|All Solutions of Type: Quantum Mechanics|
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Spin explains a lot of things.
(A) Remember orbitals? Whether a shell is full or not determines the properties of each column of the periodic table. A full shell has all electron spins paired together, while a partially filled or empty shell doesn't have that. So, spin is definitely in the Periodic Table.
(B) The specific heat of metals differs if one calculates it using the Fermi-Dirac or Bose-Einstein distributions; the first is used for fermions and the second for bosons. So, spin plays a role here.
(C) The Zeeman effect has to do with splitting caused by spin.
(D) The deflection of a moving electron is due to the magnetic field contribution to the Lorentz Force. This is a classical non-spin related phenomenon, on first analysis. This is the best choice.
(E) Fine structure has to do with splitting caused by spin.
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Quantum Mechanics}Photoelectric Effect
Perhaps one recalls the ``hot phrase" stopping potential. The used in the photoelectric equation is essentially the stopping potential (energy). This memory-recall immediately narrows the choices down to just (A) and (C). Now, to find out if the potential is positive or negative.
FYI: The Photoelectric effect equation is basically just conservation of energy. One has , where is the kinetic energy of the electron as it accelerates through the medium between the cathode and collector, is the energy to free the electron from the metal, and is the energy given by the light source. Essentially, the energy from the photon first frees the electron from a sort of (valence) electron sea it lives in while in the metal, and then the excess energy propels it from cathode to collector in order to keep the current running. The minimum kinetic energy required to get the current going is the stopping potential (energy) . (According to the fight analogy below, the photon is the one who starts the fight, socking the electron off and away across the currents...)
The electron charge is negative and the potential must be a negative quantity in order to make positive overall.
(On the lighter side... The Photoelectric effect is also related to the Compton Effect. The effects can be seen as an arena fight. Think WWWF, but with electrons and photons. In PEE, the photon knocks the electron out, while in CE, the electron retaliates! It knocks the photon off course.)
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Quantum Mechanics}Photoelectric Effect
(A) The photoelectric effect was derived before formal quantum mechanics and that angular momentum mess came along. Moreover, electron orbits don't really apply to the valance electron sea.
(B) This is true, but it doesn't help derive the photoelectric effect.
(C) Nah, there's also something weird called electron-electron annihilation. Basically, two electrons crash into each other and a photon is created. (Perhaps some other particles, too.) Also, think of your regular desktop lamp---light is emitted, but the electrons are probably not jumping between orbits.
(D) Right. Einstein won the Nobel Prize about a hundred years ago via his proposal that photons have quantized energy .
(E) As a pure-ideal theory, the photoelectric effect depends on a single photon exciting a single electron. It favors the particle view of light. Choice (E) is out.
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Quantum Mechanics}Photoelectric Effect
The quantity is also known as the work function, and it's the minimum amount of work that has to be done to free the electron from the electron sea. If one forgets whether it's the electron or photon that's being moved, then here's a useful mnemonic. Remember that the photoelectric effect has a photon knocking out an electron and the Compton effect has an electron knocking a photon off course. (For the full fight-club analogy, See Prob 32)
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Quantum Mechanics}Compton Effect
This problem can be solved via the Compton Effect equation , where is the angle of the scattered photon and is the so-called Compton Wavelength. In this case, since the particles are protons, one has .
Since the photon scatters off at 90 degrees, the equation simplifies to . This is the increase in wavelength, as in choice (D).
To a certain degree, one can hand-wave this problem via the following method:
Recall the de Broglie relation,
, where is the wavelength, is the momentum and is Planck's constant.
The momentum of the proton is , thus .
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Quantum Mechanics}Franck-Hertz Experiment
Like the typical experiment in QM, the Frank-Hertz experiment has to do with shooting a bunch of particles through some oven. Its significance is in early observations of energy levels. Also, its key conclusion is that electrons are scattered elastically.
But, if one doesn't know the above, one can always use MOE---the Method of Elimination.
(A) Elastic collision requires conservation of kinetic energy. A bit restraining, but keep the choice.
(B) Never scattered elastically? Never is too strong a word to be favored by ETS.
(C) Seems reasonable-ish. Keep it.
(D) This is, again, too restraining. It makes sense that depending on the impact angle and momentum, different amount of energy would be lost.
(E) Discrete energy lost is mentioned in (C), but again, this choice is much too restraining, stating that ``there is no energy range..."---too strong of a phrase to be favored by ETS.
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Quantum Mechanics}Identical Particles
According to Griffiths, the proof to this rule comes from QFT,
where it's for bosons and for fermions.
The given identical particle wave function contains a plus sign, so the particles must be bosons. Bosons have integer spin (while fermions have half-integer spin). Electrons, protons, and neutrons are all fermions. A positron is just a positive electron, so that is presumably, also, a fermion. Thus, the remaining choice would be deuteron---which is a boson.
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Technically, one can solve this problem without knowing anything about perturbation theory. Just remember the useful fact that the eigenfunctions of the unperturbed infinite deep well form a complete set (or complete basis). This means that any function can be represented by the old unperturbed infinite deep well. Thus, the solution to the perturbed eigenfunction should look like .
Also, since the perturbed potential is also symmetrical with respect to the origin (as the original unperturbed potential was, too), one knows that all the odd terms should go to 0. This leaves choice (B).
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The careless error here would be to just directly square the grids. When one remembers the significance of the meaning of the probability , one finds that one must square the wave function, and not the grids.
The total probability is,
The un-normalized probability from to is,
The normalized probability is thus , as in choice (E).
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Quantum Mechanics}Infinite Well
The even wave functions always have nodes in the middle, and thus the probability density for states of even n vanish. (One can deduce this by fitting curves inside a box. The first state has no nodes in the middle, but a node at each end of the well. The second state has one node in the middle. The third state has two nodes, neither of which are in the middle. The fourth state has three nodes, one of which is in the middle.)
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One can make a good stab at this problem by applying the uncertainty principle.
I. If the average momentum of the packet is 0, then one violates the uncertainty principle. See IV.
IV. True, recall the Gaussian uncertainty principle .
Since I is false, choices (A), (C), and (E) are out. Choices (B) and (D) remain. Take the conservative approach and choose (B).
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Quantum Mechanics}Probability Density Current
If one has forgotten the expression for the probability density current, then one needs not despair! If one remembers the vaguest definition of a ``probability current," one can solve this problem without the usage of the forgotten formula:
Recall that the probability current density J is the difference between incoming P(in) and outgoing P(out) probability densities. This is just the difference between the probability densities of a rightwards moving plane wave and a leftwards moving plane wave, since the probability density is related to the wave function by
The given wave function can be written in terms of plane waves,
gives the coefficient for the leftwards (negative-direction) wave, while gives the coefficient for the rightwards traveling wave. The probability density for each wave is given by,
Plugging these quantities into the formula for the probability current density above (J), one gets,
which is choice (E).
The formal expression for the probability current density can be effortlessly derived from recalling the definition of probability and Schrodinger's Equation---both of which every physics (or engineering) major should know by heart.
Probability is defined (in the Born Interpretation) as . One should recall that in general (to wit: the absolute value squared of a complex expression is itself times its complex conjugate).
The time-dependent Schrodinger's Equation is
where has the form of the familiar time-independent Hamiltonian. From this, one finds that .
Generalizing the idea of a current from classical physics to the idea of a probability current, one takes the time derivative of the probability to get , where the product-rule for baby-math derivatives has been used and the derivative has been taken inside the integral because the integral and derivative are with respect to different variables.
Plugging in the expression for from the Schrodinger's Equation, one gets
where the terms involving V's cancel out, and thus,
Rewriting , one can eliminate the integral in the probability current by applying the fundamental theorem of calculus (to wit: ),
. But, since the probability current is usually define as , one has
(Aside:) One can print-out a cool poster or decent T-shirt iron-on to remember the Schrodinger's Equation (among other miscellanai) at a site the current author made several years ago,
One can remember the general form of the probability current by recalling that it has to do with the difference of times its conjugate.
Right, so onwards with the problem:
The problem gives the wave function, so one needs just chunk out the math to arrive at the final answer,
, where one notes that the imaginary terms go to unity from the complex conjugate.
Plugging this into the probability current, one arrives at the expression for choice (E).
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Quantum Mechanics}Two-State Systems
Recall the mnemonic for remembering what a LASER is---Light Amplified Stimulated Emission Radiation.
A laser consists of two states, with a metastable-state in between the top and bottom state. Initially, one has all the atoms in the ground-state. But, photons come in to excite the atoms (through absorption), and the atoms jump into the top state; this is called a population inversion, as the ground-state atoms are now mostly in the top ``inverted" state. More photons come in to excite these already excited atoms, but instead of absorption, emission occurs, and the atoms jump to a lower meta-stable state while emitting photons (in addition to the incident photons). The atoms stay in this metastable state due to selection rules, where a transition back to the ground-state is forbidden.
One doesn't need to know all that to solve this problem. Instead, merely the idea of a laser requiring two main states and a metastable state in between would suffice. Since the question gives the bottom state as and top state as , one deduces that the metastable state must be , as in choice (B).
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Quantum Mechanics}de Broglie Wavelength
Since and the de Broglie wavelength gives , one can find the initial wavelength . This yields an expression for
When the particle enters the well, its energy becomes , and thus .
Plugging h in, one has , which is choice (E).
(This above solution is due to the lout Marko.)
Also, the current author's original approach leads to the right result, but it hand-waves the massiveness of the particle:
The initial kinetic energy of the free particle is its total energy . But, since one's dealing with de Broglie, one has . Thus, .
In the potential, its energy becomes . Replace E above with that to get . But, from the same relation above, one has . Thus, . This is choice (E).
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Quantum Mechanics}Addition of Angular Momentum
For 2 electrons, one recalls that the there are three spin triplet states and one spin singlet state (hence their names). One can apply the lowering operator multiple times to the up-up state to arrive at, respectively, the up-down state, the down-up state, and then the down-down state (the spin-singlet state).
Applying the lowering operator to choice I, one gets choice III. Applying the lowering operator to that, one gets down-down. These are the 3 spin-triplet states.
By orthogonality, the down-down state has a negative sign. So, only choices I and III are in the triplet configuration.
(David J. Griffiths vanity alert---the QM problems thus far are all straight out of his textbook, An Introduction to Quantum Mechanics.)
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Quantum Mechanics}Perturbation Theory
The energy for first-order perturbation theory of ( is the known Hamiltonian and is the perturbed Hamiltonian) is given by , where the wave-functions are the unperturbed ones.
Thus, the problem amounts to calculating . This is just raising and lowering operator mechanics.
. But, after bra-ketting, one finds that the expectation value of and are 0, since and , are orthogonal. Thus, the problem becomes,
. Applying the given eigen-equations, one finds that . For , one finds , as in choice (E).
(Note that: and and .)
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Quantum Mechanics}Bound State
Tunneling should show exponential decay for a finite-potential well, and thus choice (E) is eliminated. Choice (C) is eliminated because the wave function is not continuous. One eliminates choice (D) because the bound-state wave functions of a finite well isn't linear. The wave function for a bound state should look similar to that of an infinite potential well, except because of tunneling, the well appears larger---thus the energy levels should be lower and the wave functions should look more spread out. Choice (B) shows a more-spread-out version of a wave function from the infinite potential well.
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If one forgets the energy of an infinite well, one can quickly derive it from the time-independent Schrodinger's Equation . However, since inside, one has .
Plug in the ground-state wave function , where . Chunk out the second derivative to get . Plug in k to get .
Note that can be deduced from boundary conditions, i.e., the wave function vanishes at both ends ( and ). The second boundary condition forces the n's to be integers. Since one can't have a trivial wave function, , and thus . One finds that , since , as in choice (E).
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