PROBLEM SET 1 SOLUTIONS
DUE: October 3, 2022
CHEM 20AH
1. The lift force – 𝑳𝒍𝒊𝒇𝒕 – acting upward on an airplane depends on:
its speed – 𝒗; the area of its wings – 𝑨; and the density of air – 𝝆.
(Does it make sense to you that these are the quantities on which the lift force depends?)
Use dimensional analysis to figure out the powers in the equation 𝑳 ≅ 𝒗𝒙 𝑨𝒚 𝝆𝒛 .
Do you expect all of these powers to be positive?
For a big or small plane (e.g., a commercial passenger plane or a recreational small plane),
look up a typical wing area and a typical weight for a plane and estimate how fast it has to move
on the runway in order to lift off the ground.
First, determine dimensions of the relevant physical quantities:
(NOTE: below, […] ↔ dimensions of …)
[𝑳𝒍𝒊𝒇𝒕 ] = [𝒇𝒐𝒓𝒄𝒆] = 𝑴𝟏 𝑳𝟏 𝑻−𝟐 , [𝒗] = [𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚] = 𝑳𝟏 𝑻−𝟏 , [𝑨] = [𝒂𝒓𝒆𝒂] = 𝑳𝟐 ,
𝒂𝒏𝒅 [𝝆] = [𝒎𝒂𝒔𝒔 𝒅𝒆𝒏𝒔𝒊𝒕𝒚] = 𝑴𝟏 𝑳−𝟑 .
Then set up the equation guaranteeing that both sides of the equation/relation we’re trying to
derive have the same dimensions (here “≅" denotes “equals, apart from a multiplicative
numerical constant of order unity”):
𝑳𝒍𝒊𝒇𝒕 ≅ 𝒗𝒙 𝑨𝒚 𝝆𝒛,
𝑴𝟏 𝑳𝟏 𝑻−𝟐 = (𝑳𝟏 𝑻−𝟏 )𝒙 (𝑳𝟐 )𝒚 (𝑴𝟏 𝑳−𝟑 )𝒛 .
Equating left-side and right-side powers of M, L, and T, we have:
𝑴: 𝟏 = 𝒛
𝑳: 𝟏 = 𝒙 + 𝟐𝒚 − 𝟑𝒛
𝑻: − 𝟐 = −𝒙
Solving for x, y, and z, we have 𝑳𝒍𝒊𝒇𝒕 ≅ 𝒗𝟐 𝑨𝟏 𝝆𝟏 .
We expected all of the powers to be positive because of our expectation that the lift force will
increase with increasing speed, wing area, and air mass density.
Now, using approximate values for wing area (~10m^2) and weight (~ 2 tons, corresponding
to a gravitational force of 2 x 10^4 N -- why? how?) for a single-person small plane, and the
density of air (~1kg/m^3), I find that a speed of 45m/s (110 mi/hr) is necessary to take off....
2. “Derive” from atomic theory (i.e., from the hypotheses that elements are made up of indivisible
atoms and each compound is a fixed combination of them) the “law” of multiple proportions –
the fact that when two elements form two or more compounds the masses of one element that
combine with a fixed mass of any other are in the ratio of small whole numbers.
I’m asking here for a general “proof”, so please don’t use any particular examples.
Let’s consider the two binary compounds 𝑋𝑛1 𝑌𝑚1 and 𝑋𝑛2 𝑌𝑚2 .
1
The mass of Y combining with 1g of X in compound 1 is 𝐴𝑊
1
𝑚1
𝑛1
𝑚2
𝑋
the mass of Y combining with 1g of X in compound 2 is 𝐴𝑊
𝑋
𝑛2
𝐴𝑊𝑌 , and similarly
𝐴𝑊𝑌 .
1
1
(Here “AW” stands for atomic weight, and “𝐴𝑊 " = 𝐴𝑊
𝑋
1𝑔
𝑋
𝑔/𝑚𝑜𝑙𝑒
is the number of moles of X in
a binary compound of X and Y that contains 1g of X.)
Accordingly, the ratio of these masses is
1 𝑚1
𝐴𝑊𝑌
𝐴𝑊𝑋 𝑛1
1 𝑚2
𝐴𝑊𝑌
𝐴𝑊𝑋 𝑛2
=
𝑚1 𝑛2
𝑛1 𝑚2
= ratio of small whole numbers,
because 𝑛1 , 𝑛2 , 𝑚1 , and 𝑚2 are themselves each small whole numbers.
,
3. Carbon (C) and hydrogen (H) form a series of binary compounds with carbon mass
compositions of 92.3%, 85.7%, and 80.0%.
Compute, for each compound, the mass of hydrogen that combines with 0.67g of carbon.
Show that these compounds obey the law of multiple proportions.
What are these compounds?
The 1st compound, call it 𝐶𝑛1 𝐻𝑚1 , is 92.3% carbon by mass, and involves a carbon mass of
0.67g. Thus
0.67𝑔
0.67𝑔+𝑚𝐻,1
= 0.923, implying a hydrogen mass of 𝑚𝐻,1 = 0.056𝑔.
0.67𝑔
For the 2nd compound, we have 0.67𝑔+𝑚
𝐻,2
And for the 3rd compound, similarly,
The ratios of combining masses turn
𝑚
= 0.857, impling that 𝑚𝐻,2 = 0.112𝑔.
0.67𝑔
= 0.800 implies 𝑚𝐻,3 = 0.168𝑔.
0.67𝑔+𝑚𝐻,3
𝑚
out to be 𝑚𝐻,3
𝐻,1
𝑚
= 2,
= 3, 𝑚𝐻,2
𝐻,1
and 𝑚𝐻,3 = 32 , consistent with the law of multiple proportions.
𝐻,2
Looks like these compounds are 𝐶2 𝐻2 , 𝐶2 𝐻4 , and 𝐶2 𝐻6 − acetylene, ethylene, and ethane.
4. Consider the following hydrogen-containing compounds: 𝐶6𝐻6, N𝐻3, and 𝐻2𝑂2.
Which has the highest mass percentage of hydrogen? The lowest?
I’ll just start the calculation. For the 1st compound, 𝐶6𝐻6, benzene, the mass percentage of
hydrogen is {6(1.0)/[6(12.0)+6(1.0)]}100% = 7.69%. For the 2nd and 3rd compounds it is
17.6%and 5.88%, respectively.
5. Calculate the mass of each product that is produced from 2.24g of the boldfaced reactant,
for each of the following (not-yet-balanced) chemical reactions:
(i) 𝑴𝒈𝑵𝑯𝟒𝑷𝑶𝟒 → 𝑀𝑔2𝑃2𝑂7 + 𝑁𝐻3 + 𝐻2𝑂
(ii) 𝐶3𝐻8 + 𝑶𝟐 → 𝐶𝑂2 + 𝐻2𝑂
(iii) 𝐶𝑎𝐶𝑂3 + 𝑯𝟐𝑶 → 𝐶𝑎(𝑂𝐻)2 + 𝐶𝑂2
I’ll work part (i), where the balanced reaction reads
𝟐𝑴𝒈𝑵𝑯𝟒𝑷𝑶𝟒 → 𝑀𝑔2𝑃2𝑂7 + 2𝑁𝐻3 + 𝐻2𝑂
2.24g of 𝑴𝒈𝑵𝑯𝟒𝑷𝑶𝟒 (which has a molecular weight of 137.3) corresponds to
2.24𝑔
137.3𝑔/𝑚𝑜𝑙𝑒
=
0.0163 moles. Thus, from the balanced reaction equation we will have this many moles
of 𝑁𝐻3 product and half this number of moles of the other two products.
6. Nitrogen and oxygen form several different binary compounds. The mass percentages of
nitrogen in five of them are as follows: A, 25.9; B, 30.4: C, 36.8; D, 46.7; and E, 63.6.
What are the empirical formulas of the five compounds?
What mass of O combines with 1 g of N in each of them?
2
Do these compounds obey the law of multiple proportions?
For compound A, call it 𝑁𝑛𝐴 𝑂𝑚𝐴 , the mass percentage of N is 25.9, from which it follows that
(14.01)𝑛𝐴
= 0.259.
(14.01)𝑛𝐴 + (16.00)𝑚𝐴
𝑚
𝑚
5
Solving for 𝑛 𝐴 gives 𝑛 𝐴 = 2.51 ≈ 2, indicating the empirical formula 𝑁2 𝑂5 for compound A.
𝐴
𝐴
Similarly for compound B we find
𝑚𝐵
𝑛𝐵
= 2.00 = 2, indicating the empirical formula NO or
𝑁2 𝑂4 … for compound B.
And so on.
And once you know the empirical formula for each compound you can calculate the mass of O
that combines with 1g of N in each, and go on to show that the ratios of these combining masses
are the ratios of small integers, consistent with the law of multiple proportions.
7. Elemental sulfur (S) and methane gas (𝐶𝐻4) react to produce liquid carbon disulfide (𝐶𝑆2) and
𝐻2𝑆 gas.
What mass of elemental sulfur needs to react to yield 3.24 g of 𝐻2𝑆?
First, write the balanced reaction equation: 4𝑆 + 𝐶𝐻4 = 𝐶𝑆2 + 2𝐻2𝑆.
3.24g of 𝐻2𝑆 corresponds to 0.095 moles of 𝐻2𝑆 being produced, for which we need twice as
many moles of elemental sulfur (S) reacting.
8. A compound containing calcium, nitrogen, and carbon reacts with gaseous 𝑂2 to give 39.756g
of 𝑁𝑂2, 24.226g of CaO, and 19.024g of 𝐶𝑂2.
What can you deduce about the formula of the compound?
What is the balanced form of the reaction describing this combustion reaction?
All of the calcium in the compound ends up as 24.226g ↔ 0.432 moles of CaO and hence
0.432 moles of Ca.
Similarly, all of the nitrogen in the compound results in 39.756g ↔ 0.864 moles of 𝑁𝑂2 and
hence 0.864 moles of N.
Finally, the 19.024 g ↔ 0.432 moles of 𝐶𝑂2 imply 0.432 moles of C.
Hence we deduce the formula 𝐶𝑎𝐶𝑁2 for the compound, and write the balanced combustion
reaction
7
𝐶𝑎𝐶𝑁2 + 𝑂2 = 𝐶𝑎𝑂 + 2𝑁𝑂2 + 𝐶𝑂2
2
3
PS2 SOLUTIONS
CHEM 20AH
DUE: October 12, 2022
1. The typical kinetic energy of an electron in an atom is comparable to – i.e., is the same order of
magnitude as – its typical potential energy. And its potential energy arises from its attraction to the
nucleus. Estimate a typical electrostatic potential energy, i.e., evaluate the Coulomb energy
𝑽(𝒓) =
𝟏
𝒒𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 𝒒𝒏𝒖𝒄𝒍𝒆𝒖𝒔
𝟒𝝅𝝐𝒐
𝒓
for a typical distance from the nucleus – say, an Angstrom.
What does this imply for the typical kinetic energy of the electron, and for its speed?
For the H atom, say, the charges involved are -e and +e for the electron and nucleus. It follows that the
corresponding potential energy, calculated from the above equation, is 𝟐. 𝟑𝒙𝟏𝟎−𝟏𝟖 𝑱 = 𝟐. 𝟑𝒙𝟏𝟎−𝟏𝟖 𝒌𝒈
𝟏
𝒎𝟐
𝒔𝟐
.
Taking this to be the magnitude of the kinetic energy (𝟐 𝒎𝒆 𝒗𝟐 ) of the electron, we find (using 𝟏𝟎−𝟑𝟎 𝒌𝒈
𝒎
for the electron mass) a speed of 𝟐. 𝟐𝒙𝟏𝟎𝟔 .𝒔 . (That’s fast! A hundredth the speed of light!)
2. In problem 1 you should have found a speed that is many orders of magnitude faster than that of a
baseball pitch or a tennis serve. How does the kinetic energy of a baseball or tennis ball compare to
the typical kinetic energy of an electron?
𝒎
Even though the speed of a ball (< 𝟏𝟎𝟐 .𝒔 ) is more than 4 orders of magnitude smaller than that of an
electron, its mass (≈ 𝒌𝒈) is almost 30 orders of magnitude bigger! And, indeed,
𝒎 𝟐
(𝒌𝒈) (𝟏𝟎𝟐 ) = 𝟏𝟎𝟒 𝒌𝒈
.𝒔
𝒎𝟐
𝒔𝟐
𝟏
𝟐
𝒎𝒃𝒂𝒍𝒍 𝒗𝟐 ≈
= 𝟏𝟎𝟒 𝑱, ≫ 𝟐. 𝟑𝒙𝟏𝟎−𝟏𝟖 𝑱.
3. Estimate the mass density of liquid water (or ice) by estimating the volume occupied by a water
molecule when it is “close-packed”, i.e., surrounded on all sides by water molecules that are
similarly close-packed.
The oxygen atom in the water molecule is bonded to each of two hydrogens, with the OH bonds making
an angle of slightly more than 90° with respect to one another. The OH bond length is about an Angstrom
(as is the case for most bonds [to within a factor of 2!]), and the H atom has a radius of about half an
Angstrom. So we can approximate the space-filling volume of the water molecule as that of a sphere with
radius 1.5 Angstrom, i.e., 𝟏. 𝟒𝒙𝟏𝟎−𝟐𝟑 𝒄𝒄. Since the mass of the water molecule is 20amu, or
𝟑.𝟑𝒙𝟏𝟎−𝟐𝟑 𝒈
𝟑. 𝟑𝒙𝟏𝟎−𝟐𝟑 𝒈, we estimate a mass density of 𝟏.𝟒𝒙𝟏𝟎−𝟐𝟑𝒄𝒄 = 𝟐. 𝟑𝒈/𝒄𝒄. [Note that it’s too high (compared
with 1g/cc) because we under-estimated the effective volume of water molecules in liquid water – why,
how?]
Now estimate the mass density of water by estimating the volume of your body and assuming that
your body’s density is essentially the same as that of water.
Let’s approximate a body by a rectangular solid 2m high, 2/3m wide, and 1/6m thick. This corresponds to
a volume of 2/9 𝒎𝟑 . A mass of 90kg then implies a mass density of about 0.4g/c. [So here we seem to
have over-estimated the volume of a 90kg-person….why, how?]
4. Consider the singly-charged cation of lithium, 𝑳𝒊+ , i.e., a lithium atom that has lost one of its
1
electrons. We’re (you’re) going to estimate here the 1st and 2nd ionization energies of this atomic
species by imagining that its two electrons are 0.5Å apart from one another and from the nucleus.
(a) What is the total potential energy of each of the two electrons?
Each electron 𝑳𝒊+ has a potential energy (with charge in esu and energy in ergs)
𝒆𝟐
𝒆𝟐
𝒆𝟐
𝟐
(𝟒.𝟖𝒙𝟏𝟎−𝟏𝟎 𝒆𝒔𝒖)
−𝟑 𝟎.𝟓𝒙𝟏𝟎−𝟖 𝒄𝒎 + 𝟎.𝟓𝒙𝟏𝟎−𝟖𝒄𝒎 = −𝟐 𝟎.𝟓𝒙𝟏𝟎−𝟖𝒄𝒎 = −𝟐 𝟎.𝟓𝒙𝟏𝟎−𝟖𝒄𝒎 = −𝟗𝟔𝒙𝟏𝟎−𝟏𝟐 𝒆𝒓𝒈, and this
corresponds to the negative of the 1st ionization energy of 𝑳𝒊+ (and to the 2nd ionization energy of Li).
(b) What would be the potential energy of each electron if the other weren’t present?
𝟐
(𝟒.𝟖𝒙𝟏𝟎−𝟏𝟎 𝒆𝒔𝒖)
𝒆𝟐
Now the potential energy is given by just the first term above, −𝟑 𝟎.𝟓𝒙𝟏𝟎−𝟖𝒄𝒎 = −𝟑 𝟎.𝟓𝒙𝟏𝟎−𝟖𝒄𝒎 =
−𝟏𝟒𝟎𝒙𝟏𝟎−𝟏𝟐 𝒆𝒓𝒈. And this corresponds to the 2nd ionization energy of 𝑳𝒊+ (and to the 3rd ionization of
Li).
(c) Estimate from (a) and (b) the 1st ionization energy of 𝑳𝒊+ .
See annotation above, in parts (a) and (b)).
(d) What about the 2nd ionization energy?
Again, see annotation above. Evaluating the above 𝑳𝒊+ 1st and 2nd ionization energies in ev we find 60ev
and 88ev, as compared with the tabulated Li 2nd and 3rd ionization energies of 76ev and 122ev.
5. How does the electron affinity of an atom compare with the ionization energy of its singly-charged
anion? (The terms “cation” and “anion” refer to ions that are positive and negative, respectively.)
How does the ionization energy of the singly-charged anion compare with the ionization energy of
the neutral atom?
The EA of an atom is exactly equal (by definition!) to the IE of its singly-charged anion, as can be
shown by writing 𝑿 + 𝒆− → 𝑿− backwards.
6. In general, how do the magnitudes of electron affinities compare with those of 1 st ionization
energies? Why?
EAs are always significantly lower (e.g., by as much as a factor of 10) than IEs, because they both
correspond to electron-binding energies and the EA always involves the electron in question being
“bothered” (repelled) by an “extra” electron.
7. In Lecture #4 we estimated the bond energy of the ionic molecule KF by imagining the formation of
𝑲+ and 𝑭− from their isolated neutral atoms followed by bringing them together to a distance of the
equilibrium bond length of KF.
(a) Now perform the same calculation for KCl, whose equilibrium bond distance is 2.67Å.
𝒌𝑱
The total change in energy, as KCl is formed from 𝑲+ and 𝑪𝒍− , is the sum of 𝑰𝑬𝑲 = 𝟒𝟏𝟗 𝒎𝒐𝒍𝒆, -𝑬𝑨𝑪𝒍 =
𝒌𝑱
−𝟑𝟒𝟗 𝒎𝒐𝒍𝒆, and −
𝟐
(𝟒.𝟖𝒙𝟏𝟎−𝟏𝟎 𝒆𝒔𝒖)
𝟐.𝟔𝟕𝒙𝟏𝟎−𝟖 𝒄𝒎
= −𝟖. 𝟔𝒙𝟏𝟎−𝟏𝟐 𝒆𝒓𝒈𝒔 = −𝟖. 𝟔𝒙𝟏𝟎−𝟐𝟐 𝒌𝑱
𝟔.𝟎𝒙𝟏𝟎𝟐𝟑
𝒎𝒐𝒍𝒆
𝒌𝑱
= −𝟓𝟏𝟔 𝒎𝒐𝒍. The
𝒌𝑱
calculated/estimated bond energy is the negative of this total change in energy, hence 𝟒𝟒𝟔 𝒎𝒐𝒍.
(b) From the measured dipole moment of KCl (10 Debye), estimate the actual charge separation in
the molecule.
For the dipole moment of KCl we calculate 𝝁 = 𝒆𝑹𝒆 = (𝟏. 𝟔𝒙𝟏𝟎−𝟏𝟗 𝑪)(𝟐. 𝟔𝟕𝒙𝟏𝟎−𝟏𝟎 𝒎) =
2
𝟒𝟐. 𝟕𝒙𝟏𝟎−𝟑𝟎 𝑪 ∙ 𝒎
𝒎 = 𝟏𝟐. 𝟖𝑫𝒆𝒃𝒚𝒆 = 𝟏𝟐. 𝟖𝑫
𝟑. 𝟑𝟒𝒙𝟏𝟎−𝟑𝟎 𝑪 ∙
𝑫𝒆𝒃𝒚𝒆
Comparing this to the measured dipole moment of 10 D we conclude that the fraction () of e-charge
𝜹𝒆𝑹
𝟏𝟎𝑫
transferred is 𝒆𝑹 𝒆 = 𝜹 = 𝟏𝟐.𝟖𝑫 = 𝟎. 𝟕𝟖, i.e., the KCl bond is only 78% ionic, and the electrostatic
𝟒𝟐. 𝟕𝒙𝟏𝟎−𝟑𝟎 𝑪 ∙ 𝒎 =
𝒆
attraction energy is a factor of (𝟎. 𝟕𝟖)𝟐 = 𝟎. 𝟔𝟏 smaller.
(c) From the result obtained in (b), recalculate the bond energy of KCl, and compare your new
estimate with that found in (a).
𝟐
The −
(𝟒.𝟖𝒙𝟏𝟎−𝟏𝟎 𝒆𝒔𝒖)
𝟐.𝟔𝟕𝒙𝟏𝟎−𝟖 𝒄𝒎
𝒌𝑱
𝒌𝑱
in part (a) becomes 0.61 smaller, i.e., -315 𝒎𝒐𝒍𝒆 instead of -516 𝒎𝒐𝒍𝒆, and the bond
𝒌𝑱
𝒌𝑱
energy estimate is lowered to 245 𝒎𝒐𝒍. The measured value is 430 𝒎𝒐𝒍.
8. The force acting on a charge 𝒒𝟏 due to a charge 𝒒𝟐 at a distance r away is given by
𝟏
𝑭𝑪𝒐𝒖𝒍𝒐𝒎𝒃 (𝒓) = 𝟒𝝅𝝐
𝒒𝟏 𝒒𝟐
𝒐
𝒓𝟐
.
(Do you see how this is related to the expression for the Coulomb potential energy of interaction?)
It’s, by definition, the negative of the 1 st derivative of the potential energy with respect to distance.
Use this general expression to calculate the force on the electron in the hydrogen atom when it is a
distance 1Å from the nucleus.
(𝟏. 𝟔𝒙𝟏𝟎−𝟏𝟗 𝑪)𝟐
𝟐. 𝟑𝒙𝟏𝟎−𝟖 𝑱
|𝑭𝑪𝒐𝒖𝒍𝒐𝒏𝒃 | =
=
= 𝟐. 𝟑𝒙𝟏𝟎−𝟖 𝑵
𝟒𝝅(𝟖. 𝟖𝟓𝒙𝟏𝟎−𝟏𝟐 𝑪𝟐 /𝑱𝒎)((𝟏𝟎−𝟏𝟎 𝒎)𝟐 )
𝒎
How does the magnitude of this force compare to the force (𝒎𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 g) on the electron due to the
earth’s gravitational field?
𝑭𝒈𝒓𝒂𝒗 = 𝒎𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 𝒈 ≈ (𝟏𝟎−𝟑𝟎 𝒌𝒈) (
𝟏𝟎𝒎
𝒔𝟐
𝒎
) = 𝟏𝟎−𝟐𝟗 𝒌𝒈 𝒔𝟐 = 𝟏𝟎−𝟐𝟗 𝑵, ≪ =𝟐. 𝟑𝒙𝟏𝟎−𝟖 𝑵 = |𝑭𝑪𝒐𝒖𝒍𝒐𝒏𝒃 |
3
Problem Set 2, additional problems
Chem 20AH
DUE: Wednesday, October 12, 2022 (along with problems 1-8 assigned earlier)
9. Using a table of electron affinities, determine the first ionization energy of the singly-charged
negative ion of atomic fluorine, i.e., determine the ionization energy of F-.
IE(F-) is, by definition, the energy change associated with the “reaction” F - → F + e. But this is just
the reverse of the change F + e → F-, for which the energy change is, by definition, the negative of
the electron affinity (EA) of F. Thus IE(F-) = EA(F), which latter is equal to 328 kJ/mole, according to
Table 3.2 in the textbook and at the bottom of page 1 of Lecture 4.
10. An exact solution to the Schrodinger equation for a one-electron atomic system with a charge of
+Ze on it nucleus – i.e., for a single electron interacting with a nucleus having Z protons – yields the
following result for the quantum-mechanically-allowed energies:
1
𝐸𝑛 = −𝑍 2 (13.6𝑒𝑣) 𝑛2 , n = 1, 2, 3, 4, ….
Use this result to determine the 1st ionization energy of hydrogen, the 2nd ionization energy of
helium, the 3rd ionization energy of lithium, the 11th ionization energy of sodium, and the 12th
ionization energy of magnesium.
Note that the lowest-energy (“ground”) state of a single-electron atomic system, before it is ionized,
is the n=1 state. And that the magnitude of this energy is equal to the ionization energy of that
system.
We start be recognizing that each of the ionization energies involved here corresponds to the
ionization of a single-electron (“hydrogen-atom-like”) atomic species but with a different charge on
the nucleus: Z = 1, for H, 2 for 𝐻𝑒 + , 3 for 𝐿𝑖 2+, and … 11 for 𝑁𝑎10+ and 12 for 𝑀𝑔11+ . This is
because, for example, the 11th ionization energy of Na is the first ionization energy of 𝑁𝑎10+
– which is because 10 of the original 11 electrons have already been removed. Thus, for example,
the 3rd ionization energy of Li = the 1st ionization energy of 𝐿𝑖 2+ = the magnitude of the energy of
1
the ground state of 𝐿𝑖 2+ = 32 (13.6𝑒𝑣) 12 = 122ev, which is the value given in the table at the top of
the 2nd page of Lecture 5.
And so on for the other examples…
11. In class last week (Lecture 7), we considered an electron sitting halfway between two protons
𝑅
along their line of centers and hence a distance 𝐴𝐵
from each of them, where 𝑅𝐴𝐵 is the distance
2
between the proton on the left (A) and the proton on the right (B). In particular, we showed that the
net force on either proton, acting along their line of centers – arising from both the electron and the
other proton – pulls it towards the other proton. Now consider, instead, the position of the electron
to be somewhere along the line that is the perpendicular bisector of the internuclear line of centers:
Show that the component, along the line of centers, of the net force on each proton pulls it towards
the other proton.
Let’s consider the case where the electron is a distance
centers, making it a distance √2
right, say, is given by −
factor of 𝑐𝑜𝑠45° =
1
√2
𝑅𝐴𝐵
2
𝑒2
𝑅
4𝜋𝜖𝑜 (√2 𝐴𝐵
)2
2
𝑅𝐴𝐵
2
above the midpoint of the line of
from each proton. Thus the force exerted on the proton to the
, and its component along the line of centers is smaller by a
, resulting in a force of
2
𝑒2
√2 4𝜋𝜖𝑜 (𝑅𝐴𝐵)2
, pointing towards the other proton.
Finally, the force on the proton on the right due to the proton on the left is 4𝜋𝜖
in the opposite direction. Since
2
√2
𝑒2
2
𝑜 (𝑅𝐴𝐵 )
, but pointing
= √2 > 1, the net force on the proton on the right is pointing
towards the other proton, and hence attractive.
PS3 SOLUTIONS
Chem 20AH
DUE: October 19, 2022
Problem 1
Red light spans the wavelength range from 750-620nm, and blue light from 495-450nm,
so let’s take their typical wavelengths to be 690nm and 475nm, respectively.
For 𝜆𝑟𝑒𝑑 = 690 𝑛𝑚, then, we have
𝑐
3.00 𝑥 108 𝑚/𝑠
𝜈𝑟𝑒𝑑 =
=
= 4.35 𝑥 1014 𝑠 −1 .
𝜆𝑟𝑒𝑑
6.90 𝑥 10−7𝑚
And for 𝜆𝑏𝑙𝑢𝑒 = 475 𝑛𝑚 we have
𝑐
3.00 𝑥 108𝑚/𝑠
𝜈𝑏𝑙𝑢𝑒 =
=
= 6.32 𝑥 1014 𝑠 −1 ,
𝜆𝑏𝑙𝑢𝑒
4.75 𝑥 10−7 𝑚
and so on for the other colors.
Problem 2
2
𝑒4
𝑒2
[ 2 ] = [( ) ] = (𝑀1 𝐿3 𝑡 −2)2 = 𝑀2 𝐿6 𝑡 −4
𝜖𝑜
𝜖𝑜
[𝑚] = 𝑀1 𝐿0 𝑡 0
1
[ℎ2] = 𝑀−2 𝐿−4𝑡 +2
(see Problem 7)
Thus
𝑒4
1
[𝜖 2 𝑚 ℎ2] = 𝑀2 𝐿6 𝑡 −4 𝑀1 𝐿0 𝑡 0 𝑀−2 𝐿−4 𝑡 +2 = 𝑀1 𝐿2 𝑡 −2 = [𝑒𝑛𝑒𝑟𝑔𝑦]
𝑜
Problem 3
The 2nd ionization energy of He is the 1st ionization energy of the positive ion 𝐻𝑒 +,
a one-electron atomic species whose energies are given by
1
𝐸𝑛 = −22 (13.6𝑒𝑣) 𝑛2 , 𝑛 = 1, 2, 3, …
1
Accordingly, its ionization energy is 0 − 𝐸1 = 22 (13.6𝑒𝑣) 12 = 54.4𝑒𝑣
Problem 4
We calculate the frequency emitted with the 𝑛 = 3 → 𝑛 = 2 transtion.
1
1
1
1
5
1
1.6 𝑥 10−19 𝐽
𝜈3→2 = [𝐸3 − 𝐸2 ] = (−13.6𝑒𝑣) [ 2 − 2 ] = +
(13.6𝑒𝑣
)
ℎ
ℎ
3
2
36 (6.63 𝑥 10−34 𝐽𝑠)
𝑒𝑣
= 4.56 𝑥 1014𝑠 −1 , corresponding to an emission wavelength of 6579Å.
Problem 5
[𝑒𝑛𝑒𝑟𝑔𝑦 𝑞𝑢𝑎𝑛𝑡𝑢𝑚] = 𝑀1 𝐿2 𝑡 −2 = (𝑀1 𝐿3 𝑡 −2 )𝑥 (𝑀1 𝐿0 𝑡 0 )𝑦 (𝑀1 𝐿2𝑡 −1)𝑧
Thus
𝑀: 1 = 𝑥 + 𝑦 + 𝑧
𝐿: 2 = 3𝑥 + 2𝑧
𝑡: − 2 = −2𝑥 − 𝑧
Solving for x, y, and z gives x=2, y=1, z=-2.
Problem 6
[𝑟𝑎𝑑𝑖𝑢𝑠] = 𝑀0 𝐿1 𝑡 0 = (𝑀1 𝐿3 𝑡 −2)𝑎 (𝑀1 𝐿0𝑡 0 )𝑏 (𝑀1 𝐿2 𝑡 −1 )𝑐
Thus
𝑀: 0 = 𝑎 + 𝑏 + 𝑐
𝐿: 1 = 3𝑎 + 2𝑐
𝑡: 0 = −2𝑎 − 𝑐
Solving for a, b, and c gives a=-1, b=-1, c=2.
Problem 7
[𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚] =[rmv]= 𝐿1 ⋅ 𝑀1 ⋅ 𝐿1 𝑡 −1 = 𝑀1 𝐿2 𝑡 −1
[ℎ] = [
𝑒𝑛𝑒𝑟𝑔𝑦
]=
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑀 1 𝐿2 𝑡 −2
𝑡 −1
= 𝑀1 𝐿2 𝑡 −1
Problem 8
The potential energy is 2 times the kinetic energy in magnitude, and opposite in sign.
Problem 9
A fast ball has a speed of about 100mi/hr, or about 50 m/s; and it has a mass of about
𝑓𝑎𝑠𝑡 𝑏𝑎𝑙𝑙
one lb, or half a kg. Thus 𝜆𝑑𝑒 𝐵𝑟𝑜𝑔𝑙𝑖𝑒 ≈
6.63𝑥10−34𝑘𝑔⋅𝑚 2 ⋅𝑠−1
0.5𝑘𝑔⋅50𝑚𝑠−1
≈ 2.7 𝑥 10−35 𝑚 = 2.7 𝑥 10−25 Å
PROBLEM SET 5
CHEM 20AH
DUE: November 2, 2022
2
1. Show that 𝜓1 (𝑥) = 𝑁1 𝑥𝑒 −𝛼𝑥 satisfies the Schrodinger equation for the 1D harmonic
oscillator,
ℎ2 𝑑 2 𝜓(𝑥) 1 2
− 2
+ 𝑘𝑥 𝜓(𝑥) = 𝐸𝜓(𝑥)
8𝜋 𝑚 𝑑𝑥 2
2
𝜋
if and only if 𝛼 = ℎ √𝑘𝑚.
What is the corresponding energy of this state?
nd
The 2 -derivative (“kinetic energy”) term in the Schrodinger equation gives a sum of two terms:
ℎ2 𝛼 2 2
3 ℎ2 𝛼
− 2 𝑥 𝜓1 (𝑥) +
𝜓 (𝑥)
2𝜋 𝑚
4 𝜋2𝑚 1
𝜋
1
Substituting ℎ √𝑘𝑚 for 𝛼, the first term becomes − 2 𝑘𝑥 2 𝜓1 (𝑥), which cancels the “potential
3
1
2
2𝜋
energy” term, and the second term becomes ℎ (
𝑘
√ ) 𝜓1 (𝑥) = 𝐸𝜓1 (𝑥)
𝑚
2
𝑛 𝜋𝑥
𝑛 𝜋𝑦
𝑛 𝜋𝑧
2. Show that the functions 𝜓𝑛𝑥,𝑛𝑦,𝑛𝑧 (𝑥, 𝑦, 𝑧) = (𝐿)3/2 𝑠𝑖𝑛 𝑥𝐿 𝑠𝑖𝑛 𝑦𝐿 𝑠𝑖𝑛 𝑧𝐿 , with
𝑛𝑥 = 1,2,3, … ∞, 𝑛𝑦 = 1,2,3, … ∞, 𝑛𝑧 = 1,2,3, … ∞,
are solutions to the 3D particle-in-a-box problem, corresponding to the energies
ℎ2
𝐸𝑛𝑥 ,𝑛𝑦,𝑛𝑧 =
(𝑛 2 + 𝑛𝑦2 + 𝑛𝑧2 ), with 𝑛𝑥 = 1,2,3, … ∞, 𝑛𝑦 = 1,2,3, … ∞, 𝑛𝑧 = 1,2,3, … ∞
8𝑚𝐿2 𝑥
The “kinetic energy” terms (there are no potential terms in this Schrodinger equation) involve a
sum of 2nd-derivatives with respect to – separately – x, y, and z, each acting on only one of the
ℎ2
𝑑2
ℎ2 𝑛2
sine functions in the product function 𝜓𝑛𝑥,𝑛𝑦 ,𝑛𝑧 (𝑥, 𝑦, 𝑧). The − 8𝜋2𝑚 𝑑𝑥 2 term gives + 8𝑚𝐿𝑥2 times
ℎ2
𝑑2
ℎ2
ℎ2 𝑛2
𝑑2
the product function. And, similarly, the − 8𝜋2𝑚 𝑑𝑦 2 and − 8𝜋2𝑚 𝑑𝑧 2 terms give + 8𝑚𝐿𝑦2 and
ℎ2 𝑛2
+ 8𝑚𝐿𝑧2 times the product function. Combining the three terms we find 𝐸 = 𝐸𝑛𝑥,𝑛𝑦 ,𝑛𝑧 =
ℎ2
8𝑚𝐿2
(𝑛𝑥2 + 𝑛𝑦2 + 𝑛𝑧2 ).
3. Specify the spherical polar coordinates (𝑟, 𝜗, 𝜑) corresponding to (𝑥, 𝑦, 𝑧) =
(0,2,0)
(2,0,0)
(−2,2,0)
3 √3 1
(4 , 4 , 2 )
The corresponding (𝑟, 𝜗, 𝜑) coordinates are
𝜋 𝜋
(2, 2 , 2 )
𝜋
(2, 2 , 0)
(√8,
20
𝜋 3𝜋
2
,
4
)
1
1
√5
√3
(√16 , 𝑐𝑜𝑠 −1 ( ) , 𝑡𝑎𝑛 −1 ( )
1
Now specify the rectangular coordinates (𝑥, 𝑦, 𝑧) corresponding to (𝑟, 𝜗, 𝜑) =
(8,0,0)
𝜋
(8, 2 , 0)
𝜋 𝜋
(8, , )
2 2
𝜋 𝜋
(8, 2 , 4 )
The corresponding (𝑥, 𝑦, 𝑧) coordinates are
(0,0,8)
(8,0,0)
(0,8,0)
(4√2, 4√2, 0)
4. Use the result we derived for the infinitesimal volume element in spherical coordinates,
𝑑𝑉 = 𝑟 2 𝑑𝑟 𝑠𝑖𝑛𝜗 𝑑𝜗 𝑑𝜑, to calculate the volume of a sphere of radius R, i.e., integrate
this volume element over all 𝜗, over all 𝜑, and over all r from 𝑟 = 0 to 𝑟 = 𝑅.
𝑅
𝜋
2𝜋
𝑅3
4𝜋
Volume of sphere = ∫0 𝑟 2 𝑑𝑟 ∫0 𝑠𝑖𝑛𝜗𝑑𝜗 ∫0 𝑑𝜑 = 3 ∙ 2 ∙ 2𝜋 = 3 𝑅3
What about the volume of a spherical shell of radius R and thickness dr?
𝜋
2𝜋
Volume of shell = 𝑅2 𝑑𝑟 ∫0 𝑠𝑖𝑛𝜗𝑑𝜗 ∫0 𝑑𝜑 = 𝑅2 𝑑𝑟 ∙ 2 ∙ 2𝜋 = 4𝜋𝑅 2 𝑑𝑟
5. Show that 𝑒 −𝑟/𝑎𝑜 , with 𝑎𝑜 equal to the Bohr radius, is a solution to the hydrogen-atom
Schrodinger equation. What is the corresponding energy?
𝑑
The term in the Schrodinger equation that involves 𝑑𝑟 derivatives gives a sum of terms:
2ℎ2
ℎ2
+ 2
𝑒 −𝑟/𝑎𝑜 − 2 2 𝑒 −𝑟/𝑎𝑜
8𝜋 𝑚𝑎𝑜
8𝜋 𝑚𝑎𝑜
𝜖 ℎ2
Substituting 𝑎𝑜 = 𝜋𝑒𝑜2𝑚 into these terms gives
𝑒2
𝑚𝑒 4
+ 4𝜋𝜖 𝑟 𝑒 −𝑟/𝑎𝑜 − 8𝜖2ℎ2 𝑒 −𝑟/𝑎𝑜
𝑜
𝑜
The first term is cancelled by the potential energy term in the Schrodinger equation, and the
𝑚𝑒 4
second gives 𝐸 = − 8𝜖2ℎ2.
𝑜
6. Convince yourself that the hydrogen-atom 2𝑝𝑧 wavefunction (“orbital”),
𝑟
𝑟
−
𝜓2𝑝𝑧 = 𝐴2𝑝𝑧 ( ) 𝑒 2𝑎𝑜 𝑐𝑜𝑠𝜗,
𝑎𝑜
“lies along” (i.e., is oriented along) the z-axis.
The basic idea is that positions “hovering” around the z-axis have a polar angles (𝜗) close to 0,
and hence 𝑐𝑜𝑠𝜗 values close to 1, whereas points hovering around the x,y-plane have polar
𝜋
angles (𝜗) close to 2 , and hence 𝑐𝑜𝑠𝜗 values close to 0. That is, the wavefunction proportional
to 𝑐𝑜𝑠𝜗 “lies along” the z-axis.
7. What is the probability of finding the hydrogen-atom 2𝑝𝑦 electron within dr of the
distance r from the nucleus?
What is its most probable distance from the nucleus?
What is this electron’s average distance from the nucleus?
2
𝑟
𝑟
Here we have to integrate [(𝐴2𝑝𝑦 (𝑎 ) 𝑒
𝑟
[(𝐴2𝑝𝑦 (𝑎 ) 𝑒
𝑜
𝑜
𝑟
−2𝑎
−2𝑎
𝑜
𝑜
𝑠𝑖𝑛𝜗𝑠𝑖𝑛𝜑)2 ]𝑑𝑉 =
)2𝑠𝑖𝑛2 (𝜗)𝑠𝑖𝑛 2 (𝜑)] 𝑟 2 𝑑𝑟𝑠𝑖𝑛𝜗𝑑𝜗𝑑𝜑
over all 𝜗 and 𝜑, i.e.,
2𝜋
𝜋
𝑟
𝑟
−2𝑎 2 2
2
𝑜
𝑃2𝑝𝑦 (𝑟)𝑑𝑟 = [𝐴2𝑝𝑦 ( ) 𝑒
] 𝑟 𝑑𝑟 ∫ 𝑠𝑖𝑛 (𝜑 )𝑑𝜑 ∫ 𝑠𝑖𝑛 3 (𝜗)𝑑𝜗
𝑎𝑜
𝑜
0
2𝜋
𝜋
4
Using ∫𝑜 𝑠𝑖𝑛2 (𝜑 )𝑑𝜑 = 𝜋, and ∫0 𝑠𝑖𝑛3 (𝜗)𝑑𝜗 = 3 , we have that
𝑃2𝑝𝑦 (𝑟)𝑑𝑟 =
4𝜋
3
𝑟2
𝐴22𝑝𝑦 𝑎 2 𝑒
𝑜
−
𝑟
𝑎𝑜
𝑟 2 𝑑𝑟.
The most probable distance is the value of 𝑟 that maximizes 𝑟 4 𝑒
𝑟
−𝑎
𝑜
, i.e., the value
𝑟
−
𝑎𝑜
𝑟 = 4𝑎𝑜 where the first derivative of 𝑟 4 𝑒
vanishes.
∞
The average distance is given by ∫0 𝑟 𝑃2𝑝𝑦 (𝑟)𝑑𝑟.
3
PS 6 SOLUTIONS
1.
CHEM 20AH
DUE: November 9, 2022
The lowest-energy solution to the Schrodinger equation for the 1D harmonic oscillator
(i.e., a mass 𝑚 constrained to the 𝑥-axis, bound to a perfect spring with force constant 𝑘),
ℎ2 𝑑 2 𝜓(𝑥) 1 2
− 2
+ 𝑘𝑥 𝜓(𝑥) = 𝐸𝜓(𝑥),
8𝜋 𝑚 𝑑𝑥 2
2
2
𝜋
𝛼
is 𝜓𝑜 = 𝐴𝑒 −𝛼𝑥 with 𝛼 = ℎ √𝑘𝑚 and (the normalization factor) 𝐴 = (𝜋 )1/4.
What are the dimensions of 𝛼? Of 𝐴? Of 𝜓𝑜 ? Of (𝜓𝑜 (𝑥))2 ? And of (𝜓𝑜 (𝑥))2 𝑑𝑥?
[𝛼] = 𝐿−1 . [𝐴] = 𝐿−1/2. [𝜓𝑜 ] = 𝐿−1/2. [(𝜓𝑜 (𝑥))2 ] = 𝐿−1. [(𝜓𝑜 (𝑥))2 𝑑𝑥] = dimensionless.
1
1
𝑘
The corresponding energy is 𝐸𝑜 = 2 ℎ (2𝜋 √𝑚).
What is the probability of finding the displacement within 𝑑𝑥 of 𝑥?
𝛼
2
2
𝑃(𝑥)𝑑𝑥 = (𝜓𝑜 (𝑥))2 𝑑𝑥 = 𝐴2 𝑒 −2𝛼𝑥 𝑑𝑥 = ( )1/2 𝑒 −2𝛼𝑥 𝑑𝑥
𝜋
What is the average value of 𝑥? And the most probable value?
2
+∞
+∞
𝛼
𝑥̅ = ∫−∞ 𝑥 𝑃(𝑥)𝑑𝑥 = (𝜋 )1/2 ∫−∞ 𝑥𝑒 −2𝛼𝑥 𝑑𝑥 = 0
2
And the most probable value of x is the one that maximizes (𝜓𝑜 (𝑥))2 = 𝐴2 𝑒 −2𝛼𝑥 , 𝑖. 𝑒. , 𝑥 = 0.
2.
2
𝑛𝑦 𝜋𝑦
𝑛 𝜋𝑥
𝑛 𝜋𝑧
You have shown that the functions 𝜓𝑛𝑥,𝑛𝑦 ,𝑛𝑧 (𝑥, 𝑦, 𝑧) = (𝐿)3/2𝑠𝑖𝑛 𝑥𝐿 𝑠𝑖𝑛 𝐿 𝑠𝑖𝑛 𝑧𝐿 ,
with 𝑛𝑥 = 1,2,3, … ∞, 𝑛𝑦 = 1,2,3, … ∞, 𝑛𝑧 = 1,2,3, … ∞,
are solutions to the 3D particle-in-a-box Schrodinger equation, corresponding to the energies
ℎ2
𝐸𝑛𝑥 ,𝑛𝑦,𝑛𝑧 = 8𝑚𝐿2(𝑛𝑥2 + 𝑛𝑦2 + 𝑛𝑧2 ), with 𝑛𝑥 = 1,2,3, … ∞, 𝑛𝑦 = 1,2,3, … ∞, 𝑛𝑧 = 1,2,3, … ∞
What are the three different solutions corresponding to the energy 6
ℎ2
8𝑚𝐿2
?
(𝑛𝑥 , 𝑛𝑦 , 𝑛𝑧 ) = (2,1,1), (1,2,1) and (1,1,2)
Show that the sum of any of them is also a solution corresponding to the same energy.
2
Acting on 𝐴𝜓2,1,1 (𝑥, 𝑦, 𝑧) = 𝐴(𝐿)3/2 𝑠𝑖𝑛
2𝜋𝑥
𝑠𝑖𝑛
𝜋𝑦
𝑠𝑖𝑛
𝜋𝑧
ℎ2
𝑑2
with − 8𝜋2𝑚 𝑑𝑥 2 we find the result
𝐿
𝐿
𝐿
2
ℎ2 𝑑 2
ℎ2 𝑑 2
2 ℎ
(𝑥,
2 8𝑚𝐿2 𝐴𝜓2,1,1 𝑦, 𝑧). Similarly, acting on this same function with − 8𝜋2𝑚 𝑑𝑦 2 and − 8𝜋2𝑚 𝑑𝑦 2
ℎ2
ℎ2
gives
12 𝐴 8𝑚𝐿2 𝜓2,1,1 (𝑥, 𝑦, 𝑧) and 12 𝐴 8𝑚𝐿2 𝜓2,1,1 (𝑥, 𝑦, 𝑧), respectively. Thus,
ℎ2 𝑑 2
ℎ2 𝑑 2
ℎ2 𝑑 2
ℎ2
[− 8𝜋2𝑚 𝑑𝑥 2 − 8𝜋2𝑚 𝑑𝑦 2 − 8𝜋2𝑚 𝑑𝑧 2] 𝐴𝜓2,1,1(𝑥, 𝑦, 𝑧) = (22 + 12 + 12 ) 8𝑚𝐿2 𝐴𝜓2,1,1 (𝑥, 𝑦, 𝑧)
2
=6
In the same way we can show that
ℎ2
𝑑2
ℎ2
𝑑2
ℎ2
ℎ
𝐴𝜓2,1,1 (𝑥, 𝑦, 𝑧).
8𝑚𝐿2
𝑑2
ℎ2
[− 8𝜋2𝑚 𝑑𝑥 2 − 8𝜋2𝑚 𝑑𝑦 2 − 8𝜋2𝑚 𝑑𝑧 2] 𝐵𝜓1,2,1(𝑥, 𝑦, 𝑧) = (12 + 22 + 12 ) 8𝑚𝐿2 𝐵𝜓1,2,1(𝑥, 𝑦, 𝑧)
ℎ2
[−
ℎ2
𝑑2
8𝜋2 𝑚
𝑑𝑥 2
−
ℎ2
𝑑2
8𝜋2 𝑚
𝑑𝑦 2
= 6 8𝑚𝐿2 𝐵𝜓1,2,1 (𝑥, 𝑦, 𝑧), and
−
ℎ2
𝑑2
8𝜋2 𝑚
𝑑𝑧 2
=6
] 𝐶𝜓1,1,2(𝑥, 𝑦, 𝑧) = (12 + 12 + 22 )
ℎ2
8𝑚𝐿2
ℎ2
8𝑚𝐿2
𝐶𝜓1,1,2(𝑥, 𝑦, 𝑧)
𝐶𝜓1,1,2(𝑥, 𝑦, 𝑧).
It then follows that
1
[−
ℎ2 𝑑 2
ℎ2 𝑑 2
ℎ2 𝑑 2
−
−
] [𝐴𝜓2,1,1 (𝑥, 𝑦, 𝑧) + 𝐵𝜓1,2,1 (𝑥, 𝑦, 𝑧) + 𝐶𝜓1,1,2 (𝑥, 𝑦, 𝑧)] =
8𝜋 2 𝑚 𝑑𝑥 2 8𝜋 2 𝑚 𝑑𝑦 2 8𝜋 2 𝑚 𝑑𝑧 2
ℎ2
6 8𝑚𝐿2 [𝐴𝜓2,1,1(𝑥, 𝑦, 𝑧) + 𝐵𝜓1,2,1(𝑥, 𝑦, 𝑧) + 𝐶𝜓1,1,2 (𝑥, 𝑦, 𝑧)]
𝑟
3. Consider the hydrogen-atom 2𝑝𝑧 wavefunction (“orbital”), 𝜓2𝑝𝑧 = 𝐴2𝑝𝑧 (𝑎 ) 𝑒
𝑜
1
𝑟
−2𝑎
𝑜
𝑐𝑜𝑠𝜗.
Here 𝐴2𝑝𝑧 = (32𝜋𝑎3)1/2 is the normalization factor.
𝑜
(a) What is the probability of finding the electron in a small volume (10−2 𝑎𝑜 )3 centered at
the position 𝑟, 𝜗, 𝜑 = 2𝑎𝑜 , 0,0?
𝑜
𝑟 = 2𝑎𝑜 −𝑟=2𝑎
[𝜓2𝑝𝑧 = 𝐴2𝑝𝑧 (
) 𝑒 2𝑎𝑜 cos(𝜗 = 0)]2 (10−2𝑎𝑜 )3
𝑎𝑜
And at 𝑟, 𝜗, 𝜑 = 4𝑎𝑜 , 0,0?
𝑜
𝑟 = 4𝑎𝑜 −𝑟=4𝑎
[𝜓2𝑝𝑧 = 𝐴2𝑝𝑧 (
) 𝑒 2𝑎𝑜 cos(𝜗 = 0)]2 (10−2𝑎𝑜 )3
𝑎𝑜
𝜋
And at 𝑟, 𝜗, 𝜑 = 2𝑎𝑜 , 2 , 0?
𝑜
𝑟 = 2𝑎𝑜 −𝑟=2𝑎
𝜋
[𝜓2𝑝𝑧 = 𝐴2𝑝𝑧 (
) 𝑒 2𝑎𝑜 cos(𝜗 = )]2 (10−2 𝑎𝑜 )3
𝑎𝑜
2
𝑟=2𝑎𝑜
𝜋
You can plug in values for 𝑎 and cos(𝜗 = 2 ), etc., throughout the above and
𝑜
compute relative values for these different probabilities….
(b) What is the probability of finding the electron within 10−2 𝑎𝑜 of the distance 2𝑎𝑜 ?
𝑃(𝑟)(10−2 𝑎𝑜 ) =
𝑟
−
4𝜋 2
𝑟
𝑟 𝐴2𝑝𝑧 2 (𝑎 )2 𝑒 𝑎𝑜 (10−2 𝑎𝑜 )|
3
𝑜
4𝜋
3
𝑟=2𝑎𝑜
2 𝑟=2𝑎𝑜
2 − 𝑎𝑜
)
𝑒
(10−2 𝑎𝑜 )
𝑎𝑜
(𝑟 = 2𝑎𝑜 )2 𝐴2𝑝𝑧 (
And of 4𝑎𝑜 ?
4𝜋
3
=
𝑟=2𝑎𝑜
(𝑟 = 2𝑎𝑜 )2 𝐴2𝑝𝑧 2 (
𝑟=4𝑎
𝑟=4𝑎𝑜 2 − 𝑎 𝑜
𝑜 (10−2 𝑎 )
) 𝑒
𝑜
𝑎𝑜
4. Suppose we neglect completely the potential energy of interaction between the two
electrons in the helium atom. What is the lowest-energy state of this 2-electron atom?
And the corresponding wavefunction?
The lowest energy will be the sum of the lowest energy that each electron would have if it
were the only electron present (interacting with the nucleus of charge 𝑍𝑒) – namely,
−22
𝑚𝑒 4
8𝜖𝑜
1
2 ℎ 2 12
+ (−22
𝑚𝑒 4
1
8𝜖𝑜 2 ℎ2 12
), and the corresponding wavefunction will be the product of
the wavefunctions each of electron would have if it were the only electron present – namely,
1
1
1
2
23 2 −2𝑟
23 2 −2𝑟
𝑎0 ] [(
𝑎0 ].
𝜓(𝑟̅1 , 𝑟̅2 ) = [(
)
2
)
2
𝜋𝑎𝑜 3
𝜋𝑎𝑜 3
2
PS 7 SOLUTIONS
CHEM 20AH
DUE: November 16, 2022
We discussed in class the fact that all solutions to the Schrodinger equation for a one-electron
atomic species with charge Ze on the nucleus can be written as a product of a radial part
𝑅(𝑟) depending only on 𝑟 and an angular part 𝑌(𝜗, 𝜑) depending only on 𝜗 and 𝜑.
By squaring this product wavefunction, multiplying by 𝑑𝑉 = 𝑟 2 𝑑𝑟𝑠𝑖𝑛𝜗𝑑𝜗𝑑𝜑, and integrating
over all 𝜗 and 𝜑, we showed that the radial (distance) distribution for each state could be
written as
𝑃(𝑟)𝑑𝑟 = 𝑟 2 (𝑅(𝑟))2 𝑑𝑟.
Use the radial functions listed in the table on the 1st page of Lecture 18/19 to determine the
𝑃(𝑟) distributions for the 1s, 2s, 2p, 3s, 3p, and 3d states.
𝑍3
2
𝑟
E.g., 𝑃1𝑠 (𝑟) = 4𝑟 2 𝑎 3 𝑒 −2𝑍𝑟/𝑎𝑜 = 4(𝑍/𝑎𝑜 ) (𝑍 𝑎 ) 𝑒
𝑎𝑜
𝑍
𝑜
𝑜
𝑟
𝑟
−2𝑍𝑎
𝑜
, which implies that
𝑃1𝑠 (𝑟) = 4𝑥 2 𝑒 −2𝑥 , where 𝑥 ≡ 𝑍 𝑎 is the dimensionless distance 𝑟 and
𝑜
dimensionless radial distribution function. Plot
𝑎𝑜
𝑍
2. Now do the same for
3. And for
4. And
𝑎𝑜
𝑍
𝑎𝑜
𝑎𝑜
𝑍
𝑎𝑜
𝑍
𝑎𝑜
𝑍
𝑍
𝑃1𝑠 (𝑟) is the
𝑃1𝑠 (𝑟) = 4𝑥 2 𝑒 −2𝑥 as a function of 𝑥.
1. In the way just done above for the 1s function, determine
and plot it on at the same graph as
𝑎𝑜
𝑎𝑜
𝑍
𝑟
𝑃2𝑠 (𝑟) as a function 𝑥 ≡ 𝑍 𝑎
𝑜
𝑃1𝑠 (𝑟) = 4𝑥 2 𝑒 −2𝑥 .
𝑃2𝑝 (𝑟).
𝑃3𝑠 (𝑟).
𝑃3𝑝 (𝑟).
5. And 𝑃3𝑑 (𝑟).
𝑍
What you will have done here is to reproduce the plots shown on the 2nd and 3rd pages of
Lecture 18/19. (Please note that each of the numbers on the y-axis there should have a
decimal point before it, e.g., 5 should read 0.5, and so on.)
I’ll work things out for
The function
𝑎𝑜
𝑍
𝑎𝑜
𝑍
𝑃1𝑠 (𝑟) and
𝑎𝑜
𝑍
𝑃2𝑝 (𝑟).
𝑟
𝑃1𝑠 (𝑟) = 4𝑥 2 𝑒 −2𝑥 is positive for all 𝑥 ≡ 𝑍 𝑎 and clearly has only one point
𝑜
(𝑥 = 0) where it’s equal to zero. And there’s one point where the function has a maximum –
the point where the derivative vanishes:
𝑑 2 −2𝑥
𝑥 𝑒
= (2𝑥 − 2𝑥 2 )𝑒 −2𝑥 = 0 = (1 − 𝑥),
𝑑𝑥
4
Implying that the function has its maximum at 𝑥 = 1, where its value is 4 ∙ 12 𝑒 −2 = 𝑒 2 =
0.541. Accordingly, a plot of 4𝑥 2 𝑒 −2𝑥 involves a function that increases from zero at 𝑥 = 0 to
a maximum of 0.541 at 𝑥 = 1 and then decreases exponentially for large 𝑥, in agreement
with the 1𝑠 plot in the figure at the bottom of the 2nd page of Lecture 18/19 (when you insert
the missing decimal point before all of the numbers on the y-axis there).
For
𝑎𝑜
𝑍
1 𝑍3
𝑍𝑟
1 𝑍
𝑃2𝑝 (𝑟) we start with 𝑃2𝑝 (𝑟) = 𝑟 2 (𝑅2𝑝 (𝑟))2 = 𝑟 2 24 𝑎 3 (𝑎 )2 𝑒 −𝑍𝑟/𝑎𝑜 = 24 𝑎 𝑥 4 𝑒 −𝑥 .
𝑟
Here, again, 𝑥 ≡ 𝑍 𝑎 .
𝑜
as we did above for
𝑎𝑜
𝑍
Thus
𝑎𝑜
𝑍
1
𝑜
4 −𝑥
𝑃2𝑝 (𝑟) = 24 𝑥 𝑒
𝑜
𝑜
, and we can proceed essentially identically
𝑃1𝑠 (𝑟) = 4𝑥 2 𝑒 −2𝑥 .
𝑟
𝑥 4 𝑒 −𝑥 is positive for all 𝑥 ≡ 𝑍 𝑎 and clearly has only one point (𝑥 = 0) where it’s equal to
𝑜
zero. And there’s one point where the function has a maximum – the point where the
derivative vanishes:
𝑑 4 −𝑥
𝑥 𝑒 = (4𝑥 3 − 𝑥 4 )𝑒 −𝑥 = 0 = (4 − 𝑥),
𝑑𝑥
1
32
Implying that the function has its maximum at 𝑥 = 4, where its value is 24 44 𝑒 −4 = 3𝑒 4 =
1
0.195. Accordingly, a plot of 24 𝑥 4 𝑒 −𝑥 involves a function that increases from zero at 𝑥 = 0 to
a maximum of 0.195 at 𝑥 = 4 and then decreases exponentially for large 𝑥, in agreement
with the 2𝑝 plot in the figure at the bottom of the 2nd page of Lecture 18/19 (when you add
the missing decimal point before all of the numbers on the y-axis there).
𝑎
𝑎
𝑎
𝑎
For the other functions, 𝑍𝑜 𝑃2𝑠 (𝑟), 𝑍𝑜 𝑃3𝑠 (𝑟), 𝑍𝑜 𝑃3𝑝 (𝑟), and 𝑍𝑜 𝑃3𝑑 (𝑟), the idea is the same –
only the algebra (i.e., the finding of the zeroes and the zero-slope values of 𝑥) is more
complicated…
PS 8 SOLUTIONS
CHEM 20AH
DUE: November 23, 2022
1.
The independent-electron (“effective-field”) theory has determined the value
𝑍𝑒𝑓𝑓,1𝑠 𝐴𝑟 = 17.6 for the effective charge on the nucleus seen by 1s electrons in the Argon atom
(where the bare charge on the nucleus is 18). Use this fact to reproduce the 1s curve – the plot
𝑎
𝑟
of 𝑍 𝑜 𝑃1𝑠 (𝑟) versus 𝑎 in the figure on the 1st page of Lecture 21. In particular, determine
𝑒𝑓𝑓,1𝑠 𝐴𝑟
𝑜
the value of the curve at its maximum, and the value of 𝑥 there.
𝑎
Taking the derivative of 𝑜 𝑃1𝑠 (𝑟) = 4𝑥 2 𝑒 −2𝑥 with respect to 𝑥 we find 𝑥 = 1 ≡ 𝑍
𝑍
𝑟
1
that 𝑎 = 𝑍 = 𝑍
𝑜
1
𝑒𝑓𝑓
1
𝑟
𝑎𝑜
, implying
= 17.6 = 0.06 for Argon, in agreement with the figure.
Further, the value of
𝑎𝑜
𝑍
𝑟
𝑃1𝑠 (𝑟) = 4𝑥 2 𝑒 −2𝑥 there (at 𝑥 = 1 ↔ 𝑎 = 0.06) is 0.54.
𝑜
Multiplying by 𝑍 = 𝑍𝑒𝑓𝑓 = 17.6 we get 𝑎𝑜 𝑃1𝑠 = (17.6)(0.54) = 9.5, and multiplying by 2 (for
the 2 1s electrons) we get the value for the 1s-maximum shown in the figure.
2.
Using the plots in the figure at the bottom of the 1st page of Lecture 21, reproduce the
curve shown in the figure on the 2nd page there. (As in problem 1 above, I’m not asking you to
generate all of the numerical values of the curves involved, but rather only to convince yourself
you understand what’s being plotted – because you’re able, for example, to determine the
values and positions of the maxima that appear there….)
Here you are simply adding up all of the plots in the figure at the bottom of the 1st page of
Lecture 21, to obain the single plot on the 2nd page there (which latter curve shows three peaks,
one for each value of n…).
3.
Using “Pauli’s principle” and “Hund’s rule” – and your understanding of the
independent-electron approximation -- convince yourself that you follow the logic behind the
assignment of electrons to one-electron (hydrogen-like) “orbitals” in each of the atoms listed
(through Neon) in the table on the 3rd page of Lecture 21. Complete the table up through
Argon.
I think you know what’s going on here…Let’s talk about it if you don’t….
4.
Using the 𝑍𝑒𝑓𝑓,1𝑠 𝐻𝑒 = 1.69 value derived for the 1s electrons of Helium, calculate the
total energy of the ground (lowest-energy) state of Helium in the independent-electron
(effective-field) approximation. How does it compare with the exact, measured, value of -5.7
Rydbergs?
(1 Rydberg = 13.6ev.)
1
In the effective-field approximation, each electron has the energy −𝑍𝑒𝑓𝑓,1𝑠 𝐻𝑒 2 (1 𝑅𝑦𝑑𝑏𝑒𝑟𝑔) 12
1
−(1.69)2 12 Rydbergs , and multiplying by 2 (to get the total ground state energy of the atom)
we have −2(1.69)2 = −5.71 Rydbergs, agreeing well with the exact (measured) value of 5.81
(which I mistakenly wrote as -5.7).
1
5.
Consider the 𝑟𝐴 , 𝑟𝐵 , 𝜑 coordinate system defined by the figure at the top of the 1st page
of Lecture 22/23. What are the 𝑟𝐴 , 𝑟𝐵 , 𝜑 values of a point whose 𝑥, 𝑦, 𝑧 coordinates are
1
1
0, 𝑎𝑜 , 𝑎𝑜 ?
√2
√2
Note that the point 𝑥 = 0, 𝑦 =
1
√2
𝑎𝑜 , 𝑧 =
1
√2
𝑎𝑜 lies in the yz-plane: hence 𝜑 = 0.
And to solve this problem it’s helpful to choose a convenient value of the distance 𝑅𝐴𝐵 between
the two nuclei (protons), to simplify the trigonometry involving the distances 𝑟𝐴 and 𝑟𝐵 .
For 𝑅𝐴𝐵 = √2 𝑎𝑜 , for example, we find 𝑟𝐴 = 𝑎𝑜 and 𝑟𝐵 = 𝑎𝑜 .
And for 𝑅𝐴𝐵 = 𝑎𝑜 /√2, we find 𝑟𝐴 = 𝑎𝑜 /√2 and 𝑟𝐵 = 𝑎𝑜 .
Recall that proton B is at the origin, and proton A is a distance 𝑅𝐴𝐵 above it on the z-axis.
2
PS 4 SOLUTIONS
CHEM 20AH
DUE: October 26, 2022
𝒌
1. Show that 𝒙(𝒕) = 𝒙𝒐 𝐜𝐨𝐬⁡(√𝒎 𝒕) is a solution to Newton’s equation 𝑭(𝒙) = 𝒎
𝒅𝟐 𝒙(𝒕)
𝒅𝒕𝟐
for a perfect
spring, i.e., for a mass m that is moving in one dimension and subject to a force 𝑭 = −𝒌𝒙.
Suppose that the initial (𝒕 = 𝟎) displacement is 𝒙𝒐 .⁡
(Note that, by definition of the “initial displacement”, the initial (𝒕 = 𝟎) speed of the mass is zero.)
𝒌
Show further that 𝒙(𝒕) = 𝒙𝒐 𝐜𝐨𝐬⁡(√𝒎 𝒕) satisfies the initial conditions 𝒙(𝒕 = 𝟎) = 𝒙𝒐⁡ and
(
𝒅𝒙(𝒕)
𝒅𝒕
)
𝒕=𝟎
= 𝟎.
𝒌
What about 𝒙(𝒕) = 𝒙𝒐 𝐬𝐢𝐧⁡(√𝒎 𝒕)? Does it satisfy Newton’s equation for this problem? Does it
satisfy the initial conditions?
Calculate the potential energy of the system, and the kinetic energy, and the total energy.
𝒌
Substituting 𝒙(𝒕) = 𝒙𝒐 𝐜𝐨𝐬⁡(√𝒎 𝒕) into 𝒎
𝒌
𝒌
𝒅𝟐 𝒙(𝒕)
𝒅𝒕𝟐
we obtain
𝒌
−𝒎⁡𝒙𝒐 𝒎 𝒄𝒐𝒔 (√𝒎 𝒕) = −𝒌⁡𝒙𝒐 𝒄𝒐𝒔 (√𝒎 𝒕) = −𝒌𝒙(𝒕),
𝒌
thereby establishing 𝒙(𝒕) = 𝒙𝒐 𝐜𝐨𝐬⁡(√𝒎 𝒕) as a solution of Newton’s equation.
𝒌
And it satisfies the initial conditions because 𝒙(𝒕 = 𝟎) = 𝒙𝒐 𝐜𝐨𝐬 (√ 𝟎) = 𝒙𝒐 ,⁡and
𝒎
(
𝒅𝒙(𝒕)
𝒌
𝒌
) = −𝒙𝒐 √ 𝐬𝐢𝐧 (√ 𝒕) = 𝟎⁡𝒂𝒕⁡𝒕 = 𝟎.
𝒅𝒕
𝒎
𝒎
𝒌
Similarly, substituting 𝒙(𝒕) = 𝒙𝒐 𝐬𝐢𝐧⁡(√𝒎 𝒕) into 𝒎
𝒌
𝒌
𝒅𝟐 𝒙(𝒕)
𝒅𝒕𝟐
we obtain
𝒌
−𝒎⁡𝒙𝒐 𝒎 𝒔𝒊𝒏 (√𝒎 𝒕) = −𝒌⁡𝒙𝒐 𝒔𝒊𝒏 (√𝒎 𝒕) = −𝒌𝒙(𝒕),⁡i.e., this function also satisfies Newton’s
equation for the problem.
But it vanishes at 𝒕 = 𝟎⁡instead of equaling 𝒙𝒐.
And (
𝒅𝒙(𝒕)
𝒅𝒕
𝒌
𝒌
𝒌
) = 𝒙𝒐 √𝒎 𝐜𝐨𝐬 (√𝒎 𝒕) = 𝒙𝒐 √𝒎 ⁡𝒂𝒕⁡𝒕 = 𝟎,⁡instead of vanishing.
Finally, we have
𝟏
𝟏
𝒌
𝑷𝑬 = 𝟐 𝒌(𝒙(𝒕))𝟐 = 𝟐 𝒌𝒙𝟐𝒐 𝒄𝒐𝒔𝟐 ⁡(√𝒎 𝒕), and
𝟏
𝑲𝑬 = 𝒎(
𝟐
𝒅𝒙(𝒕) 𝟐
)
𝒅𝒕
𝟏
𝒌
𝟐
𝒎
= 𝒎𝒙𝟐𝒐
𝒌
𝟏
𝒌
𝒎
𝟐
𝒎
𝒔𝒊𝒏𝟐 (√ 𝒕) = 𝒌𝒙𝟐𝒐 𝒔𝒊𝒏𝟐 ⁡(√ 𝒕),
implying that the total energy, which is given by the sum of the PE and KE, is equal to
the sum of sine squared and cosine squared is 1.
𝟏
𝟐
𝒌𝒙𝟐𝒐 ,⁡because
1
2. Show that each of the functions 𝑨 𝐬𝐢𝐧 (
𝒉𝟐
𝒏𝝅𝒙
𝑳
) , 𝒏 = 𝟏, 𝟐, 𝟑, …. is a solution to the Schrodinger
𝒅𝟐
equation − 𝟖𝝅𝟐 𝒎 𝒅𝒙𝟐 𝝍(𝒙) = 𝑬𝝍(𝒙) for a particle in a one-dimensional box of length L centered at
𝑳
𝒙 = 𝟐, and further that it vanishes at the borders of the box.
Substituting 𝝍(𝒙) = 𝐀⁡𝐬𝐢𝐧 (
𝒏𝝅𝒙
𝑳
𝒉𝟐
+ 𝟖𝒎𝑳𝟐 𝒏𝟐 𝑨𝒔𝒊𝒏 (
And 𝝍(𝒙) = 𝐀⁡𝐬𝐢𝐧 (
𝒏𝝅𝒙
𝑳
𝒉𝟐
𝒅𝟐
𝒉𝟐
)⁡into − 𝟖𝝅𝟐 𝒎 𝒅𝒙𝟐 𝝍(𝒙) gives − 𝟖𝝅𝟐 𝒎 [−
𝒏𝝅𝒙
𝑳
𝒏𝟐 𝝅𝟐
𝑳𝟐
𝑨𝒔𝒊𝒏 (
𝒏𝝅𝒙
𝑳
)] =
𝒉𝟐
) = 𝟖𝒎𝑳𝟐 𝒏𝟐 ⁡𝝍(𝒙) = 𝑬𝝍(𝒙).
) vanishes at 𝒙 = 𝟎 and at 𝒙 = 𝑳 because 𝐬𝐢𝐧(𝟎) = 𝟎 = 𝐬𝐢𝐧(𝒏𝝅).
Determine the value of A that guarantees that each solution (wavefunction) is normalized, i.e., that
𝟐
𝑳
ensures that ∫𝟎 𝒅𝒙⁡(𝝍(𝒙)) = 𝟏.
𝟐
𝑳
𝑳
We need to have ∫𝟎 𝒅𝒙⁡(𝝍(𝒙)) = 𝟏 = ⁡ 𝑨𝟐 ∫𝟎 𝒅𝒙⁡𝒔𝒊𝒏𝟐 (
𝑳
But ∫𝟎 𝒅𝒙⁡𝒔𝒊𝒏𝟐 (
𝒏𝝅𝒙
𝑳
𝑳
𝒏𝝅𝒙
𝑳
).
𝟐
𝟐
) = 𝟐⁡implies that we must have 𝑨𝟐 = 𝑳 ,⁡and hence 𝑨 = √𝑳 .
What about the functions 𝑨 𝐜𝐨𝐬 (
𝒏𝝅𝒙
𝑳
) , 𝒏 = 𝟏, 𝟐, 𝟑, … . ? Do they satisfy the same Schrodinger
equation?
Yes, substituting 𝝍(𝒙) = 𝐀⁡𝐜𝐨𝐬 (
𝒉𝟐
+ 𝟖𝒎𝑳𝟐 𝒏𝟐 𝑨𝒄𝒐𝒔 (
𝒏𝝅𝒙
𝑳
𝒏𝝅𝒙
𝑳
𝒉𝟐
𝒅𝟐
𝒉𝟐
)⁡into − 𝟖𝝅𝟐𝒎 𝒅𝒙𝟐 𝝍(𝒙) gives − 𝟖𝝅𝟐𝒎 [−
𝒏𝟐 𝝅𝟐
𝑳𝟐
𝑨𝒄𝒐𝒔 (
𝒏𝝅𝒙
𝑳
)] =
𝒉𝟐
) = 𝟖𝒎𝑳𝟐 𝒏𝟐 ⁡𝝍(𝒙) = 𝑬𝝍(𝒙).
Do they satisfy the same boundary conditions?
No, 𝝍(𝒙 = 𝟎) = 𝐀⁡𝐜𝐨𝐬 (
𝒏𝝅𝟎
𝑳
) = 𝑨 ≠ 𝟎…
3. When the particle in the above 1D box is in its n=1 state, what is the probability that its position will lie
between L/4 and 3L/4? And between 0 and L/2, and between L/2 and L?
𝟏
𝟑𝑳/𝟒
𝟑
𝟐
𝝅𝒙
𝟐
𝟑𝑳/𝟒
𝝅𝒙
𝟐
𝟑𝝅/𝟒
𝑷𝒏=𝟏 (𝟒 𝑳 ≤ 𝒙 ≤ 𝟒 𝑳) = ∫𝑳/𝟒 𝒅𝒙 𝑳 𝒔𝒊𝒏𝟐 ( 𝑳 ) = 𝑳 ∫𝑳/𝟒 𝒅𝒙 𝒔𝒊𝒏𝟐 ( 𝑳 ) = 𝝅 ∫𝝅/𝟒 𝒅𝒚 𝒔𝒊𝒏𝟐 𝒚 < 𝟏
𝑳/𝟐
𝑳
𝟐
𝝅𝒙
𝟏
𝝅𝒙
𝑷𝒏=𝟏 (𝟎 ≤ 𝒙 ≤ ) = ∫𝟎 𝒅𝒙 𝒔𝒊𝒏𝟐 ( ) = ,⁡because 𝒔𝒊𝒏𝟐 ( ) is symmetric about the middle of the
𝟐
𝑳
𝑳
𝟐
𝑳
box and so the probability of the particle being on the left side has to equal that of being on the right
side (and the two probabilities have to add up to 1).
𝑳
𝟏
Similarly, 𝑷𝒏=𝟏 (𝟐 ≤ 𝒙 ≤ 𝑳) = 𝟐.
What about when it is in the n=2 state?
Same idea….
2
4. Consider again the problem of a point mass in a 1D box of length L. If the box is centered at x=0
instead of x=L/2, what are the solutions to the Schrodinger equation that satisfy the new boundary
conditions? What are the corresponding allowed energies?
𝝍𝒏 (𝒙) = 𝑨𝒄𝒐𝒔 (
𝑨𝒔𝒊𝒏 (
𝒏𝝅𝒙
𝑳
𝒏𝝅𝒙
𝑳
),⁡for 𝒏 = 𝟏, 𝟑, 𝟓, 𝟕, …
),⁡for 𝒏 = 𝟐, 𝟒, 𝟔, 𝟖, …,
𝒉𝟐
with corresponding energies 𝑬𝒏 = 𝟖𝒎𝑳𝟐 𝒏𝟐 , 𝒏 = 𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔, 𝟕, 𝟖, …
𝟐
5. Show that the function 𝒆−𝜶𝒙 satisfies the Schrodinger equation for a 1D harmonic oscillator, if and
𝝅
𝟏
𝟏
𝒌
𝟏
only if 𝜶 = 𝒉 √𝒌𝒎, and that the corresponding energy is 𝑬 = 𝟐 𝒉 (𝟐𝝅 √𝒎) = 𝟐 𝒉𝝂.
ℎ2
The term − 8𝜋2𝑚
−
𝑑 2 𝜓(𝑥)
𝑑𝑥 2
ℎ2
𝑑
2
in the Schrodinger equation gives − 8𝜋2𝑚 𝑑𝑥 [−2𝛼𝑥𝑒 −𝛼𝑥 ] =
ℎ2
ℎ2
𝜋
𝜋2
2
−𝛼𝑥 2
2 2 −𝛼𝑥 2
−𝛼𝑥 2
√𝑘𝑚]
(−2𝛼𝑒
+
4𝛼
𝑥
𝑒
)
=
−
(−2
[
𝑒
+
4
[
𝑘𝑚] 𝑥 2 𝑒 −𝛼𝑥 ) =
2
2
2
8𝜋 𝑚
8𝜋 𝑚
ℎ
ℎ
1
1
𝑘
2
1
2
+ 2 ℎ (2𝜋 √𝑚) 𝑒 −𝛼𝑥 − 2 𝑘𝑥 2 𝑒 −𝛼𝑥 .
1
1
2
And the term 2 𝑘𝑥 2 ⁡𝜓(𝑥) in the Schrodinger equation gives 2 𝑘𝑥 2 𝑒 −𝛼𝑥 , which cancels the 2nd
1
1
𝑘
2
term above, leaving 2 ℎ (2𝜋 √𝑚) 𝑒 −𝛼𝑥 for the left-hand side of the equation, implying that the
2
1
1
𝑘
function 𝑒 −𝛼𝑥 satisfies – is a solution of – the equation, corresponding to 𝐸 = 2 ℎ (2𝜋 √𝑚).
3