University of Alabama Department of Physics an Astronomy PH 10- / LeClair Spring 008 Problem Set 1 SOLUTIONS 1. 10 points. In the 1996 movie Eraser, a corrupt business Cyrez is manufacturing a hanhel rail gun which fires aluminum bullets at nearly the spee of light. Let us be optimistic an assume the actual velocity is 0.75 c. We will also assume that the bullets are tiny, about the mass of a paper clip, or m5 10 4 kg. (a) What is the relativistic kinetic energy of such a bullet? (b) What rest mass woul have to be completely converte to energy to supply this kinetic energy? Note that 1 kg of TNT has an euivalent energy content of about 4 10 9 J. Calculating the relativistic kinetic energy is straightforwar. Let m b be the mass of the bullet, an v its velocity. KE (γ 1) m b c! 1 p 1 v /c 1 mc «1 `5 1 10 4 kg ` 10 8 m/s 1 0.75 (1.51 1) `4.5 10 1 kg m /s.0 10 1 J 5700 kg of TNT We woul like to supply this much energy by converting some bit of rest mass m r completely to energy. Thus, the kinetic energy above shoul be euate with the rest energy of a mass m r use as propellant: KE bullet E rest, propellant (γ 1) m b c m rc m r (γ 1) m b m r (1.51 1) `5 10 4 kg m r.56 10 4 kg In other wors, in orer to supply enough energy for the bullet to reach 0.75c, we woul nee the energy euivalent of a rest mass of.56 10 4 kg, about half the mass of the bullet itself.. 10 points. Two ientical point charges + are locate on the y axis at y +a an y a. (a) What is the electric potential for an arbitrary point (x,y)? We have two charges separate by a istance a. We will choose the origin O to be precisely between the two charges, along the line connecting the charges, an we wish to calculate the electric potential at an arbitrary point P(x, y) We can reaily write own the potential at the point P it is just a superposition of the potential ue to each of the charges alone. For a single charge (call it 1) at position (0, a), the istance to an arbitrary point (x,y) is reaily foun in two imensions: x + (y a)
Given the istance an the magnitue of the charge, fining the electric potential V 1 is easy too: V 1(x, y) ke1 k e 1 x + (y a) (1) The potential ue to the secon charge has the same form, only the istance changes: V (x, y) ke r k e 1 x + (y + a) () The total potential at P(x, y) is foun by superposition - it is just the sum of the separate potentials ue to each charge. Since electric potential is a scalar, we on t even nee to worry about vectors: V tot(x, y) V 1(x, y) + V (x, y) k e 4 + x + (y + a) 1 x + (y a) (extra) How about the electric fiel? There is no trick to this one either, you just have to grin through it. First, consier one charge alone, an fin the electric fiel ue to it at point (x,y). Then fin the x an y components. Do the same for the other charge, an a the x an y components for each together to fin the total fiel. It is messy. Given the istance an the magnitue of the charge, fining E is easy too. For the first charge,: 5 E 1 ke1 r 1 k e 1 x + (y a) For the secon charge, the total fiel at (x,y) is almost the same, only the istance changes: r x + (y + a) : E ke r k e x + (y + a) The fiel ue to both charges together can be foun by superposition. Since electric fiel is a vector, we nee to fin the x an y components of E 1 an E an a them together: E x,total E x,1 + E x,. In orer to fin the x components, we nee to fin the angles θ 1 an θ that an r make with the x axis, since E x E cos θ. cos θ 1 x x x + (y a) cos θ x x r x + (y + a) Thus, the x components of the fiel are: E x,1 E 1 cos θ 1 E x, E cos θ E x,total E x,1 + E x, k e 1 x + (y a) x k e x + (y + a) x r k e 1 x + (y a) k e x + (y + a) For the y component, we nee the sin of the relevant angles: x x + (y a) x x + (y + a) k e 1x ˆx + (y a) k e x ˆx + (y + a)
sin θ 1 y y x + (y a) sin θ y y r x + (y + a) Thus, the y components of the fiel are: E y,1 E 1 sin θ 1 E y, E sin θ E y,total E y,1 + E y, k e 1 x + (y a) y k e x + (y + a) y r k e 1 x + (y a) k e x + (y + a) y x + (y a) y x + (y + a) k e 1y ˆx + (y a) k e y ˆx + (y + a) We are tol that both charges are the same, 1. Doing that, we can put the whole mess together: E k e 4 x ˆx + (y + a) x + ˆx + (y a) 5 ˆx + k e 4 y ˆx + (y + a) y + ˆx + (y a) 5 ŷ (b) A circular ring of charge of raius a has a total positive charge Q istribute uniformly aroun it. The ring is in the x0 plane with its center at the origin. What is the electric fiel (both magnitue an irection) along the x axis at an arbitrary point xb ue to the ring of charge? Hint: Consier the total charge Q to be mae up of many pairs of ientical charges place on opposite points on the ring.. 10 points. The purely electrostatic crystallization energy for a mole of crystalline soli can be foun by consiering the electrostatic potential energy of a lattice of ionic charges. We foun that it coul be expresse as a factor M times the electrostatic energy of a single pair of ions: E cryst NAz e M 4πɛ 0r where N A is Avogaro s number, z is the charge per ion in the crystal, r is the istance between ions in the crystal, an M is the Maelung constant for the lattice as iscusse in the notes. The values of M, z, an r for several common solis crystallizing in the rocksalt (NaCl) structure are given below. Compoun chg/ion M r (pm) LiF e 1.748 01.4 LiCl e 1.748 57.0 LiBr e 1.748 75.1 NaCl e 1.748 8 NaBr e 1.748 89.9 Note that 1 pm 10 1 m. As it turns out, the energy of crystallization etermine from electrostatics is essentially the same energy reuire to separate a mole of soli into a gas of its ions - we nee aitionally only a small correction factor to account for the repulsion of the ions at small istances ue to uantum effects. (a) Calculate the crystallization energy, in electron volts, for the compouns in the table above. Which compoun is the most stable?
We really just nee to use the formula above an plug in the relevant numbers - being very careful with units. Note that the charge per ion, z, is given in units of the elementary charge e - thus, z 1 for all of the compouns in the table, since they are all salts mae of monovalent alkali metals an halogens. What we really mean by z is how many units of elementary charge are there per ion, an the answer is clearly 1 for these compouns. Probably, the table shoul have just liste 1 instea of e to make this less confusing. Anyway, start like this, for LiF, noting that 1/4πɛ 0 k e: E LiF NAz e M kenaz e M 4πɛ 0r r `9 10 9 N m /C `6.0 10 mol 1 (1) `1.60 10 19 C (1.748) 01.4 10 1 m 1.1 10 6 J/mol 110 kj/mol This is not far from the accepte value of 106 kj/mol. Now we just o this for the other compouns in the table... the results are summarize below, in kj/mol an ev/ion. i The most stable compoun is the one with the lowest crystallization energy, or the one that gains the most energy by crystallizing. In this case, it is LiF, which gains 110 kj/mol by crystallizing. NaBr gains only 88 kj/mol, which makes it less stable. Remember: negative energies generally mean a boun stable state, an the more negative, the more stable. Compoun E cryst (kj/mol) E cryst (ev/ion) LiF -110-1.5 LiCl -946-9.81 LiBr -884-9.17 NaCl -86-8.94 NaBr -88-8.69 (b) Lime, CaO, is know to have the same structure as NaCl (an thus the same Maelung constant) an the ege length of the unit cell for CaO is 481 pm. Thus, Ca-O istance is 41 pm. Evaluate the energy of crystallization, E cryst for CaO. Now one thing you nee to remember from chemistry is that both Ca an O are oubly charge ions - Ca is + an O is. Thus, there are unit charges per ion, an z. Other than that, we procee as before. Since the crystal structure is the same, so is M, we nee only change z an r: E CaO NAz e M kenaz e M 4πɛ 0r r `9 10 9 N m /C `6.0 10 mol 1 () `1.60 10 19 C (1.748) 41 10 1 m 404 kj/mol 41.8 ev/ion (c) Calculate the crystallization energy for MgO, for which z an r 86 pm. Procee exactly as in (b): E MgO NAz e M kenaz e M 4πɛ 0r r `9 10 9 N m /C `6.0 10 mol 1 () `1.60 10 19 C (1.748) 86 10 1 m 400 kj/mol 5. ev/ion We note that all of these numbers are within about 10% or so of what a moern ensity functional theory calculation woul give - the 10% we miss has to o with the repulsive forces when we try to push ions too close together, which we covere briefly i In orer to get the latter, ivie the kj/mol result by N A to get J/ion, an then ivie by 1.6 10 19 J/eV to get ev/ion.
in uantum physics. The bulk of the total energy of a crystal, however, is just tie up in electrostatic energy. 4. A parallel plate capacitor has a capacitance C when there is vacuum between the plates. The gap between the plates is half fille with a ielectric with ielectric constant κ in two ifferent ways, as shown below. Calculate the effective capacitance, in terms of C an κ, for both situations. Hint: try breaking each situation up into two euivalent capacitors. (a) (b) κ κ Dielectric parallel to the plates. It is easiest to think of this as two capacitors in series, both with half the plate spacing - one fille with ielectric, one with nothing. First, without any ielectric, we will say that the original capacitor has plate spacing an plate area A. The capacitance is then: C 0 ɛ0a The upper half capacitor with ielectric then has a capacitance: () C Kɛ0A / Kɛ0A KC 0 (4) The half capacitor without then has C none ɛ0a / ɛ0a C 0 (5) Now we just a the two like capacitors in series: 1 C eff 1 KC 0 + 1 C0 (6) C eff 4KC0 KC 0 + C 0 K 1 + K C0 Dielectric perpenicular to the plates. In this case, we think of the half-fille capacitor as two capacitors in parallel, one fille with ielectric, one with nothing. Now each half capacitor has half the plate area, but the same spacing. The upper half capacitor with ielectric then has a capacitance: (7) (8) C Kɛ0 1 A Kɛ0A 1 KC0 The half capacitor without then has (9) C none ɛ0 1 A ɛ0a 1 C0 Now we just a our parallel capacitors: (10)
C eff 1 KC0 + 1 C0 1 (K + 1) C0 K + 1 C 0 (11) (1) (1) 5. 10 points. The battery in your car stores about 10 W h worth of energy at a potential ifference of 1 V. One ay while oing yar work, you ecie to run your 00 W car stereo without the engine running. In orer to start your car, you nee to eliver 500 A at 7. V for at least 1 sec. How long can you run your stereo before you can no longer start the car? This is basically a conservation of energy problem: your battery has only so much energy available, an you want to use it all up between the stereo an starting the car. The total energy of the battery shoul first be converte into W s, or Joules: E batt `10 W h (600 s/h).6 10 6 J The power reuire to start the car can be foun from the given current an voltage reuirements. Then, noting that energy is power times time, we can fin how many Joules shoul be reuire to start the car: P start (500 A) (7. V) 600 W E start P start t start 600 J The power use by the stereo is just its power times the time we are able to run it: E stereo P stereo t stereo (00 W) t stereo Now we just set up the total energy balance, an solve for our one unknown - how long we can affor to run the stereo: E batt E start + E stereo.6 10 6 J 600 J + (00 W) t stereo t stereo 18000 s 5 h Of course, the reality of things is always ifferent - can a nearly ea battery really still generate 500 A? When entering the real worl, one shoul always give these estimates a hefty safety margin. Better yet, buy an ipo. It is far more stylish an efficient, if not cheaper. 6. 10 points. How many ifferent resistance values can be constructe from a Ω, 4 Ω, an a 6 Ω resistor? (You only have one of each.) Show how you woul get each resistance value either iniviually or by combining them. There is no trick: just fin them by brute force. There are 16 ifferent values of resistance. combination number of ways resistances (Ω) single resistor, 4, 6 in series 6, 8, 10 in parallel 1., 1.5,.4 in series 1 1 in parallel 1 1.1 1 in parallel with series comb. of other 1.7,.7,.0 1 in series with parallel comb. of other 4.4, 5.5, 7.
There are 17 combinations overall, but a resistance of 6 Ω shows up twice. Thus, there are 16 ifferent resistance values. If we insist that all three resistors are use (which we i not), then of course there are only 8. 7. 10 points. An electron is moving at a spee of 0.01c on a circular orbit of raius 10 10 m aroun a proton. (a) What is the strength of the resulting magnetic fiel at the center of the orbit? (The numbers given are typical, in orer of magnitue, for an electron in an atom.) The easiest way to procee is to consier the electron in its circular orbit to be a very tiny current loop. But what is the current? Think back to the efinition of current: I Q/ t. The electron has a constant spee v in its orbit, which means that the current corresponing to this single charge is just the amount of charge e ivie by how long it takes to orbit the nucleus - the circumference ivie by the electron s velocity. I Q t e πr/v ev πr `1.6 10 19 C `0.1 10 8 m/s π (10 10 m) 7.6 10 4 A 760 µa Given this current aroun a circular loop, we know how calculate the fiel at the center: B center µ0i r µev 4πr 4.8 T (b) If the nucleus of the atom (at the center of the orbit) consists of a single proton, what woul its precession freuency be? Hint: from the nucleus point of view, it is orbiting the electron in a circular path. Recall ω B/m an ω πf. From the electron s reference frame, it is the proton that is revolving aroun it, experiencing the magnetic fiel ue to the electron. This leas to a precessional motion, for which we can calculate the freuency given the charge (+e) an mass (m p) of a proton: f ω π e B πm µe v 7 MHz 8π mr In fact, this is a crue estimate of the hyrogen proton resonance freuency. We are off by a factor of about.79, having neglecte all of uantum physics. The missing factor is basically the proton s g-factor, which is a bit too much to go in to right now... 8. 10 points. A hyrogen atom initially in its groun state (n1) absorbs a photon an ens up in the state for which n. If the atom eventually returns to the groun state, what photon energies coul the atom emit? There are two ifferent ways the electron can go from the thir to the first level: it can jump from to 1 irectly, or it can first go from to, an then from to 1. In the former case, the electron must loose an energy E 1 E E 1 by emitting a photon. In the latter case, the electron must first emit a photon of energy E E E, followe by one of energy E E E 1. This makes a total of three ifferent photons that can be emitte: 1.6 ev 1.6 ev E 1 E E 1 1.1 ev 1 1.6 ev 1.6 ev E E E 1.89 ev 1.6 ev 1.6 ev E E E 1 10. ev 1
9. 10 points. Initially, the raioactive ecay rate of a particular group of nuclei is 00 counts per secon. After 5 minutes, the ecay rate rops to about 8 counts per secon. (a) What approximately is the half life of this nucleus? (b) What will be the approximate ecay rate (in counts per secon) after.5 aitional minutes? The raioactive ecay rate (or the number of atoms in the sample, for that matter) follows an exponential law: R(t) R 0 1 «t/t1/ where R 0 is the initial rate, R(t) the rate after a time t, an T 1 per secon after 5 minutes, or 00 secons: is the half life. If the rate is reuce from 00 to 8 counts 1 8 00 8 1 00 ln 8 00 00 ln 1 T 1/ T 1/ 00 ln 1 ln 8 00 «00/T1/ «00/T1/ 100 s After an aitional.5 minutes, a total of 450 secons have elapse. Using the now known half-life with the initial ecay rate, we can fin the rate after 450 secons: «t/t1/ 1 R(t) R 0 «450/100 1 R(450 s) 00 1.5 counts/sec 10. 10 points. Calculate the bining energy in MeV of a euteron (the atom 1H), given that its atomic mass is.01410 u. Note that m p + 1.00785 u, an m n 0 1.00866 5u. The bining energy is the ifference in the mass of the bare protons an neutrons an the nucleus itself, all times c to get an energy. For a euteron, we have one neutron an one proton, so we a those two masses together, an subtract off the mass of the euteron itself: Bining Energy 6 4 X all p + an n 0 mc 7 5 matomc»`1 p + 1.00785 u p + 0.0088 u c «0.0088 u c 91 MeV u c.4 MeV «+ `1 n 0 1.008665 u c.01410 u c n 0 Note that in that last step, we use the (given) conversion of 91 MeV/u c to get from a mass ifference to an energy...