Resnick halliday solutions pdf

The Circuits chapter makes you understand about the ammeter and the voltmeter, calculating the current in a single loop circuit, multiple loop circuits, pumping charges, RC circuits, energy and EMF and the potential difference between two points.

The Magnetic Fields chapter teaches you about the magnetic force on a current-carrying wire, magnetic field, crossed fields: the hall effect, crossed fields: discovery of the electron, torque on a current loop, cyclotrons and synchrotrons, a circulating charged particle and the magnetic dipole moment.

Fundamentals of Physics Solutions by Resnick Halliday and Walker contain detailed explanations of all the concepts along with the diagrams and relevant examples. The Chapter Electromagnetic Oscillations and Alternating Current introduce you to power in alternating current circuits, damped oscillations in an RLC circuit, qualitatively, charge and current oscillations, the series RLC circuit, the electrical-mechanical analogy, transformers, LC oscillations, alternating current, angular frequencies and forced oscillations.

The chapter Images teaches you about thin lenses, plane mirrors and spherical mirrors, optical instruments, images from spherical mirrors and spherical refracting surfaces. The Diffraction chapter explains about diffraction by a double slit, intensity in single slit diffraction, single slit diffraction, X-Ray diffraction, diffraction by a single slit: locating the minima, gratings: dispersion, diffraction by a circular aperture, resolving power and diffraction and the wave theory of light.

Relativity concept introduces you to momentum and energy, the relativity of time, the simultaneity and time dilation, Doppler effect for light, measuring an event, the Lorentz transformation, the relativity of simultaneity, the relativity of length and the postulates.

Our Halliday Resnick Solutions are prepared methodically and logically by experts. The chapter More About Matter Waves teaches you about an electron in a finite well, string wave and matter waves, the hydrogen atom, energies of a trapped electron, two and three-dimensional electron traps and wave functions of a trapped electron. The All About Atoms topic gives you an idea of X-rays and the ordering of the elements, magnetic resonance, magnetic dipole moments, building the periodic table, the stern Gerlach experiment, multiple electrons in a trap, angular momentum, exclusion principle, lasers and properties of atoms.

The Conduction of Electricity in Solids Chapter educates you about semiconductors and doping, the transistors, the electrical properties of metals and the p- n junction. The Nuclear Physics Chapter introduces you to measuring radiation dosage, some nuclear properties, nuclear models, discovering the nucleus, radioactive dating, radioactive decay, beta decay and alpha decay. The chapter Energy from the nucleus explains to you thermonuclear fusion in the sun and other stars, nuclear fission, controlled thermonuclear fusion, the nuclear reactor, thermonuclear fusion and a natural nuclear reactor.

The chapter Quarks, Leptons and the Big Bang teach you about the cause for reflection, leptons, the basic forces and messenger particles, general properties of elementary particles, the Bing Bang and a summing up, the eightfold way, quarks and messenger particles and hadrons and strangeness.

Why refer to Halliday Resnick Solutions by Instasolv? The solutions have been drafted in a step by step manner for better and easy understanding of the concept. They include a detailed explanation of all the concepts, along with the appropriate diagrams and relevant examples wherever necessary. Our smart study techniques and shortcuts learning will help you grasp the concepts quickly and learn effectively. They cover a wide range of question types important from exam point of view They help you learn and revise quickly and efficiently.

Our solutions help you build a strong conceptual foundation They have been designed to clear all your doubts and improve your retention rate. They have been solved using the best methods of solving the problems. HC Verma Solutions. DC Pandey Physics. MS Chauhan Organic Chemistry. Higher Algebra by Hall and Knight. Share this: Twitter Facebook. Download the book.

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Vol 2. A physics book for university use and its solution manual is really useful! Modern Physics Pdf. With respect to its significance and applicability,we are going to launch test preparation channel over here from you can get mcqs with answers. Recorded October 15, at Stanford University. The seminal work by one of the most important thinkers of the twentieth century, Physics and Philosophy is Werner Heisenberg's concise and accessible narrative of the revolution in modern physics, in which he played a towering role.

This text presents an introduction to relativity, quantum mechanics, and statistical physics as well as applications of these theories to molecular physics, condensed matter physics, nuclear physics, particle physics, and cosmology. Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website. Please help us to share our service with your friends. College Physics 11th Edition Chapter 2 The chapters which come in class 11 and 12 under mechanical physics are: 1.

My professor only gives a few problems to work on per chapter so this has definitely helped. The mirror produces a virtual image 0. This image is upright relative to the object which formed it, which was inverted relative to the original object.

This second image is 1. So this image is actually upright. The image and object distance are the same, so the image has a magnification of 1. This image is 0. Note that this puts the final image at the location of the original object! The image is magnified by a factor of 0. In effect, these were tiny glass balls.

E a In Fig. This means that in Fig. The object should be placed 5. E Microscope magnification is given by Eq. We need to first find the focal length of the objective lens before we can use this formula.

The images of the two ends will be located at i1 and i2. Since we are told that the object has a short length L we will assume that a differential approach to the problem is in order.

Real image and objects occur when y or y 0 is positive. We can use this information to eliminate one variable from Eq. This last expression is a quadratic, and we would expect to get two solutions for o.

P a The angular size of each lens is the same when viewed from the shared focal point. Let the object distance be o1. This means it will act as an object o3 in the lens, and, reversing the first step, produce a final image at O, the location of the original object. There are then three images formed; each is real, same size, and inverted. Three inversions nets an inverted image. The final image at O is therefore inverted.

P We want the maximum linear motion of the train to move no more than 0. The object distance is much larger than the focal length, so the image distance is approximately equal to the focal length.

The size of an object on the train that would produce a 0. How much time does it take the train to move that far? Note that we were given the radius of curvature, not the focal length, of the mirror! If the interference fringes are 0. E A variation of Eq. The total is E Consider Fig.

The distance from A to the detector is m longer than the distance from B to the detector. Since the wavelength is m, m corresponds to a quarter wavelength. So a wave peak starts out from source A and travels to the detector.

When it has traveled a quarter wavelength a wave peak leaves source B. But when the wave peak from A has traveled a quarter wavelength it is now located at the same distance from the detector as source B, which means the two wave peaks arrive at the detector at the same time. They are in phase.

E a We want to know the path length difference of the two sources to the detector. If this difference is an integral number of wavelengths then we have a maximum; if instead it is a half integral number of wavelengths we have a minimum.

The right hand side becomes indeterminate, so we need to go back to the first line in the above derivation. In fact, we may have even more troubles. These maxima are located at 4. It corresponds to a point where the path length difference is 3.

It should be half an integer to be a complete minimum. E Follow the example in Sample Problem E a Light from above the oil slick can be reflected back up from the top of the oil layer or from the bottom of the oil layer. We need to find the wavelength in the visible range nm to nm which has an integer m. Trial and error might work. If we increase m to 3, then 2 1. So the oil slick will appear green. Finding the maximally transmitted wavelengths is the same as finding the minimally reflected wavelengths, or looking for values of m that are half integer.

The wave which is reflected off of the second surface travels an additional path difference of 2d. E As with the oil on the water in Ex. Consequently, 2 1. There are then bright bands. It will be bright! Well, at least we got the answer which is in the back of the book E Pretend the ship is a two point source emitter, one h above the water, and one h below the water.

The one below the water is out of phase by half a wavelength. Then I 3. Then I 2. This is because the light travels twice through any change in distance. The wavelength of light is then 2 0. How does it shift?

Since we are picking up 2. Yes, that is the equation of a hyperbola. P Follow the construction in Fig. One maxi- mum is 2 1. This is blue-violet.

All that cancels out. E a This is a valid small angle approximation problem: the distance between the points on the screen is much less than the distance to the screen. E a We again use Eq.

If the angles match, then so will the sine of the angles. We want to solve for the values of ma and mb that will be integers and have the same angle.

Since the width of the central maximum is effectively cut in half, then there is twice the energy in half the space, producing four times the intensity. For Fig. The angle against the vertical is then 0. E The smallest resolvable angular separation will be given by Eq. Now use Eq. E Remember that the central peak has an envelope width twice that of any other peak.

The fact that the answer is exactly 5 implies that the fifth interference maximum is squelched by the diffraction minimum. Then there are only four complete fringes on either side of the central maximum. Add this to the central maximum and we get nine as the answer. The separation is twice this, or 2. In the event that either shape produces an interference pattern at P then the other shape must produce an equal but opposite electric field vector at that point so that when both patterns from both shapes are superimposed the field cancel.

But the intensity is the field vector squared; hence the two patterns look identical. P a We want to take the derivative of Eq. These values will get closer and closer to integers as the values are increased.

The light is not evenly distributed over this circle. P a The ring is reddish because it occurs at the blue minimum. P The diffraction pattern is a property of the speaker, not the interference between the speak- ers.

The diffraction pattern should be unaffected by the phase shift. The interference pattern, however, should shift up or down as the phase of the second speaker is varied. There should be some sort of interference fringe, unless the diffraction pattern has a minimum at that point. E We want to find a relationship between the angle and the order number which is linear. That will be the longer wavelengths, so we only need to look at the nm behavior. Consequently, all even m are at diffraction minima and therefore vanish.

E If the second-order spectra overlaps the third-order, it is because the nm second-order line is at a larger angle than the nm third-order line. Start with the wavelengths multiplied by the appropriate order parameter, then divide both side by d, and finally apply Eq.

E Fig. E The required resolving power of the grating is given by Eq. The number of lines required is given by Eq. The value of d depends on the family of planes under consideration, but it is at never larger than a0 , the unit cell dimension.

We are looking for the smallest angle; this will correspond to the largest d and the smallest m. Then the minimum angle is 1 E We apply Eq. The four values are 26 pm, 39 pm, 52 pm, and 78 pm. The last two values are twice the first two, so the wavelengths are 26 pm and 39 pm.

E There are too many unknowns. The angles in this case are then given by 0. They are P Since the slits are so narrow we only need to consider interference effects, not diffraction effects.

There are three waves which contribute at any point. We can add the electric field vectors as was done in the previous chapters, or we can do it in a different order as is shown in the figure below.

We need to square this quantity, and then normalize it so that the central maximum is the maximum. The half width in the three slit case is smaller. P a and b A plot of the intensity quickly reveals that there is an alternation of large maximum, then a smaller maximum, etc.

P Covering up the middle slit will result in a two slit apparatus with a slit separation of 2d. We then assume incorrectly that the secondary maxima occur when the loop wraps around on itself as shown in the figures below.

Note that the resultant phasor always points straight up. P b We sketch parallel lines which connect centers to form almost any right triangle similar to the one shown in the Fig. In Fig. The two beams will then cancel out exactly because of destructive interference. The magnitude of E will be equal to the magnitude of B times c.

E Let one wave be polarized in the x direction and the other in the y direction. Consequently, there are no interference effects. The second sheet transmits according to Eq. The transmitted beam then has a polarization set by the first sheet: The second sheet is horizontal, which puts it Then the second sheet transmits cos2 E The smallest possible thickness t will allow for one half a wavelength phase difference for the o and e waves.

As such, there is no apparent change. The intensity of the transmitted light which was originally polarized is between 0 and I p , depending on the orientation of the polarizing sheet.

The stack then transmits all of the light which makes it past the first filter. Assuming the light is originally unpolarized, then the stack transmits half the original intensity.

On the other hand if the light passes first through the plate, then through the polarizer, then is reflected, the passes again through the polarizer, all the reflected light will pass through he polarizer and eventually work its way out through the plate. So the coin will be visible. These energies correspond to wavelengths between nm and 1.

Since longer wavelengths have lower energies, the bulb emitting nm must be giving off more photons per second. The number of photons is Et 1. Then the temperature would be 91 K. Then the new power level is 16 E a We want to apply Eq. This means that only cesium will work with red light. E a The stopping potential is given by Eq. Then h 6. E The change in the wavelength of a photon during Compton scattering is given by Eq.

E mc2 Note that mc2 is the rest energy of the scattering particle usually an electron , while hf 0 is the energy of the scattered photon. Then we can use the results of Exercise to get 0. E We can use the results of Exercise to get 0. P The radiant intensity is given by Eq. But energy goes both ways through the opening; it is the difference that will give the net power transfer.

As the temperature of the screen increases it will begin to radiate energy. When the rate of energy radiation from the screen is equal to the rate at which the energy from the sun strikes the screen we will have equilibrium.

We need first to find an expression for the rate at which energy from the sun strikes the screen. The temperature of the sun is T S. The radiant intensity is given by Eq. Assuming that the lens is on the surface of the Earth a reasonable assumption , then we can find the power incident on the lens if we know the intensity of sunlight at the distance of the Earth from the sun. All of the energy that strikes the lens is focused on the image, so the power incident on the lens is also incident on the image.

The screen radiates as the temperature increases. A steady state planet temperature requires that the energy from the sun arrive at the same rate as the energy is radiated from the planet. Not likely. How much too small?

If the radius of the galaxy were one meter, this distance would correspond to the diameter of a proton. Then from Eq. E This is merely a Bragg reflection problem. Increasing the potential will increase the kinetic energy, increase the momentum, and decrease the wavelength.

But the odd orders vanish see Chapter 43 for a discussion on this. E Apply Heisenberg twice: 4. E Apply Heisenberg: 6. The position uncertainty would then be 0.

Apply Eq. Keep at it. Then apply Eq. Since the real part of the left hand side must equal the real part of the right and the imaginary part of the left hand side must equal the imaginary part of the right, we actually have two equations. Note that this is an extremely relativistic quantity, so the energy expression loses validity. The energy of such an electron is considerably larger than binding energies of the particles in the nucleus. It is bouncing back and forth between the walls of the box, so the momentum could be directed toward the right or toward the left.

The least energetic state starts on E2. The least energetic state starts on E3. E The binding energy is the energy required to remove the electron. If the energy of the electron is negative, then that negative energy is a measure of the energy required to set the electron free. The initial state is only slightly higher than the final state. The six possible results are E In order to have an inelastic collision with the 6. Since the difference is The state with an excitation energy of This is the momentum of the recoiling hydrogen atom, which then has velocity p pc So it must also include some visible lines.

E We answer these questions out of order! P 2a0 2a0 e 0 0 4e E The probability is 1. It means that if we look for the electron, we will find it somewhere. E a Find the maxima by taking the derivative and setting it equal to zero. The first two correspond to minima see Fig. The results are P 0. E The probability is 5. There are 14 possible states. E There are n possible values for l start at 0! The pattern is clear, the sum is n2. But there are two spin states, so the number of states is 2n2.

The next choice is to set one of the values equal to 2, and try the set 2, 1, 1. Then it starts to get harder, as the next lowest might be either 2, 2, 1 or 3, 1, 1.

The only way to find out is to try. P a Write the states between 0 and L. This will make our lives easier later on. P Assume the electron is originally in the state n. This is the distance between the particles, but they are both revolving about the center of mass.

The radius is then half this quantity, or According to Eq. It is a property of the atom, not a property of the accelerating potential of the x-ray tube. See part b. E The The electron is now a On the third collision the electron loses the remaining energy, so this photon has an energy of E a The x-ray will need to knock free a K shell electron, so it must have an energy of at least E Remember that the m in Eq.

This means that the constant C in Eq. Or we could assume that the L shell electron is promoted to the M shell. The question that we would need to answer is which of these possibilities has the lowest energy. The answer is the last choice: increasing the l value results in a small increase in the energy of multi-electron atoms.

E Refer to Sample Problem a0 1 2 5. If there is no spin the s sub-shell would hold 1 electron, the p sub-shell would hold 3, the d sub-shell 5, and the f sub-shell 7.

Is there a pattern? The new inert gases have half of the atomic number of the original inert gases. The factor of one-half comes about because there are no longer two spin states for each set of n, l, ml quantum numbers. We can save time and simply divide the atomic numbers of the remaining inert gases in half: element 18 Argon , element 27 Cobalt , element 43 Technetium , element 59 Praseodymium.

E a Apply Eq. E a There are three ml states allowed, and two ms states. The first electron can be in any one of these six combinations of M1 and m2. The second electron, given no exclusion principle, could also be in any one of these six states. Of this total of 36, six involve the electrons being in the same state, while 30 involve the electron being in different states.

See the above discussion. Although the Bohr theory correctly predicts the magnitudes, it does not correctly predict when these values would occur. Then F y2 1. E The energy change can be derived from Eq. The corresponding wavelength is hc 6. E We need to find out how many 10 MHz wide signals can fit between the two wavelengths. The lower frequency is c 3. That would require a negative temperature.

Combining, and rearranging, 1. Only the positive answer has physical meaning. The energy of the second photon must be 5.

P Switch to a reference frame where the electron is originally at rest. Then, according to the derivation leading to Eq. Look back to Sample Problem ; we need to use some of the results from that Sample Problem to solve this problem. The factor of e2 in Eq. One last thing. There are two electrons, so we need to double the above expression. This compares well with the accepted value. P Applying Eq. E Apply the results of Ex.

E Monovalent means only one electron is available as a conducting electron. V E a The approximate volume of a single sodium atom is 0. For gold we have 3 First, the density is m 1. The Fermi energy is then! E Using the results of Exercise 19, 2f E F 2 0. E a Monovalent means only one electron is available as a conducting electron. The time between collisions is m 9. The extra factor of two is because all of the charge carriers in silicon holes and electrons are charge carriers.

E The first one is an insulator because the lower band is filled and band gap is so large; there is no impurity. The second one is an extrinsic n-type semiconductor: it is a semiconductor because the lower band is filled and the band gap is small; it is extrinsic because there is an impurity; since the impurity level is close to the top of the band gap the impurity is a donor.

The third sample is an intrinsic semiconductor: it is a semiconductor because the lower band is filled and the band gap is small. The fourth sample is a conductor; although the band gap is large, the lower band is not completely filled. The fifth sample is a conductor: the Fermi level is above the bottom of the upper band. The sixth one is an extrinsic p-type semiconductor: it is a semiconductor because the lower band is filled and the band gap is small; it is extrinsic because there is an impurity; since the impurity level is close to the bottom of the band gap the impurity is an acceptor.

E 6. Because longer wavelengths would have lower energy, and so not enough to cause an electron to jump across the band gap. As such, the photon will be absorbed, which means the crystal is opaque to this wavelength. E P We can calculate the electron density from Eq. Using the results of Ex. It should look like Fig. The ratio is then if e 0. E We can make an estimate of the mass number A from Eq. If the measurements indicate a radius of 3.

E a Since the binding energy per nucleon is fairly constant, the energy must be proportional to A. The electrostatic energy must be proportional to this. E Solve 0. The number of atoms in the sample is 0. E We will do this one the easy way because we can. The activity of the sample will fall to one-half of the initial decay rate after one half-life; it will fall to one-half of one-half one-fourth after two half-lives.

The number of decays, each of which produced an alpha particle, is 4. The difference is the amount which decayed during the two hour interval: 2. We first need to know the decay constant from Eq. Not all decay events will be picked up by a detector and recorded as a count; we are assuming that whatever scaling factor which connects the initial count rate to the initial decay rate is valid at later times as well.

Such an assumption is a reasonable assumption. But the activity of the tracer decays with time, and so without a correction factor we would record the wrong amount of phosphorus in the leaf.

R0 E The number of particles of Sm is 0. The mass of helium produced is then 0. Only reactions with positive Q values are energetically possible. E a The kinetic energy of this electron is significant compared to the rest mass energy, so we must use relativity to find the momentum.

Although the rules for standing waves in a box are slightly more complicated, it is a fair assumption that the electron could not exist as a standing a wave in the nucleus.

This is also the magnitude of the momentum of the recoiling 32 S. NA ln 2NA Then for the sample in question 3. We are told the half life, but to find the number of radioactive nuclei present we want to know the decay constant. It is helpful to work backwards before proceeding by asking the following question: what nuclei will we have if we subtract one of the allowed projectiles? Removing a proton will leave 26 protons and 33 neutrons, which is Fe; but that nuclide is unstable.

Removing a neutron will leave 27 protons and 32 neutrons, which is Co; and that nuclide is stable. Removing a deuteron will leave 26 protons and 32 neutrons, which is Fe; and that nuclide is stable. It looks as if only 59 Co n 60 Co and 58 Fe d 60 Co are possible. If, however, we allow for the possibility of other daughter particles we should also consider some of the following reactions.

Swapping a neutron for a proton: Ni n,p 60 Co. Using a neutron to knock out a deuteron: Ni n,d 60 Co. E E Shells occur at numbers 2, 8, 20, 28, 50, The shells occur separately for protons and neutrons.

E a The binding energy of this neutron can be found by considering the Q value of the reaction 90 Zr n 91 Zr which is This neutron is bound more tightly that the one in part a. If R exceeds P then nuclei will decay faster than they are produced; but this will cause N to decrease, which means R will decrease until it is equal to P. If R is less than P then nuclei will be produced faster than they are decaying; but this will cause N to increase, which means R will increase until it is equal to P.

The factor 0. There are ten protons in each water molecule. The Q value for the reaction is then 4. The difference between a and b is P a The emitted positron leaves the atom, so the mass must be subtracted. But the daughter particle now has an extra electron, so that must also be subtracted. P a Capturing an electron is equivalent to negative beta decay in that the total number of electrons is accounted for on both the left and right sides of the equation. The loss of the K shell electron, however, must be taken into account as this energy may be significant.

The total of the three is 9. E a There are 1. This requires 1. E There are 1. This amount of energy would keep a W lamp lit for 8. Gallium has 31 protons and around 39 neutrons; chromium has 24 protons and around 28 neutrons. There are then 37 protons and around 75 neutrons left over. This would be rubidium, but the number of neutrons is very wrong.

Although the elemental identification is correct, because we must conserve proton number, the isotopes are wrong in our above choices for neutron numbers. E Beta decay is the emission of an electron from the nucleus; one of the neutrons changes into a proton. The atom now needs one more electron in the electron shells; by using atomic masses as opposed to nuclear masses then the beta electron is accounted for.

This is only true for negative beta decay, not for positive beta decay. The negative value implies that this fission reaction is not possible.