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This'll be more properly organized later. Temporarily inactive as school notes are
primarily uploaded to Layman's List to be easily referenced by university friends
and people in the future. :-)
19 January 2022 WIL
Did a small break just to relax and clear my mind before school starts. Back at
it because Quantum Chemistry is hard. So, particle in a box. This is basically
a microscopic box in one dimension. The particle is "contained" by means of potential
energy - 0 inside the box, infinite outside. What makes the box useful? Well, we
can use it to determine the energy of a particle via the formula E = ((n*h_bar)^2/
8m(l)^2). m is mass and l is the length of the box - this is why large boxes that
we can actually see result in such a low energy that it is negligible, and therefore
not a part of the scope of quantum mechanics. The box also allows us to calculate
frequency and wavelength by relating the upper and lower quantized energies in the
box to the standard definition of energy E = (planck_constant*frequency) and
(frequency*wavelength) = c. Basically, the box helps us compute the properties of a
tiny particle in a tiny space.
11-12 January 2022 WIL
11-12 January 2022: Physical Chemistry Problem Solving
The next few WIL are going to just be going through a set of 33 textbook
problems step by step as a means of mastering content.
1.1: All photons have the same energy - false, since E depends on wavelength
As the frequency of light increases, wavelength decreases - true, c=lambda*v
If a light x with a wavelength 'a' does not induce the photoelectric effect in
a certain metal, then light y with wavelength 'b' (where a>b) will not induce
the effect - true, because (as determined in part a) the energy of y is lower.
1.2: Photon Energy at 984 nm wavelength: E_Photon = planck_constant*frequency
E = ((2.998 * 10^8 m/s)/ (984*10^-9 m)) * 6.626 * 10^-34 = 2.019 * 10^-19 J
Amount of photons emitted with power 8*10^6 W and duration 4*10^-8 seconds:
1W = 1J/S. E_Photon = 2.019*10^-19J. Total J = power*duration = 0.32J.
0.32J/(2.019*10^-19J) = 1.585 * 10^18 photons emitted.
1.3: Energy of 1 mole of UV photons (wavelength 300 nm). 1 mole = 6.022 *
10^23. E_Photon = ((2.998*10^8 m/s)/(300*10^-9 nm)) * 6.626 * 10^-34 =
6.622 * 10^-19J. Energy of 1 mole = 6.622*10^-19J * 6.022*10^23 = 398,751J.
1.4: Work Function of pure Na is 2.75 eV, 1 eV = 1.602*10^-19J. Maximum kinetic
energy of photoelectrons emitted from Na exposed to 260 nm: E = (planck_constant
* c)/wavelength. Maximum kinetic energy is E - work function. E = (6.626*10^-34
* 2.998*10^8 m/s)/(260*10^-9 nm) = 7.640*10^-19J. Work Function = 1.602*10^-19J *
2.75 = 4.406*10^-19J. MaxKE = 7.640*10^-19J - 4.406*10^-19J = 3.234*10^-19J or
2.02 eV. Longest wavelength that will curve photoelectric effect in Na: this will
happen when E = 4.406*10^-19J. Working backwards, 4.406*10^-19J = (6.626*10^-34
* 2.998*10^8 m/s)/(x nm), x = (6.626*10^-34* 2.998*10^8 m/s)/(4.406*10^-19J) =
451 nm.
1.5: Wien's law - I = av^3/e^(bv/T), a and b are constants.
Plack's equation - I = av^3/(e^(bv/T)-1). At high frequencies, Wien's law is a good
approximation, since as v grows e^(bv/T)-1 can be approximated to e^(bv/T), because
e^(bv/T) >> 1. Rayleigh-Jeans law - I = 2(pi)v^2(kb)T/(c^2). Taylor expansion of e^x is
sum (from k=0 to infinity) of (x^k)/(k!). Taylor expansion of e^(bv/T) is basically
an expansion of e^(cx) (c=b/T, since both are just numbers). This results in sum
(from k=0 to infinity) of (cx)^k/(k!), and since x is small, this can be approximated
to 1 + cx, or 1 + bv/T. So Planck's equation can be approximated to I = av^3/(bv/T),
and through algebraic manipulation, this can be expressed as I = a(v^2)T/b. While not
identical to the original law, it is important to remember that a and b are numerical
values just like 2pi, kb, and c^2, so by applying the proper constants, the formulas
match.
1.6: Calculate de Broglie wavelength of an electron moving at speed c/137: wavelength
= planck_constant/(mass_electron*velocity_electron). This leads to the expression
6.626*10^-34/(9.109*10^-34*2188321.168) = 3.324*10^-10, which is 3.324 angstrom or
0.3324 nm.
1.7: Integrate the function dx/dt = -gt + gt_0 + v_0 to find x as a function of time:
Multiply both sides by dt then integrate both sides. Integral(dx) = Integral((-gt+gt_0
+v_0)dt), integrals are definite (from x_0 to x), (from t_0 to t). This results in
x - x_0 = (-1/2)g(t-t_0)^2 + v_0(t-t_0).
1.8: psi = ae^(-ibt)e^(-bmx^2/h_bar). Find potential energy function V - this can be
done via time-dependent Schrodinger eq (t is present in equation). The relevant eq is
(-h_bar/i)(partial_psi(x,t)/partial_t) = (-(h_bar^2/2m))(2partial_psi(x,t)/2partial_x)
+V(x,t)psi(x,t). Goal is to find V(x,t). partial_psi/partial_t = -ibae^(-ibt)e^(-bmx^2/
h_bar). 2partial_psi/2partial_x = ae^(-ibt)(2bm/h_bar^2)(2bmx^2-h_bar)e^(-bmx^2/h_bar).
Simplifying, we end up with (h_bar)bae^(-ibt)e^(-bmx^2/h_bar) =
ae^(-ibt)(-b)(2bmx^2-h_bar)e^(-bmx^2/h_bar) + V(x,t)ae^(-ibt)e^(-bmx^2/h_bar). So V(x,t)
= ((h_bar)bae^(-ibt)e^(-bmx^2/h_bar) - ae^(-ibt)(-b)(2bmx^2-h_bar)e^(-bmx^2/h_bar) /
ae^(-ibt)e^(-bmx^2/h_bar)). V(x,t) = 2mb^2x^2
10 January 2022 WIL
10 January 2022: Physical Chemistry
In Physical Chemistry, I've started solving problems from the book as a means to
better understand what exactly I'm learning. The first (and simplest) questions
are on probability. However, before I delve into it, some backstory is required.
Probability is crucial in the field of quantum mechanics, a large reason being
the Heisenberg Uncertainty Principle, which basically dictates that we can never
know both the exact position and velocity of a tiny particle (like an atom). So
as a way to somewhat get around this, probability is used to see how likely some
outcome is in a quantum mechanics problem (such as the odds that the particle
lies in some region of space from a to b).
So, some things about the actual calculations. They solve one-particle, one-
dimensional systems (so they're pretty basic). They utilize something called a
wavefunction, but it must be modified first. We're dealing with probabilities,
so the "Born Interpretaion," created by Max Born, is used. Born says that the
product of conjugate wave function and the normal wavefunction results in the
probability density, which is another function. This can be simplified down to
the absolute value of the wavefunction squared results in probability density.
The conjugate is just used to show that the math holds for complex numbers.
Now what can we do with this probability density? We can take its derivative to
determine the probability that a particle is between x and x+dx, where dx is a
tiny value. We can also state that the integral from -infinity to +infinity of
the probability density function is 1. This is called normalization, and basically
tells us that the particle must be somewhere on the x axis (1 is 100%).
With this information, we can now solve numerical problems. The book goes through
one example that I will transcribe. The wavefunction (also known as psi) is equal
to a^(-1/2)*e^(-|x|/a) at t = 0. a = 1.0000 nm (DO NOT express this as 1*10^-9 m).
Objective 1 is to find the probability that the measured value (the particle) lies
between 1.5000 and 1.5001 nm. Objective 2 is to find the probability between 0 nm
and 2 nm. Objective 3 is to verify that the probability function will obey
normalization.
Objective 1 utilizes a unique solution. Since the interval is tiny (0.0001 nm) a
derivative is not needed. You can just use 0.0001 nm as the dx. The computation is
pretty straightforward - square the wavefunction (and use the absolute value), then
plug in the numbers. "e" is Euler's number.
Objective 2 uses integration. Use the squared wavefunction from (1) and integrate
from 0 to 2. Remember that "a" is a constant, which means you can extract it. Then
evaluate.
Objective 3 is akin to Objective 2 regarding how you solve the problem, but there's
one catch. Since you're using absolute values, split the integral into two - one
from -infinity to 0, the other from 0 to +infinity. Then evaluate.
As a bonus exercise, attempt Objective 1 but with the following wavefunction:
(32a^3/pi)^(1/4)*xe^(-ax^2). For the purpose of proper units, a is 1.0000 nm^-2. Also
use x between 1.2000 and 1.2001 nm. Happy solving!