Download 36 POINTS - University at Albany

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Physics 240, Exam #3
Tuesday, Nov. 24, 2015 (4:15-5:35 p.m.)
36 POINTS (PICK 3 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 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 of the last problem, or request additional sheets from me.
IF YOU DO ALL FOUR PROBLEMS, ONLY YOUR TOP THREE WILL BE SCORED.
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1. CONCEPTUAL QUESTIONS (12 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. (2 points)
(b.) Name and explain one of the few features of quantum mechanics that still follows
“common sense.” (2 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. (2 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? (2 points)
(e.) Why are chemical elements on the periodic table grouped and ordered the way they
are? What causes some different elements on it to possess similar properties? (2 points)
(f.) Describe one of the properties of a laser beam that distinguishes it from light normally
found in nature. (2 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.) By similar properties; usually the numbers of electrons in the outermost (“valence”)
shells cause similar chemical properties by group. Ordered by increasing Z downward vert.
(f.) spatial/angular coherence (tight beam in 1 direction), phase coherence (waves in same
phase), monochromatic/monoenergetic (all photons exactly the same energy or color), can
be used to make holograms (ordinary light can illuminate holograms but not generate
them), powerful ones can cause physical damage to materials through energy transfer
2. BASIC QUANTUM MECHANICS RAPID-FIRE QUESTIONS (12 points)
(a.) How slow does an electron have to be moving to have an effective wavelength of exactly
1 mm? (2 points)
(b.) Demonstrate that the group velocity of a photon is always the vacuum speed of light c.
(2 points)
(c.) For how long can a violation of conversation of energy occur that is the same amount as
the seventh excited state of a hydrogen atom? (2 points)
(d.) If we are talking about a proton in part (c.) then what is its uncertainty in position, and
how small is that compared to the Bohr radius? (3 points)
(e.) Show that A * exp [ i * ( k * x - ω ∗ t) ] satisfies the time-dependent Schrodinger wave
equation. (3 points)
(a.) Wavelength = Planck’s constant / momentum, so momentum is Planck over wavelength
p = h / λ = 6.626e-34 J-s / 1 mm = 6.626e-34 J-s / 1e-3 m = 6.626e-31 kg*m/s
p = m*v (assume non-relativistic) so v = p / m = 6.626e-31 kg*m/s / 9.109e-31 kg = 0.7 m/s
(b.) v_group = dω / dk = dE / dp but E = p * c for a photon, so v_group = d(pc)/dp = c
(c.) Seventh excited state means n = 8 (ground state is n = 1). | E_n | = 13.6 eV / 8^2 =
(13.6 / 64) eV = 0.2125 eV = 3.4e-20 Joules
delta-E times delta-t is greater than or equal to h-bar over 2, so
Δt <= h-bar / (2*ΔE) = [h / (2 * pi)] / (2 * ΔE) = 6.626e-34 J-s / (4*3.14*3.4e-20 J) = 1.5e-15 s
(d.) Convert from energy into momentum. E = p^2/2m, so p = sqrt(2mE), classically (E<<m)
p = sqrt ( 2 * m_p * 3.4e-20 J ) = sqrt ( 2 * 1.67e-27 kg * 3.4e-20 J ) = 1.066e-23 kg * m/s.
Call this delta-p; product of delta-x and delta-p must be greater than or equal to h-bar/2
Δx <= h-bar / (2*Δp) = [h / (2 * pi)] / (2 * Δp) = 6.626e-34 J-s / (4*3.14*1.066e-23 kg * m/s)
= 5e-12 m or 5 pm (picometers) = almost exactly 1/10 of Bohr radius
(e.) Please refer to Example 6.2 on pg. 204 of Chapter 6, Quantum Mechanics II. Section 6.1
3. ADVANCED QUANTUM MECHANICS QUESTIONS (12 points)
(a.) Determine the expectation value of p2 for a particle in an infinite square well for the
third excited state. You can use either the operator method or energy equation. (4 points)
(b.) Normalize the wave function for the second excited state of the quantum simple
harmonic oscillator. (5 points)
(c.) What is the %-chance of success for a water molecule (mass of 3 x 10-26 kg) with energy
0.9 eV to surmount a potential barrier of potential energy 1 eV? (3 points)
(a.) Third excited state means n=4 (because n=1 is ground state. n=0 means no
wavefunction). Method 1: Momentum operator p-hat = (- i * h-bar) * (d/dx) partial deriv.
So, p^2 = (- i ) ^ 2 * (h-bar)^2 * (d/dx)^2 = - h-bar^2 * d2/dx2.
Ψ 4 = sqrt ( 2 / L ) * sin ( 4 *pi * x / L ) where n=4
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 ( 4 *pi * x / L ) (-h-bar^2) * d2/dx2 [sqrt ( 2 / L ) * sin ( 4 *pi * x / L ) ]
= ( -2 / L ) * h-bar^2 * sin ( 4 *pi * x / L ) * -sin ( 4 *pi * x / L ) * ( 4 * pi / L ) ^ 2
= ( 32 * pi ^ 2 / L^3 ) * h-bar^2 * sin 2 ( 4 *pi * x / L ) = 16 * pi ^2 * h-bar^2 / L ^ 2
Method 2 (the easy way):
En = ( n^2 * π^2 * h-bar^2 ) / ( 2 * m * L^2 ) so for n=4
E4 = 16 * π^2 * h-bar^2 / ( 2 * m * L ^2) but if E = p^2 / 2m just multiply by 2m; you have
16 * π^2 * h-bar^2 / ( L ^ 2 ), the exact same result as with method #1 above. Both methods
OK for full credit. Note this is extremely similar to Example 6.8 in your textbook. This is
also the same as 4 * h^2 / ( L ^ 2 ) substituting for h-bar=h/(2*pi) or using the energy
formula for particle in classical box (Ch. 5) instead of an infinite square well, same thing
(b.) The second excited state is n = 2 since n = 0 is the ground state in this case.
Ψ 2 = A * (2 * a * x^2 – 1) * exp (- a * x^2 / 2), using the second Hermite polynomial. The
integral from –infinity to infinity of Ψ 2 ^2 must be 1
Integral from –infinity to infinity of A^2 * (2 * a * x^2 – 1) ^ 2 * exp ( - a * x^2 ) OR A^2 *
[4 * a^2 * x^4 * exp ( - a * x^2 ) – 4 * a * x^2 * exp ( - a * x^2 ) + exp ( - a * x^2 )]. Using the
integral tables (more specifically both equation A6.2 and page A-11, Appendix 6) it’s A^2 *
[8 * a^2 * 3 / (8*a^2) * sqrt ( π/a ) – 4 *a*(2 / 4a)*sqrt ( π/a ) + (2/2) *sqrt ( π/a ) ] = (3 – 2 +
1 ) * sqrt ( π/a )*A^2
So A^2 * 2 * sqrt ( π/a ) = 1 and A^2 = sqrt ( a / π) / 2 => A = (a / π) ^ (1/4) * (1/sqrt(2)),
which is perfectly consistent with p. 224 in Chapter 6, top right (c.) κ = sqrt ( 2m * ( V0 – E ) ) / h-­‐bar (from second side of formula sheet near top)
= sqrt ( 6e-­‐26 kg * ( 1 eV – 0.9 eV ) ) / 1.05e-­‐34 J-­‐s = 2.94e11 m^-­‐1 or 1/m (1 eV=1.602e-­‐19 J) T = [ 1 + V0^2 * sinh2 ( κ * L ) / ( 4 * E * ( V0 – E ) ) ] ^ -­‐1
I neglected to provide “L” so as long so you solved for κ and wrote the correct formulae
down then you got as far as you possibly could with the information I provided and get full
credit
If for example L = 0.01 nm = 0.01e-9 m = 1e-11 m (my original intention) and you plug in
for V0 = 1.0 eV and κ as calculated above and L = 1e-11 m 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)!
4. THE HYDROGEN ATOM (12 points)
(a.) Demonstrate that the radial wavefunction for the M shell (hint: convert that to n) and d
subshell is normalized. (2 points)
(b.) What are all of the possible values of the quantum numbers (n, l, ml, ms) for this state?
How many different possible states total are thus represented here? (2 points)
(c.) Name a transition to either a lower or higher energy state from this one that follows the
selection rules and is thus allowed. (2 points)
(d.) What is the probability of the radius being between the Bohr radius and twice the Bohr
radius for this scenario? (2 points)
(e.) An external magnetic field is now turned on. To calculate the energy splitting you need
the Bohr magneton. What is its value? (2 points)
(f.) What is the change in energy for the maximum value that ml can take on, for a magnetic
field of 15 Tesla? What is the corresponding wavelength for this energy? (2 points)
(a.) If you look at your formula sheet, you will see this was a simple lookup exercise. The M
shell means n=3 and the d subshell means that l = 2. So we need to look at hydrogen radial
wavefunction from Table 7.1 on p. 247 Section 7.2 of
R32(r) = (1/a03/2) * (4/(81*sqrt(30)))* (r2/a02) * exp ( -r / 3a0)
Integral over dr from 0 to infinity (there is no such thing as negative 3D radius of course) of
r^2*R^2 = r^2 * (1/a03) * (16/(6561*30))* (r4/a04) * exp ( -2r / 3a0)
= ( r^6 / a0^7 ) * ( 8 / 98415 ) * exp ( -2r / 3a0). Use p. A-5 from Appendix 3, Math Relations
= ( 1 / a0^7 ) * ( 8 / 98415 ) * 6! * [( 3 / 2 ) * a0] ^ 7
= ( 1 / a0^7 ) * ( 8 / 98415 ) * 720 * ( 2187 / 128 ) * a0 ^ 7 = 1
(b.) This is a 3d state so n=3 and l=2. Thus ml = -2,-1,0,1,2 and ms is +/- ½ as always. There
are 10 total states possible (5 ml’s times spin up or spin down electron). Can also calculate
that with 2*(2*l+1) = 2*(2*2+1) = 10 (equation from bottom of p. 274 in table under ‘Total’)
(c.) The change in n can be anything, so 1,2,4,… and delta-l must be +/-1, so from d either p
or f (l = 1 or l = 3). (No need to quote ms, ml, mj, j, or, s here for full credit. Optional.)
(d.) “This scenario” is supposed to refer all the way back up to (a.) but if you misunderstood
and used part (c.) then you still get full credit. Let’s do it the way I intended with part (a.) as
then we have a unique solution.
P (r=a0 to 2*a0) = Integral from a0 to 2*a0 of r^2*R^2 = r^2 * (1/a03) * (16/(6561*30))*
(r^4/a04) * exp ( -2r / 3a0) = ( r^6 / a0^7 ) * ( 8 / 98415 ) * exp ( -2r / 3a0) copying from part a.
above. Use first equation from upper right of A-5, Appendix 3, or symbolic integration on
calculator to get 4.6e-4 = 4.6e-2%
(e.) See equation 7.32 on p. 255 of Chapter 7. µ B = e * h-­‐bar / (2me-­‐) = (1.602e-19 C * 1.05e34 J-s) / (2*9.109e-31 kg) = 9.27 e-24 J/T or A-m^2 (Joules per Tesla is same as Amperes of
current times m^2 of surface area in SI units)
(f.) ΔE = VB = µ B ml B = 9.27e-24 J/T * (2) * (15 tesla) = 2.7822e-22 J = 2e-3 eV. E = h*c/λ,
so λ = h*c / E = (6.626e-34 J-s) * (299,792,458 m/s) / (2.78e-22 J) = 7e-4 m ~ 0.7 mm (border
between infrared and microwaves). Full credit if you solved instead for change in λ which
based on homework problem #21 on p. 269 is: Δλ = λ0^2 ∗ µ B * B / (h*c) where the
original energy is for 3d so 13.6 eV / 3^2 = 1.51 eV so λ0 = (h*c) / (1.51 eV) = 8.2e-7 m, so
Δλ = 4.71e-10 m ~ 0.5 nm which is an x-ray (alternative interpretation of question)