CHEM 120A PS1
January 2021
1
1(a)
We can compute the energy of a single photon with a wavelength of 760 nm:
E = hν =
hc
(6.626 ∗ 10−34 Js)(3 ∗ 108 m/s)
=
= 2.614 ∗ 10−19 J
λ
760 ∗ 10−9 m
We can then divide the total energy of the pulse by the energy per photon
to find the number of photons:
Nphotons = Epulse /Ephoton =
10 ∗ 10−3 J
= 3.826 ∗ 1016 photons
2.16 ∗ 10−19 J
1(b)
We know the number of pulses per second, as well as the number of photons per
pulse, so we can compute the number of photons per second for the pulse laser:
3.826 ∗ 1016 photons 1000 pulse
∗
= 3.826 ∗ 1019 photons / s
1 pulse
1s
We can find the energy of a single photon in the continuous laser:
E = hν =
hc
(6.626 ∗ 10−34 Js)(3 ∗ 108 m/s)
=
= 3.748 ∗ 10−19 J
λ
530 ∗ 10−9 m
We can then find the number of photons per second since we know how many
Joules per second the laser outputs:
5J
1 photon
∗
= 1.334 ∗ 1019 photons / s
1 s 3.748 ∗ 10−19 J
The ratio (pulse/cont.) of photons per second:
3.826 ∗ 1019 photons / s
= 2.868
1.334 ∗ 1019 photons / s
1
Let’s move onto power. We can compute the power of the pulse laser since
we know the energy per pulse, as well as the pulses per second:
P =
10 ∗ 10−3 J 1000 pulse
∗
= 10 W
1 pulse
1s
The ratio (pulse/cont.) of power:
10 W
=2
5W
The ratios do not match up as the lasers have different wavelengths, i.e. the
energy per photon of each laser is different.
1(c)
We can compute the energy in Joules:
E = hν =
(6.626 ∗ 10−34 Js)(3 ∗ 108 m/s)
hc
=
= 6.621 ∗ 10−20 J
λ
3 ∗ 10−6 m
We can then convert to other units:
6.621 ∗ 10−20 J = 3333.33 cm−1 = 6.621 ∗ 10−13 ergs = 0.413 eV
1(d)
Assuming a rectangular pulse, the power is equivalent across the entire pulse.
We can just divide the pulse energy by the length of the pulse to get power:
P =
10 ∗ 10−3 J
= 4 ∗ 1011 W
25 ∗ 10−15 s
1(e)
We can divide the peak power by the beam area to get peak intensity. Assuming
the area of the beam is rectangular:
(100 ∗
4 ∗ 1011 W
2
= 4.000 ∗ 1015 W/(cm )
cm)(100 ∗ 10−4 cm)
10−4
Or, if the area of the beam is assumed to be circular:
4 ∗ 1011 W
2
= 5.093 ∗ 1015 W/(cm )
π ∗ (50 ∗ 10−4 cm)(50 ∗ 10−4 cm)
2
2(a)
The plot can be seen in figure 1. Note the axis units are Joules and Hz.
2
1e15
3.0
Frequency (Hz)
2.5
Cs
Zn
y = 1.509e+33 x + 8.942e+14
2.0
1.5
1.0
y = 1.509e+33 x + 4.592e+14
0.5
0.0
0.2
0.4
0.6
0.8
1.0
Kinetic Energy (J)
1.2
1.4
1e 18
Figure 1: The incoming photon frequency vs the emitted electron kinetic energy
for both Cs and Zn.
2(b)
We can get the work function based on our linear fits of the plot. We are
looking for the highest frequency of light where electrons are no longer emitted
(i.e. the y-intercept). This is the point where we have just enough energy to
strip an electron from the metal, and any extra energy will be converted to
kinetic energy. We then convert that frequency to an energy:
ΦCs = hν = (6.626 ∗ 10−34 J s)(4.592 ∗ 1014 s−1 ) = 3.043 ∗ 10−19 J = 1.899 eV
ΦZn = hν = (6.626 ∗ 10−34 J s)(8.942 ∗ 1014 s−1 ) = 5.925 ∗ 10−19 J = 3.698 eV
2(c)
Once our incident photon has enough energy to strip an electron from the metal,
all remaining energy is converted to kinetic energy of the emitted electron. So
the KE of the electron should equal:
KEemit = hνincid − Φ
So if we fit a line to a plot of the electron KE vs incident light frequency,
the slope should be h. In our case, we plotted the inverse, so our slope should
3
be
1
h:
1. Cs:
h=
1
1
=
= 6.629 ∗ 10−34 J s
m
1.509 ∗ 1033 Hz/J
h=
1
1
=
= 6.627 ∗ 10−34 J s
m
1.509 ∗ 1033 Hz/J
2. Zn:
2(d)
Cs has a much lower work function than Zn. If you remember your periodic
table trends, you’ll realize that Cs has a much lower ionization energy, i.e. it is
much easier to lose an electron. Cs is a group 1 element so it is quite easy to
remove its one valence electron.
In a PMT, an incoming photon hits the photocathode, which then releases
an electron which is then multiplied into many electrons. Since Cs has such a
low work function, it will release an electron when hit by light even in the visible
range. This makes it a good option for a photocathode.
2(e)
We can first compute the kinetic energy of the emitted electron:
KEemit
KEemit
hc
− ΦZn
λincid
(6.626 ∗ 10−34 Js)(3 ∗ 108 m/s)
− 5.925 ∗ 10−19 J
=
150 ∗ 10−9 m
=7.319 J
KEemit =
We can then compute the velocity of an electron with this kinetic energy:
1
KE = mv 2
2
1
7.319 J = (9.109 ∗ 10−31 kg)v 2
2
v =1.268 ∗ 106 m/s
v =1.268 ∗ 108 cm/s
4
3
3(a)
We can find the uncertainty in the momentum first:
∆p = m(∆v) = (0.025 kg)(0.00001 m/s) = 2.5 ∗ 10−7 kg m/s
We can then use the position momentum uncertainty principle to compute
the minimum uncertainty in position:
∆x =
h̄
1.055 ∗ 10−34 J s
=
= 2.11 ∗ 10−28 m
2∆p
2(2.5 ∗ 10−7 kg m/s)
So the uncertainty in position for this bullet is negligible and not measurable.
3(b)
We can find the energy variation by converting the minimum and maximum
wavelengths to energy:
hc
(6.626 ∗ 10−34 Js)(3 ∗ 108 m/s)
=
= 2.614 ∗ 10−19 J
λ
760 ∗ 10−9 m
(6.626 ∗ 10−34 Js)(3 ∗ 108 m/s)
hc
=
= 2.365 ∗ 10−19 J
= hν =
λ
840 ∗ 10−9 m
∆E = Emax − Emin = 2.489 ∗ 10−20 J
Emax = hν =
Emin
Then, assuming the uncertainty in time is simply the pulse length, we can
find the pulse length:
∆t =
h̄
1.055 ∗ 10−34 J s
=
= 2.118 fs
2∆E
2(2.489 ∗ 10−20 J)
3(c)
Now we know the uncertainty in time, so we can solve for the uncertainty in
energy:
∆E =
h̄
1.055 ∗ 10−34 J s
=
= 5.273 ∗ 10−23 J = 2.65 cm−1
2∆t
2(10−12 s)
4
4(a)
Some possible expressions are listed in table 1. Notice that the equations for
matter include mass, whereas the equations for light do not. The de Broglie
wavelength explains how matter(something with mass) behaves like a wave.
Light/photons do not have mass and cannot be used with these equations.
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Light
c = λν
E = hν
E = hc
λ
E = h̄ω
etc.
Matter
KE = 12 mv 2
λ = hp
p = mv
h
λ = mv
etc.
Table 1: Expressions for light and matter.
4(b)
First we can find the KE of the electron by finding the energy change of an
electron in a 1000 V potential:
KE = qV = (1.602 ∗ 10−19 C)(1000 J/C) = 1.602 ∗ 10−16 J
We can then find the velocity from this KE:
1
KE = mv 2
2
1
1.602 ∗ 10−16 J = (9.109 ∗ 10−31 kg)v 2
2
v =1.876 ∗ 107 m/s
Using this velocity, we can then compute the de Broglie wavelength:
λ=
h
mv
λ=
(6.626 ∗ 10−34 Js)
(9.109 ∗ 10−31 kg)(1.876 ∗ 107 m/s)
λ =3.878 ∗ 10−11 m
λ =3.878 ∗ 10−2 nm
4(c)
We will need electrons with a wavelength of max 100nm to resolve features of
a similar size. We can use this de Broglie wavelength to find the velocity of the
electrons:
h
λ=
mv
(6.626 ∗ 10−34 Js)
100 ∗ 10−9 m =
(9.109 ∗ 10−31 kg)v
v =7273.895 m/s
We can then use this velocity to compute the KE:
KE =
1
1
mv 2 = (9.109 ∗ 10−31 kg)(7273.895 m/s)2 = 2.409 ∗ 10−23 J = 1.504 ∗ 10−4 eV
2
2
6
Indeed this is significantly smaller than the 20 keV used in real electron
microscopes. Unlike a typical microscope, which diffracts light in order to magnify an image of the sample, an electron microscope works by focusing a beam
of electrons onto a very small area, which is then scattered by the regions of
high electron density in the sample. The scattered electrons therefore give information about which areas of the sample have high electron density, allowing
a map/image of the sample to be constructed. In order to focus the beam of
electrons onto such a small area, electron microscopes use of many lenses and
apertures, which slow down some of the electrons. Thus, higher energy electrons
are used to counteract this. Additionally, using higher kinetic energy electrons
will allow for better resolution - a 20 keV microscope would give a resolution of
0.08 angstroms, which would have no trouble resolving chemical bonds which
are typically on the order of 1 angstrom.
5
5(a)
Following the Bohr atom quantization condition (i.e. the circumference must
be a multiple of the wavelength), along with the de Broglie wavelength we can
find the allowed velocities of an electron:
2πr =nλ
nh
2πr =
mv
nh
v=
2πmr
nh̄
v=
mr
To find the radius of orbit associated with a given v, we can equate the
centrifugal force to the coulombic force. Li2+ has 3 protons compared to the 1
of H. This will affect the coulombic force (which will be 3x stronger):
mv 2
3e2
=
2
4π0 r
r
3e2
=mv 2 r
4π0
3e2
=r
4π0 mv 2
We know the allowed velocities already, so we can plug that expression in to
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find the allowed radii:
3e2
4π0 mv 2
3e2 m2 r2
r=
4π0 mn2 h̄2
3e2 mr2
r=
4π0 n2 h̄2
4π0 n2 h̄2
r=
3e2 m
r=
Now that we know the allowed radii and velocities, we can simply compute
the allowed energies:
E =KE + V
3e2
1
E = mv 2 −
2
4π0 r
2
1
nh̄
3e2
E= m
−
2
mr
4π0 r
n2 h̄2 1
3e2 1
−
2m r2
4π0 r
n2 h̄2 9e4 m2
3e2 3e2 m
=
−
2m 16π 2 20 n4 h̄4
4π0 4π0 n2 h̄2
9e4 m
9e4 m
−
=
2
2
32π 2 0 n2 h̄
16π 2 20 n2 h̄2
9e4 m
=−
32π 2 20 n2 h̄2
9e4 m
=− 2 2 2
80 n h
E=
E
E
E
E
The new Rydberg constant will be everything except the n2 :
R0 =
9e4 m
= 1.96188512 ∗ 10−17 J = 987636 cm−1
820 h2
This makes the new Rydberg formula:
1
1
0
∆E = R
− 2
n21
n2
5(b)
The energy level diagram can be seen in figure 2.
8
0
2000
n=17
n=16
n=15
n=14
E (cm 1)
4000
n=13
6000
n=12
8000
n=11
10000
n=10
Figure 2: The energy level diagram for Li2+
5(c)
Using our formula from the previous question:
1
1
∆E =R0
−
n2
n22
1
1
1
0
− 2
∆E =R
102
12
1
1
−1
∆E =(987636 cm )
−
100 144
∆E =3017.776 cm−1
This transition is labelled in blue in figure 2.
9
5(d)
Using our formula from the previous question:
1
1
0
∆E =R
− 2
n2
n2
1
1
1
0
− 2
∆E =R
132
12
1
1
−1
∆E =(987636 cm )
−
169 144
∆E = − 1014.583 cm−1
This transition is labelled in orange in figure 2.
5(e)
The Lyman series consist of transitions from the ground state, which must be
the 1s. Possible transitions are therefore:
1s →2p
1s →3p
1s →4p
The Balmer series consists of transitions from the first excited state, which
could be the 2s or 2p state. Possible transitions are therefore:
2s →3p, 2p → 3s, 2p → 3d (all degenerate)
2s →4p, 2p → 4s, 2p → 4d (all degenerate)
The Paschen series consists of transitions from the second excited state,
which could be 3s, 3p, or 3d states:
3s →4p, 3p → 4s, 3p → 4d, 3d → 4p, 3d → 4f (all degenerate)
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