Download 50 POINTS - University at Albany

Name: SOLUTIONS
Physics 240, Exam #3
Monday, Nov. 28, 2016 (2:45-4:05 p.m.)
50 POINTS (must do all 5 problems)
Instructions: Complete all questions as best you can, and show all of your work. If you just
write down an answer without explaining how you got it, you will not receive any points.
Do not use any device to retrieve text, equations, or algorithms during the exam. You can
use a scientific calculator but no tablets nor smartphones. Circle or box your final answers.
You may also use the one 2-sided formula sheet provided, but do not consult with or discuss
this exam with anybody else. Good luck, Happy Thanksgiving, and budget enough time.
Use the space below if you run out of room to answer the questions on any upcoming sheet.
You can also use the back side of each problem and/or request an additional sheet from me.
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1. CONCEPTUAL QUESTIONS (10 points) Answer ONLY TWO. If you answer a third, it
can count towards extra credit (maximum 5 points)
(a.) Name and explain one of the counter-intuitive features of quantum mechanics that
violates your intuition about how the universe is or should work. (5 points)
(b.) Name and explain one of the few features of quantum mechanics that still follows
“common sense.” (5 points)
(c.) Come up with at least two different reasons why energy levels based only on n split for
an atom, either with or without a magnetic field. (5 points)
(d.) Since electrons do not have well-defined positions due to the uncertainty principle, then
what do the shapes (orbs, lobes, etc.) of the various shell/orbitals represent? (5 points)
(e.) Name and explain what at least two of the different quantities mean within the Bragg
equation of x-ray scattering. (5 points)
Briefly justify answers, mentioning equations if need be.
(a.) Can’t measure position and speed or energy or momentum at same time (Heisenberg
uncertainty), energy conservation can be violated for short periods of time (ditto), particles
are both particles and waves at the same time and can go through two slits simultaneously
even as single entities (wave-particle duality, and double-slit experiment), quantum
entanglement “spooky action at a distance” conveying information faster than light
apparently (EPR “paradox”), universe probabilistic not deterministic, can’t know
everything if knowing all initial conditions, can “quantum tunnel” through energy barriers,
even barriers lower than a particle’s kinetic energy can have effect including reflection,
universe is discrete and quantized like E=h*f for photon energies, with a smallest possible
value of certain things like harmonic oscillator energy and non-zero electron angular
momentum in an atom. Lastly, photon energy depends on frequency or wavelength not on
“amplitude,” and even matter is a wave (any/all of the above; list not complete/exhaustive)
More: quantum superposition of states -- Schrodinger’s cat both dead and alive until it is
observed, electron is both spin up and down, and in multiple places at same time. Parallel
universes possible (Everett’s Many Worlds Interpretation). The Copenhagen interpretation
(wavefunction collapse after measurement)
(b.) Wavefunctions are continuous and differentiable, and probability still conserved and
single-valued and capped at 1. Math, specifically differential equations, still describe nature.
(c.) Fine structure (spin-orbit coupling due to internal electron magnetic field), hyperfine
structure (coupling of electron to nuclear spin caused by internal proton magnetic field),
Zeeman effect (magnetic dipole from electron orbit in external field), anomalous Zeeman
effect (dipole for e- spin in external field). Relevant equations accepted. Also: Heisenberg
(d.) Probabilistic distributions in space of likely positions for different average-E levels
(e.) n is the “order” of the peak (n = 1 is the first bright spot/area/region, etc.), while
lambda is the wavelength of the incoming radiation (usually x-ray photons). ‘d’ is the
spacing between planes of atoms inside of the crystal you are scattering your particle/waves
off of, and theta is the angle of reflection of the rays or waves off of the lattice planes.
2. De Broglie and Heisenberg, plus the Hydrogen Atom (10 points)
(a.) How fast or slow does a proton have to be moving to have an effective wavelength of
exactly 1 nm? (3 points)
(b.) If the uncertainty in that speed is ±15% then what is the uncertainty in position, in units
of position (not percent)? (3 points)
(c.) How small or large is that compared to the Bohr radius? (1 point)
(d.) Translating that uncertainty in speed to an uncertainty in energy and one which
happens to violate conversation of energy for a short period of time, for how long can such a
violation occur? (2 points)
(e.) At what value of n in a hydrogen atom would an electron have an energy similar to (a.)
in absolute value? (1 point) Hint: It’s a very large n and that’s OK!
(a.) This is too slow to require a relativistic calculation, but since you don’t know that in
advance it is OK either way. Wavelength = Planck’s constant / momentum, so momentum is
Planck over wavelength p=h/λ=6.626e-34 J-s/1 nm=6.626e-34 J-s/1e-9 m = 6.626e-25 kg*m/s
p = m*v so v = p / m = 6.626e-25 kg*m/s / 1.67e-27 kg ~ 400 m/s
(b.) Product of delta-x and delta-p must be greater than or equal to h-bar/2, where delta-p
is equal to (1.15 * 6.626e-25) kg*m/s minus (6.626e-25 / 1.15) kg*m/s, using part (a.) (Also is
acceptable if doing / and * 0.85. Very similar results).
Δx <= h-bar / (2*Δp) = [h / (2 * pi)] / (2 * Δp) = 6.626e-34 J-s / (4*3.14*1.858e-25 kg * m/s)
= 2.838e-10 m or ~0.3 nm or ~3 A (Angstroms). Report in any units or any # of sig figures.
(c.) 2.84e-10 m / 5.29e-11 m = almost exactly 5 times the Bohr radius
(d.) E_high = p_high^2/2m = ( 1.15 * 6.626e-25 )^2 / ( 2 * 1.67e-27 kg ) = 1.738409461E-22 J
E_low = p_low^2/2m = ( 6.626e-25 / 1.15 )^2 / ( 2 * 1.67e-27 kg ) = 9.939412517E-23 J
delta-E = E_high – E_low = 1.738409461E-22 J - 9.939412517E-23 J = 7.444682093E-23 J
~0.5 milli-eV (no need to convert). delta-E x delta-t is greater than or equal to h-bar/2, so
Δt <= h-bar / (2*ΔE) = [h / (2*pi)] / (2 * ΔE) =6.626e-34 J-s / (4*3.14*7.44e-23 J) = 7.08e-13 s
(e.) -13.6 eV / n^2 = -p^2 / 2m = -(6.626e-25 kg*m/s)^2 / ( 2 * 1.67e-27 kg ) =-1.31449E-22 J
-13.6 eV / n^2 = -1.314487E-22 J / 1.602e-19 (J/eV) = -0.0008205289047 eV
=> 13.6 eV = (0.00082 eV) * n^2 => n^2 = 13.6 / 0.00082 (eV’s cancel) => n = √16,585 ~ 129
3. Wave Properties: The Different Types of Speed, Group Versus Phase (10 points)
(a.) Demonstrate that the phase velocity of a relativistic particle is c / β . (5 points)
Hints: Use lambda*f for the phase velocity, E=gamma*m*c^2 from SR, E=h*f like for photons
since particle is very relativistic, and the de Broglie equation
(b.) Is this faster than the vacuum speed of photons? (2 points)
(c.) Does this show special relativity is incompatible with quantum mechanics? (3 points)
(a.)
Here,theGreekletternu
(ν)whichlookslikeav
standsforthefrequency,f
v_phase = c^2 / v = c * ( c / v ) = c * ( 1 / beta ) = c / beta, where beta = v / c and v is v_group
(b.) Yes, because beta is always less than 1.0 (beta = v/c and v < c at all times)
(c.) NOO; even if they are, this doesn’t show it, because it’s the group velocity that must be
< c, and group velocity represents the speed of travel for a particle/wave, or of information
4. Mesoscopic Quantum Tunneling, Through “Impassable” Obstacles (10 points)
(a.) What is the %-chance of success for a water molecule (mass of 3 x 10-26 kg) with energy
0.9 eV to reflect from a potential barrier of potential energy 1 eV? (5 points) Assume that the
barrier is 0.01-nm wide, and that complete absorption is impossible (or in other words, T+R=1)
(b.) What would the reflection coefficient have been classically, and why? (2 points)
(c.) How much larger than a water molecule in mass would an object have to be in order for
it to have a transmission probability of 1 in a million for this quantum barrier? (3 points)
(a.) κ=sqrt(2m*(V0–E))/h-bar(fromsecondsideofformulasheetneartop)
= sqrt(6e-26kg*(1eV–0.9eV))/1.05e-34J-s=2.94e11m^-1or1/m(1eV=1.602e-19J)
T=[1+V0^2*sinh2(κ*L)/(4*E*(V0–E))]^-1
If L = 0.01 nm = 0.01e-9 m = 1e-11 m, and you plug in for V0 = 1.0 eV and
κ as calculated above and E = 0.9 eV then T = 0.0055 = 0.55% which surprisingly is not that low for
this situation (since water molecule small and energy close)! T+R=1 => R=1-T = 99.45%
(b.) Zero%, because not enough energy to surmount barrier means you can never make it,
like a marble trapped in a parabolic well.
(c.) Guessing and checking different kappa’s (equation cannot be solved analytically in
closed form) shows that kappa ~ 7e11 m^-1 would yield a T of ~1e-6. Solving for mass, m, in
the equation for kappa κ above (and converting eV into Joules) we get this value occurring
when the mass is somewhere around 1.5e-25 kg, or 50 times more massive
5. A Particle in a Box or an Infinite-Potential Square Well (10 points) 2pts.each
(a.) Consider a small but macroscopic particle of mass m = 1 micro-gram confined to a 1dimensional box with L = 1 micro-meter, e.g., a tiny bead on a very short wire. Compute the
bead’s minimum kinetic energy.
(b.) What are the corresponding momentum^2 and speed?
(c.) Find the expectation value of the momentum^2 in the ground state (n = 1) thus using the
operator method to check on (b.)
(d.) If the particle is in the n = 3 state now instead, show that its wavefunction is normalized
(e.) In that same state as (d.) what is the probability of being between 3L/4 and L?
hint: You may need to use one of the integrals below.
∫ sin2 x dx = -(1/2)cos(x)sin(x)+(1/2)x = (1/2)x–(1/4)sin(2x)
∫ sin3 x dx = -(1/3)cos(x)(sin2x+2)
∫ x sin2x dx = x2 / 4 – x sin(2x)/4 – cos(2x)/8
∫ x2 sin2 x dx = x3 / 6 – (x2 / 4 – 1 / 8) sin(2x) – x cos(2x) / 4
(a.) En = ( n^2 * π^2 * h-bar^2 ) / ( 2 * m * L^2 ) so for n=1 (n=0 makes energy 0) we get
E_1 = ( 1^2 * π^2 * h-bar^2 ) / ( 2 * 10^-6 g * (10^-6 m)^2 ) =
( π^2 * h-bar^2) / (2 * 10^-9 kg * (10^-6 m)^2 ) = 5.5e-47 J or ~4e-28 eV, super-tiny!
(b.) Convert from energy into momentum. E = p^2/2m, so p^2 = 2mE =
2 * 1e-9 kg * 5.5e-47 J ~ 1.1e-55 J-kg, whereas p would be √ of this, or 3.3e-28 kg*m/s
Extremely non-relativistic, so p = m * v => v = p / m = 3.3e-19 m/s
(c.) Momentum operator p-hat = (- i * h-bar) * (d/dx) partial derivative
So, p^2 = (- i ) ^ 2 * (h-bar)^2 * (d/dx)^2 = - h-bar^2 * d2/dx2
Ψ 1 = sqrt ( 2 / L ) * sin ( 1 *pi * x / L ) = sqrt ( 2 / L ) * sin ( pi * x / L ) when n = 1
So the value of <p^2> =
Integral from 0 to L assuming same bounds as example from text and lecture (not –infinity
to infinity since wavefunction 0 outside of well. Common sense check: Wouldn’t work
anyway out to infinities because sin(e) function has infinite number of wiggles, over that full
range) of
sqrt ( 2 / L ) * sin ( pi * x / L ) (-h-bar^2) * d2/dx2 [sqrt ( 2 / L ) * sin ( pi * x / L ) ]
= ( -2 / L ) * h-bar^2 * sin ( pi * x / L ) * -sin ( pi * x / L ) * ( pi / L ) ^ 2
= ( 2 * pi ^ 2 / L^3 ) * h-bar^2 * sin 2 ( pi * x / L ) = pi ^2 * h-bar^2 / L ^ 2, which is (a)*2m
Plugging in the numbers yields same answer as part (b.) ~ 1.1e-55 J-kg or kg^2*m^2/s^2
(d.) Ψ 3 = sqrt ( 2 / L ) * sin ( 3 *pi * x / L )
Ψ 3* = sqrt ( 2 / L ) * sin ( 3 *pi * x / L ), same thing because no imaginary numbers present.
Thus, the normalization simplifies into ∫ |Ψ|^2 = ∫ Ψ^2 = ∫ (2 / L) * sin2 ( 3 *pi * x / L ) dx =
= (2 / L) * ∫ sin2 ( 3 *pi * x / L ) dx, pulling the constant out of the integral. This looks like
the first equation from above, so we use that, and our integration limits are 0 and L (not ∞).
A u-substitution of sin^2 u du where u = 3 *pi * x / L and thus du = 3 *pi * dx / L and dx =
du*L / (3 *pi) => ∫ Ψ^2 = ( 2 / L ) * ( L / 2 ) = 1 (dimensionless)
(e.) Here we need to integrate ∫ Ψ^2 again, but with bounds of 3L/4 to L.
∫ Ψ^2 dx = (2 / L) ∫ sin2 ( 3 *pi * x / L ) dx, so we use the first integral down in the list and
get (2 / L) * L * (3*pi + 2) / (24 * pi) = (3 pi + 2)/(12*pi) ~ 30%, not quite the 25% classical