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All Solutions of Type: Advanced Topics
 0 Click here to jump to the problem! GR8677 #16 Advanced Topics$\Rightarrow$}Particle Physics The properties of a muon most closely resembles that of an electron. In the usual $\beta$-decay, an electron is emitted along with either a neutrino or an antineutrino. In a wildcard $\beta$-decay, a muon is emitted instead of an electron. Click here to jump to the problem!
 1 Click here to jump to the problem! GR8677 #17 Advanced Topics$\Rightarrow$}Radioactivity $X^A_Z\rightarrow ? \rightarrow ? \rightarrow Y^{A-4}_{Z-1}$ by natural decay Recall the following decay processes, and recall that the subscripts and superscripts have to balance on both sides of the arrow. Beta ($\beta^-$) Decay: $X^A_Z\rightarrow X'^A_{Z+1} + \beta^0_{-1} +\bar{\nu}, $ where $\bar{\nu}$ is the symbol of an anti-neutrino. (The $\beta^+$ decay is accompanied by a neutrino, $\nu$, instead of an anti-neutrino.) Alpha Decay: $X^A_Z\rightarrow X'^{A-4}_{Z-2}+He^{4}_{2}$ Gamma Decay (no change in A or Z): $X^A_Z\rightarrow X^A_Z\rightarrow + \gamma $ Deuteron Decay (rare): $X^A_Z\rightarrow X^{A-2}_{Z-1}+H^2_1$ (A) $X^A_Z\rightarrow X^A_{Z+1} + \beta^0_{-1} \rightarrow X'^{A-4}_{Z-1} +He^4_2 \rightarrow Y^{A-4}_{Z-1}$. This works out. (B) $\beta^-$ decay always occurs with an antineutrino. (C) $\gamma$ emission doesn't change the atomic number. (D) Although the numbers work out here, deuteron decay is rare, thus not natural. (E) The subscripts and superscripts don't add up. Click here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #23 Advanced Topics$\Rightarrow$}Solid State Actually, one can figure this one out with only knowledge of lower div baby physics. (A) Electrical conductivities for conductors, semiconductors, and insulators go (in general), like this $k_c> k_s > k_i$. Thus, copper should be more conductive than silicon. This is true (but one is trying to find a false statement). (B) The resistivity $\rho = \rho_0 (1 + \alpha \Delta T)$, thus the conductivity, $\sigma =1/\rho\propto 1/(\Delta T)$. As $T$ increases, $\sigma$ decreases. This is true. (C) Silicon is a semi-conductor, and thus it probably does not follow the same relations as (B). In fact, semiconductors have negative $\alpha$ temperature coefficient of resistivities. Thus, $\rho < \rho_0$, which implies that $\sigma > \sigma_0$ for temperature increase. (D) Doping a conductor like copper will just make it cheap. Think of cheap wire. (E) Doping a semiconductor, however, can make it more conductive. Click here to jump to the problem!
 3 Click here to jump to the problem! GR8677 #38 Advanced Topics$\Rightarrow$}Logical Circuit (A) This is the OR gate. Triangle with fat-end on input side denotes OR. (B) Triangle with fat-end closer to output side denotes AND. (Pointy tip points to each input.) (C) A 2 bit-adder involves more operations than this... (D) A flip-flop is a sort of sandal that flips and flops. It might also flip the floating point. (E) A fan-out describes the maximum number of outputs a circuit can excrete. (Fan-in would be inputs.) Click here to jump to the problem!
 4 Click here to jump to the problem! GR8677 #40 Advanced Topics$\Rightarrow$}Radioactivity Radioactive counting rates (number of decay per unit time) follow the Poisson Distribution. (Recall that the Poisson Distribution describes the results of experiments in which one counts events that occur independently, thus at random, but at a definite average rate.) In the PD, $\sigma=\sqrt{\bar{x}}$, i.e., the standard deviation is approximately the square-root of the average. Suppose $9934\approx 10000$ is the average. The square root of that is $100$, hence (A) is the answer. (Depending on whether one is in the mood for rolling the celestial self-loaded 5-sided dice, one can look for the most obvious relationship between numbers in problems. $\sqrt{x}, x^2$ are two commonly used relationships.) The Normal (Gaussian) distribution describes the distribution of values for any measurement subject to many sources of error that are all random and small. But, the Poisson Distribution fits this case more closely. Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #50 Advanced Topics$\Rightarrow$}Solid State Physics Recall $R_H=-\frac{1}{nec}$, where the sign shows that the charge carrier is negative. ($n$ is the electron density, $e$ electron charge, $c$ speed of light) The Hall effect has to do with a bar of metal placed in a magnetic field. An electric current runs through it perpendicular to the field direction. However, the magnetic field term in the Lorentz force creates a force perpendicular to both the magnetic field and electric field. This pulls" the current astride from its original straight path. Charge starts accumulating on one side of the slab from this force, even as current continues to flow perpendicular to the accumulation direction. In equilibrium, the magnetic force balances the electric force from this charge accumulation. Apply some basic EM theory, and one can derive the Hall coefficient ($R_H$) defined above. Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #85 Advanced Topics$\Rightarrow$}Nuclear Physics Pair production only occurs above a certain energy on the order of MeV. Thus, all except choices (B) and (C) remain. The photoelectric effect is dominant for low energies, so its cross-section must be line (1). Choose choice (B). Click here to jump to the problem!
 7 Click here to jump to the problem! GR8677 #10 Advanced Topics$\Rightarrow$}Particle Physics Recall that in gamma-ray production, the excited nucleus jumps to a lower level and emits a photon $\gamma$. In internal conversion, however, an orbital electron is absorbed and ejected along with an X-ray. Click here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #29 Advanced Methods$\Rightarrow$}Dimensional Analysis The current author is fortunate enough to have taken a String Theory course as an undergraduate, and thus know by heart that the Planck length is $\sqrt{G \hbar /c^3}$. However, the problem can also be solved via dimensional analysis (A) $G\hbar c$ has units of $\left(m^3/(kgs^2)\right) (kg m/s^2 m s) (m/s) =m^5/(s^3)$, which doesn't have the units of $m$. (B) $\left(m^3/(kgs^2)\right) (kg m/s^2 m s)^2 (m/s)^3=m^{10} kg/s^7$, which doesn't have the units of $m$. (C) $\left(m^3/(kgs^2)\right)^2 (kg m/s^2 m s) (m/s)=m^9/(kg s)$, which doesn't have the units of $m$. (D) $\left(m^3/(kgs^2)\right)^{1/2} (kg m/s^2 m s)^2 (m/s)=\sqrt{kg}m^{4.5}/s^3$, which doesn't have the units of $m$. (E) This is the last one. Take it! Click here to jump to the problem!
 9 Click here to jump to the problem! GR8677 #30 Advanced Topics$\Rightarrow$}Fluid Mechanics Equipotential leads to equipressure in a fluid. Thus, take the pressure at the base of the dark fluid and set it equal to the pressure (of the lighter-colored fluid) at a horizontal-line across on the right-hand side of the U: $P_{dark}=\rho_4 g (5) = P_{light}= \rho_1 g (h_2 - (h_1-5))=\rho_1 g (h_2-h_1+5)$. The initial total height of the columns is 40, thus after the darker liquid is added, the total height is 45. Plug $h_1+h_2=45$ into the equation above to get $h_1=15$, $h_2=30$, and therefore $h_2/h_1=2/1$, as in choice (C). (Ah, one should remember that the fluid pressure at a point is due to all the water on top of it, thus $P=\rho g h$, where $h$ is the height of the water on top of the point.) Click here to jump to the problem!
 10 Click here to jump to the problem! GR8677 #34 Advanced Topics$\Rightarrow$}Particle Physics Regularly, electrons are emitted in any direction. Thus, there is infinite symmetry. In the case of a magnetic field, electrons are more likely to be emitted in a direction opposite to the spin direction of the decaying atom. Place the atom in an x-y plane, with its spin-direction pointing along the z-axis. If the electron is mostly emitted in the -z axis, then reflection symmetry is violated since it's not (mostly) emitted in the +z axis, i.e., not mirrored across the x-y plane. Choice (D). (This is due to Joe Bradley.) Click here to jump to the problem!
 11 Click here to jump to the problem! GR8677 #53 Advanced Topics$\Rightarrow$}Particle Physics One can ignore baryon numbers and lepton numbers and all that and just deal with spin conservation. For the positronium-electron spin singlet state, one has, initially, $s_i=0$. The decay must conserve spin. Thus, one must have the final spin as $s_f=0$. Since a photon is its own antiparticle (and antiparticles have the negation of the usual particle's quantum number), the photon has spin $s_p=1$ and the antiphoton (just another photon) has spin $s_{\bar{p}=-1$. Thus, two photons are emitted to conserve spin. (Wheee... can one get more ad hoc than the Standard Model?) Click here to jump to the problem!
 12 Click here to jump to the problem! GR8677 #63 Advanced Topics$\Rightarrow$}Particle Physics If one knows little about particle physics, one can make (at least) the following deductions, wherein one recalls the composites of each particle: (A) A muon is a lepton. Leptons, along with quarks, are considered the fundamental particles. (B) Pi-Meson consists of a quark and its antiparticle. (Contribution to this part of the solution is due to user danty.) Moreover, a pi-meson is a hadron. Hadrons interact with the strong-force, and all of them are composed of combinations of quarks. (The fundamental particles are classified as quarks and leptons.) (C) A neutron is made up of 3 quarks. (D) A deuteron consists of a proton and a neutron. (tritium is two neutrons and a proton, while regular Hydrogen is just an electron and proton) (E) An alpha particle consists of electrons and protons and neutrons. Choice (A) remains. Choose that. If one has some time, one might want to remember the elementary particles involved in the Standard Model. There are six quarks and six leptons. Three of the leptons are neutrinos and the other three are the electron, the tau, and the muon. (Also, in a decay similar to beta-decay, a muon is emitted instead of an electron. Charge conservation works since a muon is like an electron except it is about 200 times more massive.) Wikipedia has a good reference on this: \begin{verbatim} http://en.wikipedia.org/wiki/Fundamental_particle \end{verbatim} Click here to jump to the problem!
 13 Click here to jump to the problem! GR8677 #64 Advanced Topics$\Rightarrow$}Nuclear Physics In symmetric fission, the change in kinetic energy is just the change in binding energy. The change in binding energy for a $N-nucleon$ heavy nucleus is the difference in energy between the initial un-fissioned heavy nucleus and the final 2 medium-sized nuclei, $\Delta E = 2\times 0.5 N \times 8MeV/nucleon - N \times 7MeV/nucleon = N \times 1MeV/nucleon$. For a heavy nucleus, one has $N\approx 200$, and thus one arrives at choice (C). (This is due to David Schaich.) Click here to jump to the problem!
 14 Click here to jump to the problem! GR8677 #67 Advanced Topics$\Rightarrow$}Schwarzchild Radius Using quite a bit of handwaving, the current author has seen an astrophysicist derive the Schwarzchild Radius (radius at which the curvature of space mooches and eats up light completely) via $1/2 m c^2 = G M m/r$. (The guy totally neglected relativity, assuming that kinetic energy has the same form in the relativistic regime, but anyway...) Handwaving like a good astrophysicist, one finds that $r = 2 G M/c^2 \approx 2\times 7E-11 \times 6E24/9E16 = 1 cm$. (Note: the author is currently declared as an astrophysics major, and if the above comment is to be interpreted as pejorative towards astrophysicists, then she has thus implicitly insulted herself.) Click here to jump to the problem!
 15 Click here to jump to the problem! GR8677 #78 Advanced Topics$\Rightarrow$}Solid State Physics A n-type semiconductor is a material with negative-charge carriers, such as electrons. A p-type semiconductor is a material with positive-charge carriers, such as holes (positrons). In band theory, n-type semiconductor impurities are (electron) donors, while p-type semiconductor impurities are (electron) acceptors. The setup is as follows: Impurities add in more levels to the energy bands. Without impurities, one has just a valance band and a conduction band with an energy gap in between. The impurities supply an extra energy level in between the conduction and valance bands. In an n-type semiconductor, the material becomes conducting when there are electrons in the conduction band; the impurity helps the material become conducting by supplying it with electrons. Essentially, one starts with a lattice of pure semiconductor atoms, say Silicon. Silicon has four valance electrons and forms a decent crystal lattice. Pluck out a few silicon atoms and replace them with some impurities, like Arsenic, which five valance electrons. The extra electron from each impurity atom is free to roam around. In fact, these extra electrons act as donors to the conduction band. This is choice (E). Click here to jump to the problem!
 16 Click here to jump to the problem! GR8677 #95 Advanced Topics$\Rightarrow$}Solid State Physics The specific heat of a superconductor jumps at the critical temperature (c.f. with its resistivity jump). Ordinarily, the specific heat of a metal is $c=aT+bT^3$. When it is superconducting, the first term, the electronic-contribution, is replaced by $\approx e^{-cT}$. The revised plot of the specific heat has a jump from an exponentially increasing specific heat to a much lower value somewhere in the range for positive T. Reference: Ibach and Luth p 270ff Click here to jump to the problem!
 17 Click here to jump to the problem! GR8677 #19 Advanced Topics$\Rightarrow$}Astrophysics It takes 4 H's to create a Helium nucleus in the sun's primary thermonuclear reactions. One either remembers this or can derive it from conservation of mass. The atomic mass of Hydrogen is 1 (since it has just 1 proton), while the atomic mass of Helium is 2 (2 protons, 2 neutrons). Click here to jump to the problem!
 18 Click here to jump to the problem! GR8677 #52 Advanced Topics$\Rightarrow$}Solid State Physics Recall the definition of a primitive cell to be the unit cell divided by the number of lattice points in a Bravais lattice. A Bravais lattice is just a lattice that looks isotropic from any point---everywhere the same no matter the point-perspective. Simple cubic has 1 lattice point to generate its Bravais lattice. Body-centered cubic has 2 lattice points to generate its Bravais lattice. (One can keep on tesselating the a lattice point on one corner and the lattice point on the body center to generate the whole BCC lattice.) Face-centered cubic has 4 lattice points to generate its Bravais lattice. (One can keep on tesselating the lattice point on each face surrounding a corner lattice point.) So, anyway, from the above, one finds choice (C) for BCC's unit cell. Click here to jump to the problem!
 19 Click here to jump to the problem! GR8677 #53 Advanced Topics$\Rightarrow$}Solid State Physics From Ibach and Luch, one finds that the resistivity of a semiconductor varies as $1/T$. (Also, from elementary electrodynamics, one recalls the resistivity equation $\rho(T_2) = \rho(T_1)(1+\alpha \Delta T)$. Semiconductors have a negative coefficient of resistivity $\alpha$, and thus the resistivity should decrease with increasing T. The only graph that shows this behavior of decreasing resistivity with T is (B).) Click here to jump to the problem!
 20 Click here to jump to the problem! GR8677 #66 Advanced Topics$\Rightarrow$}Radioactivity The problem supplies the details for $\gamma$-emission and $\beta$-emission to be: $\frac{d\gamma}{dt}=-\frac{ln(1/2)}{24}\gamma$ $\frac{d\beta}{dt}=-\frac{ln(1/2)}{36}\beta$ (Why? Well, one starts with the equation $\frac{dA}{dt}=-kA \Rightarrow ln A/A_0 = -k t \Rightarrow A = A_0 e^{-kt}$, where one integrates both sides. Plugging in the condition for half-life, one has $1/2 = e^{-kT} \Rightarrow k=ln(1/2)/T$, where T is the half-life decay time.) The total decay is given by $\frac{dR}{dt} = \frac{d\gamma}{dt} + \frac{d\beta}{dt}=-\frac{ln(1/2)}{24}\gamma-\frac{ln(1/2)}{36}\beta = -\frac{ln(1/2)}{\tau}R$. Thus, $1/24 + 1/36 = 1/T$. Solve for T to get choice (D). Click here to jump to the problem!
 21 Click here to jump to the problem! GR8677 #67 Advanced Topic$\Rightarrow$}Binding Energy The binding energy is basically the energy that keeps a nucleus together; and makes its mass slightly different than its constituent particles. (If its mass were exactly the same as its constituent particles, its binding energy would be 0, and it would be unstable.) The initial binding energy is $U_i = 238(7.6 MeV) \approx 240(7.6MeV)$, which is the number of nucleons times the binding energy per nucleon. The final binding energy is $U_f = 120 X + 120X =240X$, where the daughter nuclei have half the number of nucleons from the given fact that the original nucleus splits into 2 equal fragments. The difference in binding energy is equal to the kinetic energy. $U_i - U_f = 2\times -100 MeV \Rightarrow 240(7.6MeV - X) = -200 MeV$. Solving, one finds that $X\approx 8.5$, as in choice (E). Click here to jump to the problem!
 22 Click here to jump to the problem! GR8677 #68 Advanced Topics$\Rightarrow$}Decay Since Lithium has one less electron than Be, one might think this decay is just $beta$-decay. However, $beta$-decay always occurs with a neutrino or anti-neutrino, and since none of the choices show this. Beta-decay emits either an electron or positron with an antineutrino or neutrino: $_A^Z N \rightarrow _{A-1}^Z M + \beta^{+} + \nu$ $_A^Z N \rightarrow _{A+1}^Z M + \beta^{-} + \bar{\nu}$ (The bit on antineutrinos and neutrinos has to do with conservation of Electron Lepton number $L_e$. Namely, electrons and neutrinos have $L_e=1$, while positrons and antineutrinos have $L_e = -1$.) (Also, $\alpha$-decay emits a Helium atom with 4 neutrons and 2 protons, so the numbers won't work out here.) The remaining choice is (E). One can check that it's right by noting that it is the only choice that conserves the electron-lepton number. Click here to jump to the problem!
 23 Click here to jump to the problem! GR8677 #76 Advanced Topics$\Rightarrow$}Solid State Physics Elimination time: (A) Electrons have less degrees of freedom than free atoms, since electrons are bound to potential wells. (B) Why not? (C) The electrons do indeed form a degenerate Fermi gas, since the ratio of the fraction of electrons in the ground-state is given by $N_0/N = 1-(T/T_F)^{3/2}$, where $T_F \approx 30,000$ K for most metals. (D) The electrons in metal travel at a drift velocity of about $10^{-6} m/s$. Not quite fast enough to be relativistic. (E) Electron interaction with phonons has nothing to do with their mean kinetic energy. Click here to jump to the problem!
 24 Click here to jump to the problem! GR8677 #78 Advanced Topics$\Rightarrow$}Particle Physics A few lines from the Particle Physics chapter of my upcoming book (title tentative, to be publicized on this website eventually), Reviewing Forgotten Physics for the GRE, Prelim's and Qualifier's: Baryons and mesons are hadrons. (Hadrons are particles that interact with the strong nuclear force.) Baryons have a baryon number of $B=\pm 1$, while mesons have $B=0$. Nucleons (which are baryons) have $B=1$, while antinucleons (antineutrons and antiprotons) have $B=-1$. Leptons (electrons, neutrinos, muons, and tau's) interact with the weak nuclear force. There are three kinds of lepton numbers. There is the electron lepton number $L_e$ and there is a muon lepton number, as well as a neutrino lepton number. Electrons $\beta^{-}$ and the electron neutrino $\nu_e$ have $L_e=1$, while positrons (antielectrons) $\beta^{+}$ and the electron antineutrino $\bar{\nu}_{e}$ have $L_e=-1$. All other particles have $L_e=0$ Conservation of baryon number or lepton number is just summing up either the baryon number or the lepton number on both sides of the reaction. Conservation of the numbers above explain for why a reaction like $\mu \rightarrow \beta^{-} + \nu_\mu + \bar{\nu}_e$ must occur in lieu of $\mu \rightarrow \beta^{-} + \bar{\nu}_e$ or $\mu \rightarrow \beta^{-} + \nu_e$ or $\mu \rightarrow \beta^{-} + \bar{\nu}_{\mu}$ or $\mu \rightarrow \beta^{-} + \nu_{\mu}$. For the first reaction, the electron Lepton numbers are $0 \rightarrow 1+0-1$, and thus $L_e$ is conserved. (Muon Lepton numbers are $1 \rightarrow 0+1+0$) For the second reaction, the muon lepton numbers are $1 \rightarrow 0 + 0$, but that does not add up, and so $L_\mu$ is not conserved. For the third reaction, the electron Lepton numbers are $0 \rightarrow 1+1$, but that does not add up, and so $L_e$ is not conserved. As an exercise, one can calculate the fourth and fifth reaction. (If you're beginning to feel like this is more alchemy than physics, then note that the Standard Model of particle physics is just a transitory theory, like alchemy was to chemistry. It is to be replaced soon with a less ad-hoc theory, perhaps by you.) With respect to this question on the GRE exam, one sees that lepton number would not be conserved for any of the possible cases of the latter reactions. Click here to jump to the problem!
 25 Click here to jump to the problem! GR8677 #25 Advanced Topics$\Rightarrow$}Particle Physics Choice (A) and (C) involve atoms, which are quite massive. Choice (B) involves protons, which are also pretty massive. Massiveness eliminates three choices, leaving just (D) and (E). Neutrinos are massless, but muons aren't. Both positrons and electrons have the same mass. Massiveness has lost its charm. (No pun intended.) According to David Schaich, the Super-Kamiokande and the Antarctic Muon and Neutrino Detector Array (AMANDA) are both located deep down underground to avoid interaction with other particles. Thus, with the hindsight of this bit of trivia, choice (D) is correct. Click here to jump to the problem!
 26 Click here to jump to the problem! GR8677 #39 Advanced Topics$\Rightarrow$}Fourier Series There's no need to go through the formalism of integrating out the coefficients. One can tell by inspection that the function is odd. Thus, one would use the Fourier sine series. This leaves choices (B) and (A). Choice (A) is trivially zero since for all integer n, $\sin(n\omega t)=0$. Choice (B) remains. Click here to jump to the problem!
 27 Click here to jump to the problem! GR8677 #49 Advanced Topics$\Rightarrow$}Scintillator The maximal speed the muons can travel at is slightly less than c. Thus, since the distance is $x=3m$, the time required would be $c=x/t \Rightarrow t=x/c=1E-8$. The largest scintillator time is the one closest to this, which is 1 ns, as in choice (B). Click here to jump to the problem!
 28 Click here to jump to the problem! GR8677 #75 Advanced Topics$\Rightarrow$}Binding Energy The binding energy for heavy atoms ($>200 nucelons$) is about $8MeV/nucleon$. The change in binding energy is the kinetic energy, thus the Helium atom has a kinetic energy of $(235-231)8MeV$. (The binding energy of He is ignored.) This is much larger than the kinetic energy of the He nucleus. Click here to jump to the problem!
 29 Click here to jump to the problem! GR8677 #91 Advanced Topics$\Rightarrow$}Strangeness Elimination time: (A) Only muons, neutrinos and electrons are leptons. Moreover, the pi-meson is a meson, which is a hadron with baryon number 0. (Hadrons interact with the strong nuclear force, while leptons interact with the weak nuclear force, em force, and possibly even the gravitation force.) (B) The lambda has spin 1/2, as do most baryons. (The mesons have spin 0, but positive strangeness numbers.) (C) Lepton number is already conserved, since none of the particles involved have non-zero lepton numbers. Thus, introducing a neutrino would violate (electron) lepton number conservation. (D) No reason why... (E) Only hadrons have non-zero strangeness (strangeness was proposed when strong particles interact as if weak particles---i.e., instead of having super-fast decay times characteristic of strong-force particles, their decay times appeared as if weak-force decays). Protons have 0 strangeness, as do pi-mesons, even though they are both hadrons. However, the lambda has -1 strangeness. Thus, strangeness is not conserved.Click here to jump to the problem!
 30 Click here to jump to the problem! GR8677 #95 Advanced Topics$\Rightarrow$}Dimensional Analysis The final units must be $cm^2/steradians$. One is given $10^{12}protons/s$ $10^{20}nuclei/cm^2$ $10^2protons/s$ $10^{-4}steradians$ The combination $(10^2/10^{20})(1/10^{20})(1/10^{-4})$ gives the right units as well as answer choice (C). Click here to jump to the problem!
 31 Click here to jump to the problem! GR8677 #97 Advanced Topics$\Rightarrow$}Solid State Physics This is a result one remembers by heart from a decent solid state physics course. It has to do with band gaps, which is basically the core of such a course. Then again, one can easily derive it from scratch upon recalling some basic principles: $E=p^2/(2m)=\hbar^2 k^2/(2m)$, $p=\hbar k=mv$, where k is the wave vector, E is the energy, m is the mass, and p is the momentum. From the above, one has $dv/dt = \frac{1}{m}\frac{dp}{dt}=\frac{\hbar}{m}\frac{dk}{dt}$. $dE/dk = \hbar^2 k/m = \hbar p/m = \hbar v \Rightarrow v = \frac{1}{\hbar}\frac{dE}{dk} \Rightarrow dv/dt = \frac{1}{\hbar}\frac{d^2E}{dtdk}$ Set the two $dv/dt$'s equal to get $\frac{\hbar}{m}\frac{dk}{dt}=\frac{1}{\hbar}\frac{d^2E}{dtdk}$. Cancel out the $dt$'s to get $\frac{\hbar^2}{m}dk=\frac{dE}{dk} \Rightarrow m = \hbar^2/(\frac{dE}{dk})$, after differentiating with respect to k on both sides. Alternatively, one can try it Kittel's way: Start with $\hbar v_g=dE/dk$. Then, $dv_g/dt = \hbar^{-1}(d^2E/dt^2dk/dt)= \hbar^{-1}(d^2E/dt^2F/\hbar)$. Thus, the effective mass is defined by $F=\hbar^2/(d^2E/dk^2) dv_g/dt=mdv_g/dt$. Click here to jump to the problem!

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