2
3. Once again, only 1 quantum #, n =1, 2, 3, 4 … ∞ , is needed to describe a 1-dimensional standing
wave. There are n patterns of vibration, where n = # of half-wavelengths confined between 2 end
points, and there are n-1 nodes. In theory, there are an infinite number of possible harmonics, since n
ranges from 1 to ∞.
1.5
1
0.5
0
-0.5
-1
-1.5
0
90
180
y=sin x, n = 1
y=sin2x, n=2
y=sin4x, n=4
y=sin32x, n=32
3b. Take a look at the following. For a 1-dimensional harmonic standing wave on a string vibrating
along a fixed length L. As n, the quantum number, increases, the frequency increases.
L = n λ/2, where λ is the wavelength, and L is the fixed length.
c = λ ν , where c is the speed and ν is the frequency, so λ = c / ν
L = nc /2ν
nc/2L = ν. L and c are constant, so
ν α n that is, the higher the quantum number, the higher the frequency.
n=4
1
0.5
0
0
180
n = 4 L = 2 λ, ν = 4ν
-0.5
-1
n=3
1
0.5
0
0
180
-0.5
-1
n = 3 L = 3/2 λ. ν = 3ν
n=2
1
0.5
0
0
-0.5
180
n = 2 , λ = L, ν = 2ν
-1
n=1
1
0. 9
0. 8
0. 7
0. 6
0. 5
0. 4
0. 3
0. 2
n = 1 λ= 2L , ν = ν
0. 1
0
0
180
7. An electronic orbital is the region of space where the electron hangs out most of the time. We
don’t know exactly where it is, but its probably around there. The quantum numbers are:
1. The PRINCIPAL quantum number, n, which can take on values 1,2,3,4 … ∞ .
n gives you the energy level of the electron, and roughly how far away from the nucleus it is.
Energy α – 1/n2
2. The AZIMUTHAL quantum number, l, depends on n. l takes on values 0, 1,2, …n-1.
l establishes the shape of the orbital. l values are often given letter designations rather than
numbers.
l = 0 the electron lives in a spherical region of space around the nucleus.
As n gets larger, the diameter of the sphere increases. The l qn = 1 is usually given the letter
designation ‘s”
n = 1, l = 0
n = 2, l = 0
l = 1; the electron lives in a dumbbell-shaped region of space. This is usually given the letter
designation”p”
l = 2 ; the electron lives in a double dumbbell-shaped region of space. This is usually given
the letter designation”d”
l = 3; the electron lives in a rather complex geometrical region of space. This is usually given
the letter designation”f”
For more orbital viewing options, download the atomic orbital views from Salzman at U
Arizona. http://www.chem.arizona.edu/~salzmanr/orbitals.html Its pretty cool!
3. The MAGNETIC quantum number, ml, takes on integer values form –l to l. The magnetic
qn gives the orientation in space of the electronic orbital.
For example, if n = 2 and l = 1, ml can take the values -1, 0, +1
ml =
-1
0
+1
l
8. The Heisenberg Uncertainty Principle states that simultaneously measuring position (x, where
the electron is right now) and velocity or momentum (p, where the electron is going) is subject
to error:
ΔxΔp ≥ h, where h is Planck’s constant. If you measure the position of the e- precisely, then Δx = 0.
What Happens to the error in measurement of momentum (Δp, where the e- is going)?
ΔxΔp ≥ h,  Δp ≥ h/Δx ; if Δx = 0, then Δp = ∞ !
9. The description of a 1 electron orbital (Ψ) is a solution to the Shrödinger equation ĤΨ(x) = E
Ψ(x), for a single electron moving about a single proton. It represents one of the electron’s
allowed quantum states, its energy and angular momentum. Ψis a wave function,
characterized by a set of n, l and ml quantum numbers corresponding to a particular orbital.
If Ψ is evaluated for some point (x,y,z) in space, Ψ gives the the probability amplitude for
finding the electron there, and Ψ2 is proportional to the probability itself.
10. n = 1,2,3,4… l = 0(s), 1(p), 2(d), 3(f)… n-1.
And what are the shapes of these guys?
subshell
n
l
3p
2s
4p
1s
2p
3
2
4
1
2
1
0
1
0
1
subshell
n
l
3d
4d
3s
4s
4f
3
4
3
4
4
2
2
0
0
3
n
l
subshell
3
2
4
1
2
1
3
0
3d
2p
4f
1s
11.
12.
14.
n
l
1
0
s
2
0
s
1 p
3
0
s
1 p 2 d
4
0
s
1 p 2 d 3 f
5
0
s
1 p 2 d 3 f
4 g
20. a. 1d corresponds to n=1, l = 2, which is forbidden (l cannot be >= n)
b. 5d corresponds to n=5, l = 2, which is allowed.
c. 5f corresponds to n=5, l = 3, which is allowed.
d. 5h corresponds to n=5, l = 5, which is forbidden (l cannot be >= n)
e. 6h corresponds to n=6, l = 5, which is allowed.
f. 8d corresponds to n=8, l = 2, which is allowed.
21. Subshells are defined by n & l only. ml tells you how it is oriented in 3-space.
n
l
subshell
3
2
4
1
2
1
3
0
3d
2p
4f
1s
n
l
subshell
2
6
3
3
0
5
1
0
2s
6h
3p
3s
22.
26. The radius of an orbital increases as n increases. The probability is greater that the electron
will be found farther from the nucleus (the atom “expands”).
27. For a 1 electron atom (and that’s the best we can do so far), the orbital energues are given by:
En = -R∞ Z2/n2, where
n is the principal quantum number
Z is the atomic number (nuclear charge)
R∞ is the Rydberg constant, 2.1798741 x 10-18 J.
a. Plugging in: E2 – E1 = -R∞ Z2/n22 - -R∞ Z2/n12 .
n
1
2
R (J)
2.1798741E-18
2.1798741E-18
Z
n1 is the ground state.
En(J)

1
-2.17987E-18 1.63491E-18 J
1
-5.44969E-19
b. The energy of a photon of light is given by the Planck-Einstein equation, which we know and
love from chapter 4:
E = hν = hc/λ
h = 6.6260755 x 10-34 J sec
c = 2.99792458 x 108 m/sec.
E = 1.63491 x 10-18 J
λ = hc/E = (6.6260755 x 10-34 J sec)( 2.99792458 x 108 m) = 1.2150 x 10-7 m = 122 nm. This is UV light.
(1.63491 x 10-18 J)
(
sec
)
c. Ek = ½ mv2.
The mass of a He atom is: 4.0026 g/mole(1 mole/6.022 x 1023 particles)
= 6.647 x 10-24 g = 6.647 x 10-27 kg
Ek = ½ (6.647 x 10-27 kg)[100 m/sec]2 = 3.323 x 10-21 kg m2/sec2 (J). Not enough energy.
28. Once again using the Rydberg equation:
For an H electron jumping from the 4th to the 5th level:
n
4
5
R (J)
2.1798741E-18
2.1798741E-18
Z
En(J)

1
-1.36242E-19 4.90472E-20 J
1
-8.7195E-20
An He+ electron making the same jump has to pull away from a +2 nuclear charge. You’d predict it
would take more energy:
n
R (J)
Z
En(J)

2.1798741E-18
2
-5.44969E-19 1.96189E-19 J
2.1798741E-18
5
2
-3.4878E-19
And it does, by a factor of 1.96 x 10-19J/4.905 x 10-20J = 4/1
4
29. We know that energy is related to frequency…the Planck Einstein equation gives us the
relationship:
E = hν
The available energy we have for goosing the electron is:
Egoose = 6.626 x 10-34 J sec )(6.169 x 1014/sec) = 4.088 x 10-19 J.
We can plug this value into the Rydberg eqn:
ΔE = Efinal – Einitial = -R∞(1/n2final – 1/n2initial)
4.088 x 10-19 J = -R∞(1/n2final – 1/n2initial)
4.088 x 10-19J/-R∞ = 4.088 x 10-19J/2.1799 x 10-18 J = 0.1875 = 1/n2f – 1/n2i
By trial and error, if nf = 4 and ni = 2, it works.
30. Use the Rydberg equation to calculate ΔE for the transition, and then plug ΔE into the PlanckEinstein equation for the frequency.
a. 1s 2p is an n1  n2 jump: This takes energy (absorption)
n
1
2
R (J)
2.1798741E-18
2.1798741E-18
Z
En(J)

h (J sec)
frequency =E/h
2.47E+15
1
-2.17987E-18 1.63491E-18 J
6.63E-34
1
-5.44969E-19
6.63E-34
sec-1
b. 5f to 4d will be a smaller jump (energy levels get scrunched together as n increases). It will
emit energy, since the electron is falling toward the nucleus.
n
5
4
R (J)
2.1798741E-18
2.1798741E-18
Z
En(J)

h (J sec)
frequency =E/h
-7.40E+13
1
-8.7195E-20 -4.90472E-20 J
6.63E-34
1
-1.36242E-19
6.63E-34
sec-1
Indeed, is much smaller, and the frequency is lower.
c. 3d  ∞ a long jump…the electron leaves the atom, leaving a cation behind.
Energy will be absorbed. I can’t plug infinity into a spreadsheet, so I used a really big number (10100)
This is an IONIZATION ENERGY, the amount of energy it takes to remove an e- completely.
n
R (J)
2.1798741E-18
1.00E+100 2.1798741E-18
3
Z
En(J)

h (J sec)
frequency =E/h
3.66E+14
1
-2.42208E-19 2.42208E-19 J
6.63E-34
1
-2.1799E-218
6.63E-34
sec-1
30 (continued)
d. ∞  1s …the electron jumps into the atom, making an anion and emitting energy. This is the
electron affinity.
n
R (J)
2.1798741E-18
1.00E+100 2.1798741E-18
1
Z
En(J)

h (J sec)
frequency =E/h
3.29E+15
1
-2.17987E-18 2.17987E-18 J
6.63E-34
1
-2.1799E-218
6.63E-34
sec-1
For a lark, check out the Molecule of the month – this time, garlic! Keep the vampires away.
http://www.chem.ox.ac.uk/mom/allicin/allicin.html