1. Física I – Sears, Zemansky, Young & Freedman.
  2. Sears' and Zemansky's University Physics With Modern Physics 1
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Available in the Pearson eText and in the Study Area of MasteringPhysics. * Indicates .. Sears and Zemansky's university physics: with modern physics. -- 13th. SearS and ZemanSky'S univeRSitY PHYSiCS WitH moDeRn PHYSiCS 14tH eDition HugH D. Young RogeR A. FReeDmAn University of California, Santa. WitH moDeRn PHYSiCS. 14tH eDition. SEARS AND ZEMANSKY'S. 1. 08/11/14 AM.

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Physics with Calculus. Contribute to RandyMcMillan/PHY University Physics with Modern Physics 14ed [].pdf.. 3 years ago. · Changes to. Where can I download a free PDF of Sear & Zemansky's University Physics How do I download a PDF of the “University Physics 11th Edition”. MARK W. ZEMANSKY was born in New York City in , graduated from edition of College Physics and their University Physics in During his long.

Young RogeR A. He was an undergraduate at the University of California campuses in San Diego and Los Angeles and did his doctoral research in nuclear theory at Stanford University under the direction of Professor J. Dirk Walecka. Freedman has taught in both the Department of Physics and the College of Creative Studies a branch of the university intended for highly gifted and motivated undergraduates. He has published research in nuclear physics elementary particle physics and laser physics. In recent years he has worked to make physics lectures a more interactive experience through the use of classroom response systems and pre-lecture videos. In the s Dr.

Instead we invent a simplified version of the problem.

Física I – Sears, Zemansky, Young & Freedman.

We ignore the size and shape of the ball by representing it as a point object or particle. We ignore air resistance by making the ball move in a vacuum and we make the weight constant. Now we have a problem that is simple enough to deal with Fig. We will analyze this model in detail in Chapter 3. We have to overlook quite a few minor effects to make an idealized model but we must be careful not to neglect too much.

If we ignore the effects of grav- ity completely then our model predicts that when we throw the ball up it will go in a straight line and disappear into space. A useful model simplifies a problem enough to make it manageable yet keeps its essential features. Direction of motion Direction of motion Treat the baseball as a point object particle. No air resistance. Baseball spins and has a complex shape. Air resistance and wind exert forces on the ball. Gravitational force on ball depends on altitude.

Gravitational force on ball is constant. This model works fairly well for a dropped cannonball but not so well for a feather. Idealized models play a crucial role throughout this book. Watch for them in discussions of physical theories and their applications to specific problems. Experiments require measurements and we generally use numbers to describe the results of measurements.

Any number that is used to describe a physical phenomenon quantitatively is called a physical quantity. For example two physical quanti - ties that describe you are your weight and your height. Some physical quantities are so fundamental that we can define them only by describing how to measure them.

Such a definition is called an operational definition. Two examples are measuring a distance by using a ruler and measuring a time interval by using a stopwatch. In other cases we define a physical quantity by describing how to calculate it from other quantities that we can measure.

Thus we might define the average speed of a moving object as the distance traveled measured with a ruler divided by the time of travel measured with a stopwatch.

When we measure a quantity we always compare it with some reference stan - dard. When we say that a Ferrari Italia is 4. Such a standard defines a unit of the quantity. The meter is a unit of distance and the second is a unit of time. To make accurate reliable measurements we need units of measurement that do not change and that can be duplicated by observers in various locations.

Appendix A gives a list of all SI units as well as definitions of the most fundamental units. Time From until the unit of time was defined as a certain fraction of the mean solar day the average time between successive arrivals of the sun at its highest point in the sky.

The present standard adopted in is much more precise. It is based on an atomic clock which uses the energy difference between the two lowest energy states of the cesium atom Cs. When bombarded by microwaves of precisely the proper frequency cesium atoms undergo a transition from one of these states to the other. One second abbreviated s is defined as the time required for cycles of this microwave radiation Fig.

Length In an atomic standard for the meter was also established using the wavelength of the orange-red light emitted by excited atoms of krypton 1 86 Kr2. From this length standard the speed of light in vacuum was measured to be ms. In November the length standard was changed again so that the speed of light in vacuum was defined to be precisely ms. These measurements are useful for setting standards because they give the same results no matter where they are made.

Light source Cesium atom Cesium atom Microwave radiation with a frequency of exactly cycles per second An atomic clock uses this phenomenon to tune microwaves to this exact frequency.

It then counts 1 second for each cycles.

Light travels exactly m in 1 s. This modern definition provides a much more precise standard of length than the one based on a wave- length of light. An atomic standard of mass would be more fundamental but at present we cannot measure masses on an atomic scale with as much accuracy as on a macroscopic scale. The gram which is not a fundamental unit is 0.

Other derived units can be formed from the fundamental units. For example the units of speed are meters per second or ms these are the units of length m divided by the units of time s. Unit Prefixes Once we have defined the fundamental units it is easy to introduce larger and smaller units for the same physical quantities. In the metric system these other units are related to the fundamental units or in the case of mass to the gram by multiples of 10 or 1 10 Thus one kilometer 11 km2 is meters and one centi- meter 11 cm2 is 1 meter.

We usually express multiples of 10 or 1 10 in exponential notation: 10 3 1 10 -3 and so on. With this notation 1 km 10 3 m and 1 cm 10 -2 m. The names of the additional units are derived by adding a prefix to the name of the fundamental unit. Table 1. Fun - damental particles are the smallest things in the universe and cosmology deals with the biggest thing there is—the universe itself. The development of high-energy accelerators and associated detectors has been crucial in our emerging understanding of particles.

We can classify par - ticles and their interactions in several ways in terms of conservation laws and symmetries some of which are absolute and others of which are obeyed only in certain kinds of interactions. In about b. This idea lay dormant until about when the English scientist John Dalton — often called the father of modern chemistry discovered that many chemical phenomena could be explained if atoms of each element are the basic indivisible building blocks of matter.

The characteristic spectra of elements suggested that atoms have internal structure This image shows a por- tion of the Eagle Nebula a region some light-years away where new stars are forming.

Looking back at … In Rutherford made an additional discovery: When alpha particles are fired into nitrogen one product is hydrogen gas. He reasoned that the hydrogen nucleus is a constituent of the nuclei of heavier atoms such as nitrogen and that a collision with a fast-moving alpha particle can dislodge one of those hydrogen nuclei.

Thus the hydrogen nucleus is an elementary particle that Rutherford named the proton. Physicists were on their way to understanding the principles that underlie atomic structure. Atoms and nuclei can emit create and absorb destroy photons see Section Considered as particles photons have zero charge and zero rest mass.

In particle physics a photon is denoted by the symbol g the Greek letter gamma. Experiments by the English physicist James Chadwick in showed that the emitted particles were electrically neutral with mass approximately equal to that of the proton. Chadwick christened these particles neutrons symbol n or 1 0 n.

This is the principle of the cloud chamber described below. Because neutrons have no charge they are difficult to detect directly they interact hardly at all with electrons and produce little ionization when they pass through matter. However neutrons can be slowed down by scattering from nuclei and they can penetrate a nucleus. Hence slow neutrons can be detected by means of a nuclear reaction in which a neutron is absorbed and an alpha particle is emitted.

Later experi - ments showed that neutrons and protons like electrons are spin 1 2 particles see Section The discovery of the neutron cleared up a mystery about the composition of the nucleus.

Before the mass of a nucleus was thought to be due only to protons but no one understood why the charge-to-mass ratio was not the same for all nuclides. It soon became clear that all nuclides except 1 1 H contain both protons and neutrons. Hence the proton the neutron and the electron are the building blocks of atoms. However that is not the end of the particle story these are not the only particles and particles can do more than build atoms.

The photograph was made by Carl D. Anderson in Positron track Lead plate 6 mm thick The positron follows a curved path owing to the presence of a magnetic field. The track is more strongly curved above the lead plate showing that the positron was traveling upward and lost energy and speed as it passed through the plate. Figure The chamber contained a supercooled vapor a charged particle passing through the vapor causes ionization and the ions trigger the condensation of liquid droplets from the vapor.

The cloud chamber in Fig. The particle has passed through a thin lead plate which extends from left to right in the figure that lies within the chamber. The track is more tightly curved above the plate than below it showing that the speed was less above the plate than below it. Therefore the particle had to be moving upward it could not have gained energy passing through the lead.

The thickness and curva- ture of the track suggested that its mass and the magnitude of its charge equaled those of the electron. But the directions of the magnetic field and the velocity in the magnetic force equation F S qY S : B S showed that the particle had positive charge. Anderson christened this particle the positron.

To theorists the appearance of the positron was a welcome development. In Section One of the puzzling features of the Dirac equation was that for a free electron it predicted not only a continuum of energy states greater than its rest energy m e c 2 as expected but also a continuum of negative energy states less than -m e c 2 Fig.

That posed a problem. The exclusion principle see Section A vacancy in a negative-energy state would act like a positive charge just as a hole in the valence band of a semiconductor see Section Initially Dirac tried to argue that such vacancies were protons. Furthermore the Dirac energy-state picture provides a mechanism for the creation of positrons. When an electron in a negative-energy state absorbs a photon with energy greater than 2m e c 2 it goes to a positive state Fig.

The vacancy that it leaves behind is observed as a positron the result is the creation of an electron—positron pair. Similarly when an electron in a positive-energy state falls into a vacancy both the electron and the vacancy that is the positron disappear and photons are emitted Fig.

Thus the Dirac theory leads naturally to the conclusion that like photons electrons can be created and destroyed. While photons can be created and destroyed singly electrons can be produced or destroyed only in electron—positron pairs or in association with other particles. Creating or destroying an electron alone would mean creating or destroying an amount of charge -e which would violate the conservation of electric charge.

In the situation posed in this question, the potential energy increases monotonically. Once the particle reached the height where its kinetic energy is zero, everything above is classically forbidden, 6. That is, the group velocity is smaller than the phase velocity. When the central value of k is large, these two quantities are equal, and so are the phase and group velocities.

Thus, only in these regions does the group velocity not exceed the phase velocity. Wherever the dispersion relation follows the parabolic plot, its second derivative is the same constant.

Only for k values near the band gap does it deviate. It will definitely reflect. We might suppose that there should be a smooth match to a solution in the region beyond the downward step. But the fact that the potential energy is infinite means that the derivative is discontinuous at the drop.

The wave function is simply a standing wave to the left of the step, zero at the step, and zero beyond the step. Now plugging back in: The wave is zero at the step. The wave is maximum at the step and there is much penetration. Thus the two will interfere destructively, leading to no overall reflection, when the extra distance traveled by the one reflecting off the back, 2t, is an integral number of wavelengths: The window for resonant transmission can be rather small.

Dividing the 4 condition by the 2 eliminates F. This qualifies as a wide barrier. At small T, it is more than 30 orders of magnitude more likely to tunnel at the higher energy. The speed of a 4. A typical alpha particle hits the wall every 10 s! Thus time particle is given by 6. At one decay every 10 seconds, the mean lifetime would be near 10 seconds or 44 about 10 years. Setting this equal to 9 7 time 2rnuc 6. However, the order of magnitude is most important, and this is set by the value of the exponential.

A change of roughly a factor of 2 in our model potential energy alters the mean life by more than thirty orders of magnitude. That the tunneling goes maximum at separations differing by one-half wavelength certainly suggests that a resonance condition is at play.

This is infinite, implying the same thing. Adding a second barrier would be superfluous. For a total distance of 8km, 4. The phase velocity varies from one k to another. Equivalently we may say that the dispersion relation is not a linear function of k. Its second derivative is not zero. In the region 0 6. The tunneling probability vanishes. For this to hold, E must be no greater than zero.

This is the case if the energy is between 0 and e. The time it would last would be this time divided by the tunneling probability. Quite often it arises from imposing physical conditions, such as continuity or normalizability, on mathematical solutions of a governing differential equation.

Striking a smaller mass causes the target to move very fast in the initial direction of motion and the projectile to continue moving, though slower, in the direction it initially moves. In fact, the colder it gets, the more atoms will be in this level, so upward transitions from there are to be expected. A circle has no outside. A constant, meeting itself smoothly after a complete circle, is perfectly acceptable.

The angular momentum vector would have only two directions in which it could point. This still seems rather restricted. The momentum, so long as it is radial, could be anything and still give zero angular momentum. One might argue that they are a probability per unit solid angle, but a solid angle, like a simple angle, has no dimensions.

Multiplying the square of the radial function by r , the radial probability will have —1 dimensions length. This too makes sense, for it is a probability per unit distance in the radial direction.

However, in multiple dimensions, it is possible to have different standing waves that still have the same frequency energy. A square two-dimensional membrane, for instance, could have a wave with one bump along x and two along y, but the wave with two along x and one along y would by symmetry have to have the same frequency. Angularly, the d states have multiples antinodes, while the s state has no angular variations at all.

The complexities of the radial and angular parts appears to be inversely related. A cubic box is not rotationally symmetric—flat walls should interfere with the establishment of well defined angular momentum states. The cubic well definitely looks more linear, and the quantized values of kx, ky, and kz, if not representing well defined linear momentum the direction, at least, is still uncertain , are related to kinetics energies along these three perpendicular axes.

It also gives the energy of that state. It also gives the energy of any given state in that subset. This causes the lifetime of the 2s state to be unusually long—known as metastable.

The dimensions work the same way. Three states have the same 2. Three also correspond to 3. How many different sums of three squares equal 27? Thus, there are four. The single energy be decreased to It could be destroyed by making Ly and Lz unequal.

The region includes one third of the well. The other two probabilities are large because the slice is centered on an antinode. Consider x: Two particles may have this function. With two particles in each, we add six to the existing two. This adds another six to the existing eight—a total of fourteen.

But precisely because of this, and the fact that they are equally probable, the probability of finding the particle is greatest at the center, where all directions have maxima. Each orbit would be nearly circular.

Multiplying by dr would give the time for the radius to change by that much. From Figure 7. So what is ninitial? As calculated in Example 7. Its velocity would be m 1.

But with a high enough applied voltage, they at some point have enough energy to cause a jump, so they lose a great deal of energy and have to speed up again. A smaller average current flows. As the applied voltage is further increased, they acquire more energy after exciting the jumps, and the current again increases. If the real D must be an integer, at which values the imaginary part is zero.

Lz is strictly less than L. They only way they can be eliminated is if either A or B is zero. A and B cannot both be zero, or the solution itself would be zero. Adding the equal contributions on the other two dimensions give the expected result.

This is the correct ground-state energy. Given this, there are five different allowed mA values: We must sum these values for all allowed values of A: There is no angular variation. The angular functions are constants. Thus, there is no rotational kinetic energy. All solutions have some radial curvature, and thus antinodes. They are not constants. From Table 7. A computer is a better idea. There is still a rather large probability of finding the electron not very near its most probable radius.

The probability per unit volume is thus maximum at the origin, as indicated in Figure 7. Quite a bit! The value is greater than r for the 3d. We are told that the two energies are equal: But it moves faster has the greater kinetic energy where its potential energy is lower, when r is smaller. It can take on more than one value because r may take on more than one value.

Find P r first. It is more like a circular orbit, at a less indefinite radius.

Sears' and Zemansky's University Physics With Modern Physics 1

Classically, there should be no uncertainty in r for a circular orbit, so this agrees with the classical expectation. Energy differences with nf is 3 are: All that remains is the radial equation. Consider the first term in Therefore, the probability is the same as in Example 7.

They will all thus change sign. They are unchanged. It amplitude is 6. From Tables 7. The frequency is the same as in Example 7. The character of the 4. Since these all differ, so do their angular probabilities. We need only substitute the reduced mass for the electron mass. This matches the high-n limit perfectly. For Doppler: At perihelion, kg 6. At aphelion, kg 6. In such an elliptical orbit, the motion will be largely radial when halfway between perihelion and aphelion.

Sines and cosines or complex exponentials result if the constant is negative, and these functions are all periodic. Therefore the constant C must be negative. This implies that ei 2 an integer. This is the ground state wave function. Rearranging gives the form shown.

All other sets of C and mA clearly diverge. If the plots are squared they match the 1s, 2s, 2p, 3s, 3p, and 3d plots of Figure 7. The change in angular momentum is therefore perpendicular to the field.

Suppose the dipole moment vector is in the xy-plane and the field is along x. If the dipole initially has no angular momentum, it gains an angular momentum along the z-axis, perpendicular to the field, as it aligns. But it if already had an angular momentum in the xy-plane, simple alignment with the field represent a change in angular momentum in the xy-plane, i. Were they to countercirculate, their angular momenta would cancel, but each would represent a current in the same direction, so they would produce a magnetic moment.

The Stern-Gerlach experiment sends atoms through a nonuniform magnetic field, which exerts a force proportional to the magnetic moment. Both are quantized. The orbital and intrinsic magnetic moments are also the same physical quantity and are quantized like their respective angular momenta, for each is simply proportional to its respective angular momentum, though by slightly different proportionality constants.

As a simple example, countercirculating plus and minus charges have a net charge of zero but constitute a current and thus a magnetic dipole moment. The spin-down beam would do the same. The probability of finding particle 1 in a given state must therefore be the same as the probability of finding particle 2 in that state. This symmetry in the probability can be ensured by either adding to or subtracting from the two-particle state the same state with the particle labels swapped—the square, which gives the probability, is symmetric either way.

Fermions by nature assume the case with the minus sign. If the state of particle 1 and the state of particle 2 are the same, then swapping the labels gives exactly the same two-particle state, and subtracting this labels-swapped state would thus give zero—not allowed.

For the difference to be nonzero, the two individual-particle states must be different, for then swapping the labels gives a different two-particle state. The requirement that the individual particle states be different is the exclusion principle, and it applies to fermions. The electron is a fermion, and so no two electrons sharing the same space—as in an atom—can occupy the same individual particle state.

Thus an odd number of neutrons means an odd total number of fermions, and thus fermionic behavior for the unit.

An even neutron number gives an even total, and bosonic behavior. Li-7, with an even total number of fermions, behaves as a boson. The atoms of Li-7 are bosons, so they could of course behave as bosons. Opinions differ on where to draw the line. In a 4-electron atom, there are, besides 4 electron-nucleus interactions, six electron-electron interactions. The next electron has to go to the next shell, at a much higher energy. This electron is easily stolen in a chemical reaction.

Further electrons would begin to fill up this shell, which meanwhile is itself dropping in energy, as more protons attract it to the nucleus. When it is nearly full, its electrons are very tightly bound and it is more likely that any space left in the shell will steal electrons from other atoms than that any electrons in this tightly bound shell would be stolen by other atoms.

After this shell fills, the next electron has to go to the next higher shell, at a considerably higher energy, and the process repeats itself. The ordering of levels does not depend on n alone. Its energy is lower than — Removing this first electron allows the remaining second electron to settle into a lower energy state itself, for there is no longer a repulsion. If the second electron gives away some energy, then the first one can have a rather low negative energy, but I will not have to expend energy of that large a magnitude to remove it.

In essence, wherever an appears we should replace it m m 11 2e eB 1. If the spins acquire a downward angular momentum, the only way to conserve the initial zero angular momentum is for the cylinder to acquire an upward angular momentum, rotating counterclockwise when viewed from above.

Sensibly, the probabilities add to 1. This is also sensible, for up along the first apparatus would be down in the inverted second. For a continuous wave function there are different probabilities for finding the particle at any of a continuum infinite number of x values.

The symmetric state tends to have particles closer together, so there is a greater than normal probability of finding them on the same side; the antisymmetric state tends to separate particles. Thus the integral gives 2. If this is to be 1, the A had better be. Letting x2 equal x1, this is zero! Due to the symmetry of the integrands about the origin, each of the two 0 restricted integrals must be exactly half the full integral, so the quotient is 0.

Now we must carry out the Gaussian integrals. An electron in the left atom would be in an entirely different spatial state, A — B, so it would have no bearing on the occupation of states in the right atom. Electrons in the left atom would be in C — D and would again have no bearing on the exclusion principle for electrons in the right atom. Total When two columns of a matrix are identical, the determinant is zero.

When rows are switched, the sign of the determinant is switched. This is neither the same nor the opposite of I, so its exchange symmetry is neither. Both have neither symmetric nor antisymmetric exchange symmetry. If the arrows are swapped, it does change—it is antisymmetric. If both are swapped, it changes sign, so it is antisymmetric. An electron in a morecircular 2p state will orbit essentially that whole cloud.

The more-elliptical 2s, for which the orbit has a larger probability of being very close to the origin, would pierce the inner cloud and partly feel the attraction to more protons. Thus, 0. Z Z b There are 54 other electrons, and an s-state is rather elliptical, so it pierces the cloud of these other electrons. Apparently the 4s electrons are already fairly close in energy to the 3d. The 5s, 4d, 5p trend is the same. It is not quite so good for the valence electron. From equation 7. Yes, they are all considerably smaller than the quoted atomic radius.

Despite increased screening at higher n, energy levels still tend to get closer together as n increases, so the 2 to 1 jump is bigger than the 3 to 2. Therefore, to eject an electron from a hole this deep, the incoming electron would have to be accelerated through at least 11kV. Inserting Z values: The function obtained via the model is plotted.

Agreement is good! If and s are equal, Jmin is zero. We must sum these over the allowed value of jT. The first sum is zero. For the latter, jT can be only. Thus, jT can be either or. Here the splitting in the upper state is one-third that in the lower. The middle two are separated by twice this amount, or 3. As noted in Section 8. Using law of cosines, Now law of sines: Were this Law of sines: The separation would be twice this: There will be four lines.

Though there is a hole in the middle 3 3 3 3 2me Differences are: The sodium doublet splitting is about It is much larger than the Zeeman splitting because it is due to the internal field, which is much stronger than the 0.

The rules of angular momentum addition could be satisfied by any of the allowed values of T: In fact, T is zero for ground-state nitrogen. Total number: We see that relativity is more likely to be a factor with large Z and small n. For no value of n 1. Inner shell electrons of high-Z elements are where relativistic effects become most prominent. In Example 7. In reality, then, This must be the repulsive energy of the electrons. Thus we have The value here is roughly a diameter.

Multiplying the two to —25 1. The energy 2 1. The Z plot makes sense. The 2s is clearly the lower energy, due to its piercing of the inner electron cloud—less screening. The energy obtained is —0. If they were really composed of many particles internally, they could qualify as a thermodynamic system and they could absorb KE. To specify the microstate we would need to specify the state of each particle—position and velocity for a classical system, or quantum state for a quantum system.

Each macrostate can be produced in multiple microscopic ways, so the number of microstates is larger. Distributions are based on averages over all the microscopic ways of obtaining the overall macroscopic state.

Accordingly, the number occupying a lower-energy state must never be less than the number if a higher-energy state. A density of states is a number of states at a given energy. It would not be needed if already summing or integrating over all the states, but in a sum or integral over energies, it could be used to account for all these states.

Usually we use it to replace a sum over states by an integral over energies. Overall, no units. This is sensible because the denominator is the total number of particles in the system. Permutations of particle labels are not relevant to the Bose-Einstein distribution, which assumes indistinguishable particles, so, relatively speaking, the division in which three particles have energy 0 is more likely in the Bose-Einstein than in the Boltzmann.

In other words, the state with higher occupation of the lowest energy is less likely in the Boltzmann than in the Bose-Einstein. The average energy of these charges determines the amount of energy and the most probable frequencies in the photon gas.

An equal number of downward transitions is possible because stimulated emission is augmented by spontaneous emission. This allows establishment of a population inversion, which ensures more stimulated emission than stimulated absorption. A nonmetastable state would not lead to a population inversion. The plastic alters the wavelength between the mirrors and thus destroys the standing wave condition. They may also rotate about the single axis, adding a third degree of freedom.

The particles will be evenly divided. Thus, the result is , as it should be. Well, at least the dimensions agree: It is overwhelmingly more likely to be found in the state where the energy is less unevenly distributed.

True equilibrium would of course be even more probable. Equation gives the probability. The sum in the denominator can be simplified: We see that a larger T implies a smaller probability.

The other route is to use To find the number at a given energy level we simply multiply the probability by the total number. The other route is to use 9. The temperature has to be many times the jump between levels before the ground state is so depleted. No particle is special, so we drop the subscript i on the n. As the temperature increases, the occupation of the higher energy state increases, but it is at most equal to that of the lower energy, so the average reaches a maximum of half the upper energy.

In the limit of high temperature, particles are just as likely to be antialigned as aligned. The number of particles with energy En is the number of states times the Boltzmann occupation number: Thus 0.

Taking into account ionized atoms would change the whole picture. Differentiating both sides gives: We wish to express this in terms of E, rather than n.

Lower by about 0. The denominator is kBT mg. Setting this to zero, we get escape speed. Escape velocity is slightly more than twice vrms. Hydrogen should escape. Nitrogen should remain. Meanwhile, vrms is 0. For hydrogen, the rms speed exceeds the escape speed, so it will escape with ease.

The probability is somewhere between 10 and 10 , so even it has an excellent chance. The total number is the combinatorial factor in which 6 objects are divided so 3 that two are in each of three categories. Using J-5 from Appendix J, 6! All distributions have an energy of 6E0. The distributions with higher occupation of the ground state would be proportionally underweighted.

Correspondingly, the Bose-Einstein indistinguishable accounting would weight the distributions in a and b proportionally higher—equal to that in b —suggesting a larger probability of occupying the ground state. Thus, the fermion distribution gives much higher probability to distributions in which fewer particles are in the ground state, as in c.

Reinserting gives 9. Together we see that the occupation number would always be much less than 1 under this condition. The Bose-Einstein, as expected, has even more, while the Fermi-Dirac, as always, cannot have more than one particle per state.

It drops by only 0. Occupation of the ground state is more probable for distinguishable than for fermion, and much more probable for boson than for distinguishable. It has nothing to do with how many particles might be available to fill those states, so it should not depend on N. In the 2D case, it seems almost constant. With one conduction electron per atom, the number of conduction electrons per unit volume equals the number of atoms per unit volume.

Equation would not even be close! Since air N 1. But these 1 are the spread out conditions in which we would expect classical behavior. Thus, classically speaking, is small B and terms in the series would get progressively smaller. Utotal follows the same way now as N.

The change in energy per change in T is the derivative. We may ignore it. To find at what wavelength this is maximum we differentiate.

Plugging in the wavelength Plugging in Thus, six different wavelengths would fit the condition. Thus, with springs in two dimensions there are four degrees of freedom. At low temperature, the heat capacity should go to zero as thermal energy would be insufficient to excite the oscillators above their ground states.

Thus, K is 0. Most of the fermions in the bulge are still just filling up the lower states as they did before.

If the particles available to gain energy become fewer as the temperature decreases, then the heat capacity should go to zero as T goes to zero. The number excited, i. Only when kBT approaches Eu is there a significant change in energy with increasing temperature. Protons repel one another, while neutrons do not. While N and V would be comparable, me is only about 1 of the neutron and 2, proton mass.

Thus its minimum energy is higher. In Exercise 40 the distribution predicted by exact probabilities is shown to be 7,3,1,0,0,0. It is most like a smooth, exponential fall-off, and it can be produced in by far the greatest number of ways. Pretty close. The result of the numerical integration, shown below, agrees with Figure 9. The wavelengths—and therefore the momenta and kinetic energies—of the lower pair of functions within each well are essentially the same as that of the ground state in one of the wells.

Similarly, the wavelengths inside the well of both functions in the upper pair are essentially the same as that of the first-excited state in a single well. If that axis is, say, the x-axis, a py or pz state qualifies, but an s-state has its electron cloud centered in the molecular axis, so it does not qualify. In nitrogen and fluorine, electrons occupy molecular states in spins-opposite pairs, canceling their intrinsic magnetic dipole moments.

Atomic states with any directional nature are not involved. The covalent bond by nature involves the combining of electron clouds sticking out in different directions. In an ionic solid, the positive ions lose valence electrons, which are added to the negative ions.

The charge density should thus alternate between positive and negative. In a metallic solid the valence electrons move more or less freely through the bulk.

The density of valence electrons should be roughly constant. Atoms or molecules in a molecular solid do not share their valence electrons with other atoms molecules.

The density should thus be largest at the locations of the atoms molecules themselves. They would be unable to form covalent bonds to multiple surrounding atoms in a lattice. This source of disorder would not diminish as the temperature decreases, and that the resistance also falls off slowly further reinforces the conclusion.

Nevertheless, as temperature increases, irregularities due to thermal motion of the ions increase, leading to higher resistance.

This is naturally present in a semiconductor but is outweighed by the promotion of more charge carriers across the valence-conduction gap. The larger gap in silicon ensures fewer minority carriers. Increasing the temperature, however, can produce more minority carriers, excited across the valence-conduction gap, and these love to flow from one side to the other in a reverse-bias condition.

What had been a repulsion would then be an attraction. However, the bottom disk is not a permanent magnet—its behavior is to exclude field lines, whichever direction they point. Therefore, it would do no good for the top magnet to flip over; the surface currents in the disk below would switch directions, to exclude the now-opposite field lines, and the magnet above would still be repelled. The heavier the lattice ions, the smaller their response, and the more difficult is the establishment of a superconducting state i.

The pairs behave somewhat as bosons and so are able to occupy the same quantum-mechanical state and thus move as one— completely ordered.

The Type-II alone can form networks of nonsuperconducting vortices through which field lines pass, and they generally exhibit higher critical temperatures and critical magnetic fields.

In this order they resemble the ground, first-excited, and next-excited states in a single well. Also, as the atoms draw close, penetration of the classical forbidden region would be more pronounced. Wave I would be entirely above the horizontal axis, essentially a single antinode with a half-wavelength stretching from one end of the unit to the other. Wave II would still have a node in the middle. Giving a shorter wavelength, and wave III would still have two nodes.

Its negative would be negative at less than 2a0. When these are added, we get something resembling the diagram below. We would like to put them in a 1. This is 5. The molecule in which the atoms are closer will be the one in which electrons are more likely to be shared between atoms. Thus, it must be that the HF is considered to be covalent, and NaF ionic. The other leg is along a diagonal, of which it is half the length: Now from Tables 7.

The ratio is negative one. Its value is intimately related to the rotational inertia of the molecule. At room temperature, N2 behaves as a rigid rotator: It stores rotational energy, but not vibrational. At K, kBT is 1. This is much larger than the rotational jump, so rotations should be active, but much less than the vibrational, so vibrations should be inactive.

This evidence suggests that vibrations do not participate in energy storage, which 8. Because in either case the quantity in brackets is proportional to the larger of the two values, it cannot be zero. Thus, it can be any nonzero integer, positive or negative, and equation follows.

There will be many wavelengths clustered around a photon energy of 0. The excited curve also has its minimum, 2 the atomic separation, at a larger a. Molecules and Solids b Let the charge in question be a positive charge at the origin. Thus, the appropriate width is the atomic spacing. Thus the atomic separation is roughly 8. In the collision time it would travel 1.

This is about times the atomic spacing. Evaluated at the e top limit, this is zero, and evaluated at the bottom, it is the result shown in equation This is very close to the ratio. This is about half a percent of the 0. Molecules and Solids c Because kBT at room temperature is only about eV , while Egap is 1.

Inserting this is returns 2 2m 2m m dk 2m simply m. The exponential factors are: Even with a pool —5 of potential carriers only 10 times as large, the number of carriers from the donor band should significantly outnumber carriers promoted from the valence band. About eV.

The diode represents a huge resistance, so all of the supply voltage is across it. The diode continues to block current until the dogleg is evened out in forward bias, which appears to be somewhere around 2.

At this bias the applied voltage shifts the bands by 2. Once the diode is on, the remainder of the supply voltage is across the resistor, and the larger the voltage across the resistor, the larger the current in the circuit. The photon energy is approximately the gap energy. Therefore, current falls to a negligible value. Once the dogleg is completely smoothed out, i.

It will oscillate above and below by 0. But rather than an electrical contact as in the bipolar, it is the electric field produced by the gate that effects the shift. The only significant current is the flow of conduction electrons through the conduction bands holes through the valence bands in the pnp from source to drain. One half cycle later, current from the bottom high-V side of the Input flows through the lower-right diode to the top of the Output, returning to the top low-V side of the Input through the upper-left diode.

For the potential drop to total 1V, each much me 0. In Section 9. Suppose instead that the area of an oscillating ion is the predominant factor.

The equipartition theorem says that the average potential energy, 12 kA2 , of an oscillator 1 is proportional to T, so the average area would vary as T. This corresponds better to the experimental observations.

But this is a contradiction, for the net force would then be to the left. The object therefore must accelerate to the left. This requires the leftward frictional force to exceed FT, but the net torque can still be counterclockwise, as it would have to be, because the smaller force has the larger moment arm. We see that this equals the work done by FT. A change in microscopic order should alter the electrical resistance.

The light red plot seems to change its character in this temperature range, so it must be the one with the change in spin ordering.

The black curve must be the other metal, consistently losing resistance as temperature drops. The gray curve is probably a semiconductor, whose resistivity increases as the temperature drops and the conduction band empties. The dark red one seems to be a rather normal conductor to a point, where its resistivity appears to vanish.

Probably a hightemperature superconductor. The first is very close; the highest is 2 quite a bit below 14 0. The main difference is that in the separated-atoms case the energies separate into bands. Thus there should be a big energy jump from the former to the latter.

The energies are 2. The scatter plot is shown below with later ones. It clearly shows bands. Energies are similar if the other well is changed. The scatter plot shows two energy levels are split off above each band. The scatter plot shows two energy levels are split off below each band.

The deeper wells are analogous to pentavalent impurities. Impurities with a higher valence create donor states occupied by some electrons, which can easily jump up to the conduction band slightly above. If the objects attract each other, then energy can be extracted from the system in letting them draw near. A lower internal energy implies a smaller mass. A system in a state of high binding energy requires more energy for separation, and is thus in a lower energy state. The most stable configuration of orbiting electrons is helium, but this is an entirely different criterion.

However, as accounted for in the negative second term, this increase is partially offset because the extra proton also adds some surface area—proportional to the volume and number to the two-thirds—where nucleons are not surrounded. This repulsion reduces the binding energy, which is accounted for in the third term. Lastly, the protonneutron imbalance was initially 4, and afterward it is 3, so the last term will be smaller.

A smaller negative is a relative increase in binding energy, because it moves closer to the equal-numbers situation favored by the exclusion principle. No matter how short the half life, each isotope along the way will always be around so long as there are thorium and uranium atoms around.

Lighter nuclei have closer to equal numbers, so freeing the neutrons results in a lower energy. If two particles become only one, the lone final particle in that frame would have to be at rest. This could only represent a loss of kinetic energy. Only if there are multiple final particles could the final kinetic energy exceed the initial.

If merely displaced by two protons and four neutrons, thorium must absorb a neutron and beta decay twice to uranium Not much difference! Therefore, the correct binding energy is Thus, the formula is Figure This is twelve times the number of bonds per nucleon in the deuteron. The binding energy per nucleon for the deuteron is about 1MeV, so the maximum binding energy per nucleon would be 12, if all were surrounded and not exhibiting any repulsion.

Given that not all nucleons are surrounded there, and that Coulomb repulsion must be factored in, the result is sensible. Nitrogen is more tightly bound, by 3. As Z increases there should be a trend toward less stability binding energy per nucleon.

They differ too in the asymmetry term. While the carbon has equal numbers of protons and neutrons, boron and nitrogen, on either side, have two more of one than the other.

They should be equal on this account, less stable than carbon. By these arguments, nitrogen must be the least well bound. The carbon prediction is low. The actual value is considerably higher because of the effect, ignored in the semiempirical binding energy formula, in which the binding is tighter when the numbers of protons or neutrons are even. In carbon, both are even. Still, the trend predicted by the formula agrees with the actual trend.

The actual mass is slightly smaller because factors ignored in the semiempirical binding energy formula, such as the advantage of having even numbers of neutrons and protons, are significant for carbon.

Instructor solutions manual Sears and Zemansky's University physics

The lower binding energy of the surface nucleons is a very Clearly, the fact that not all nucleons are surrounded is less of a factor for larger nuclei. On the pernucleon data, all have the same ideal surrounded energy per nucleon, the destabilizing area term is most significant for the smaller nuclei, and the destabilizing Coulomb and asymmetry terms are dominant factors for larger nuclei.

Iron is the best balance between too much area and too much repulsion. All the main factors involved in nuclear binding change in the same way, except Coulomb repulsion.

It takes less energy to extract something that is repelled by what remains. The volume term is constant; the surface term actually decreases; and the asymmetry term is constant. But in equation we see that the volume of an arbitrary nucleus is proportional to the number of nucleons. If the level of the highest occupied state is the same no matter what the value of A, then the spacing between the levels must decrease proportionally to the number of nucleons.

If both the proton number and neutron number are odd, the net magnetic moment can still be nonzero, but if both are even, the net moment is zero.

They simply have N and Z switched. Given that the slope of the curve here is greater than unity, this too would move the nucleus toward the curve. Does mass decrease? The mass of a nucleus is the mass of its parts minus the absolute value of its binding energy—equation rearranged. Thus, we can see if mass decreases by determining if binding energy increases.

For energy to be released, we must end up with more tightly bound products; the binding energy must increase. It would move toward equalizing the neutron and proton numbers, but not enough to offset the increased coulomb repulsion. The surface energy does increase as smaller nuclei are created, but Coulomb repulsion of the separated fragments from a distance comparable to the fragment separation still accounts for most of the energy released.

Energy would have to be put into the system—the binding energy would end up smaller. The increase in binding energy per nucleon of the nucleons that actually remain bound is insufficient to offset the lost binding energy of the neutrons freed.

Setting their kinetic energy sum to the 4. With 30min per 1. We need the initial decay rate. Of these, 1.

PDF Sears and Zemansky's University Physics with Modern Physics 13th…

By the time the number of radioactive nuclei drops to such a small value, the fluctuations become a factor. About a hundred years. There is a net loss in KE. Kinetic energy is lost, and the unit is unstable against reseparation. Another alpha needs to come along quickly to form carbon. This 26 22 implies that there are 0.

With 2MeV released per atom, the 22 22 10 yield is 3.

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