Start with a day free trial, then just. 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.
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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. This is a short and interactive conditionals type one and type two review practice. Cells Chapter 4. Physics is one of the most fundamental scientific disciplines, and its main goal is to understand how the universe behaves.
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All right reserved. Professional youtube idot. Physics, Volume 1 book. Read 5 reviews from the world's largest community for readers. Jarrell told me that the bullet that killed Jim Eber was a thirty-eight. That was all he told me, just that. But I knew why he told me, it was because his own gun is a thirty-eight. He has always kept it in a drawer of his desk.
Physics — Halliday Resnick Krane — 5th Edition Physics Volume 2 edition 5, by Halliday, Resnik, Krane University Physics Ppt Newton's laws of motion Physics — Halliday Resnick Krane — 5th Edition In classical mechanics , Newton's laws of motion are three laws that describe the relationship between the motion of an object and the forces acting on it.
E We can use Eq. I E dt We start with the left hand side of this expression. The problem has cylindrical symmetry, so the induced electric field lines should be circles centered on the axis of the cylindrical volume. Now for the right hand side. The path of integration is chosen so that if our right hand fingers curl around the path our thumb gives the direction of the magnetic field which cuts through the path. Since the field points into the page a positive electric field would have a clockwise orientation.
Since B is decreasing the derivative is negative, but we get another negative from the equation above, so the electric field has a positive direction. Now for the magnitude. P a We are only interested in the portion of the ring in the yz plane. Your right b From c to b. This means use only one loop to maximize the emf. Since the magnetic field is directed out of the page, a positive emf would be counterclockwise hold your right thumb in the direction of the magnetic field and your fingers will give a counter clockwise sense around the loop.
But the answer was negative, so the emf must be clockwise. The negative sign indicate a decreasing radius. This is identical to the rate of change of gravitational potential energy. First we will find an expression for the emf.
Since B is constant, the emf must be caused by a change in the area; in this case a shift in position. This emf causes a current. Let us assume that it is constrained to that region of the disk. This electric field will result in a force on the free charge carries electrons? P Assume that E does vary as the picture implies.
E a Using the right hand rule a clockwise current would generate a magnetic moment which would be into the page. So that must be the answer. E a The electric field at this distance from the proton is 1 1. E a Look at the figure.
I think both fields are far beyond our current abilities. E Since 0. E Using Eq. The negative indicates inward. E The total magnetic flux through a closed surface is zero. There is inward flux on faces one, three, and five for a total of -9 Wb.
E The stable arrangements are a and c. The torque in each case is zero. The disks can be sliced into rings. Your fingers curl in the direction of the current in the wire loop. Hence, the net force is directed to the left. P b Point the thumb or your right hand to the left. Hence, the net force is directed to the right. E a Eq. Leq L1 L2 b If the inductors are close enough together then the magnetic field from one coil will induce currents in the other coil. Then we will need to consider mutual induction effects, but that is a topic not covered in this text.
The we can invert Eq. E I When the switch is just closed there is no current through the inductor or R2 , so the potential difference across the inductor must be 10 V. The potential difference across R1 is always 10 V when the switch is closed, regardless of the amount of time elapsed since closing.
II After the switch has been closed for a long period of time the currents are stable and the inductor no longer has an effect on the circuit. Then the circuit is a simple two resistor parallel network, each resistor has a potential difference of 10 V across it. Using Eq. The magnetic energy density is found from Eq.
The four possible capacitances are then 2. It requires one-quarter period for the capacitor to charge, or 0. E a An LC circuit oscillates so that the energy is converted from all magnetic to all electrical twice each cycle. It occurs twice because once the energy is magnetic with the current flowing in one direction through the inductor, and later the energy is magnetic with the current flowing the other direction through the inductor.
The period is then four times 1. Note that maximum frequency occurs with minimum capacitance. We want to charge a capacitor with one-ninth the capacitance to have three times the potential difference.
Closing S1 and S2 will not work, because the energy will be shared. Instead, close S2 until the capacitor has completely discharged into the inductor, then simultaneously open S2 while closing S1. The inductor will then discharge into the second capacitor. E The damped angular frequency is given by Eq. Substitute this in and we arrive at Eq.
P Choose the y axis so that it is parallel to the wires and directly between them. As you have been warned so many times before, learn these differentials!
U0 qm For small enough damping we can expand the exponent. It does this through the configuration of the magnets and coils of wire. One complete turn of the generator will could? Not only does this set the frequency, it also sets the emf, since the emf is proportional to the speed at which the coils move through the magnetic field. E a The reactance of the capacitor is from Eq.
This is discussed on page , although it is slightly hidden in the text of column one. But what does this really mean? It means that the inductor plays a major role in the current through the circuit while the capacitor plays a minor role. The more inductive a circuit is, the less significant any capacitance is on the behavior of the circuit. For frequencies below the resonant frequency the reverse is true. The impedance is equal to the resistance, and it is almost as if neither the capacitor or inductor are even in the circuit.
The net x component is R. R E Yes. E a The voltage across the generator is the generator emf, so when it is a maximum from Sample Problem , it is 36 V. For a resistive load we apply Eq. Since this factor of 2 appears in all of the expressions, we can conclude that if the rms values are equal then so are the maximum values. Focus on the right hand side of the last equality.
Then, according to Eq. So everything is V. The power factor is one for a system in resonance. Finally, the rms current is E rms When only one device is between the two points the impedance is equal to the reactance or resistance of that device.
E Apply Eq. Np 65 E a Apply Eq. The same possibilities are true for the secondary connections. Ignoring the one-to-one connections there are 6 choices— three are step up, and three are step down. The step down ratios are the reciprocals of these three values. This is the amount by which the supply voltage must be increased. From Eq. In position 1 the circuit is an RLC circuit, but the capacitance is equal to the two capacitors of C in parallel, or 2C. In position 2 the circuit is a simple LC circuit with no resistance.
We can then choose any two wires and expect by symmetry to get the same result. We choose 1 and 2. We need to add these two sine functions to get just one. Resonance circuits have a power factor of one.
P P Use Eq. But the resistor would get hot, while on average there is no power radiated from a pure inductor. Then the two values of r are 2. H d B E Substitute Eq. E a Consider the path abef a. I E dt b Do everything above again, except substitute B for E. This multi-valued result would be quite unphysical. E a Consider the part on the left. It has a shared surface s, and the other surfaces l. Note that dA Consider the part on the right.
It has a shared surface s, and the other surfaces r. Magnetic dipole because the current loop acts like a magnetic dipole. E The electric and magnetic field of an electromagnetic wave are related by Eqs. E Use the right hand rule. It is in the positive x direction. E a Intensity is related to distance by Eq. E a The electric field amplitude is related to the intensity by Eq. E Radiation pressure for absorption is given by Eq.
N m Am N Cm Am sm m m s E We can treat the object as having two surfaces, one completely reflecting and the other completely absorbing. A c E We can treat the object as having two surfaces, one completely reflecting and the other completely absorbing. Add these two energy densities to get the net energy density outside the surface. Solving for H, 8P 8 4. The value for I is in Ex.
P For the two outer circles use Eq. Energy only passes through the yz faces; it goes in one face and out the other. We can then deal with a scalar, instead of vector, integral, and we can integrate it in any order we want. Then we are done! R d Read part b above. If the thumb is in the direction of current then the fingers of the right hand grip ion the direction of the magnetic field lines. S is found from the cross product of these two, and must be pointing radially inward.
If we assume that this is the data transmission rate in bits per second a generous assumption , then it would take days to download a web-page which would take only 1 second on a 56K modem! E a We refer to Fig. The limits are approximately nm and nm. E a 2 4. E This is a question of how much time it takes light to travel 4 cm, because the light traveled from the Earth to the moon, bounced off of the reflector, and then traveled back.
Note that I interpreted the question differently than the answer in the back of the book. The path of the ray can be projected onto the xy plane, the xz plane, or the yz plane. If the projected rays is exactly reflected in all three cases then the three dimensional incoming ray will be reflected exactly reversed. But the problem is symmetric, so it is sufficient to show that any plane works. Now the problem has been reduced to Sample Problem , so we are done.
There is no loss of generality in doing so; we had to define our coordinate system somehow. The choice is convenient in that any normal is then parallel to the z axis.
Furthermore, we can arbitrarily define the incident ray to originate at 0, 0, z1. Lastly, we can rotate the coordinate system about the z axis so that the reflected ray passes through the point 0, y3 , z3. The point of reflection for this ray is somewhere on the surface of the mirror, say x2 , y2 , 0.
The only point which is free to move is the reflection point, x2 , y2 , 0 , and that point can only move in the xy plane. We do this minimization by taking the partial derivative with respect to both x2 and y2. But we can do part by inspection alone. Any non-zero value of x2 can only add to the total distance, regardless of the value of any of the other quantities.
We are done! The normal is parallel to the z axis, so it also lies in the yz plane. Everything is then in the yz plane. E Refer to Page of Volume 1. We can use these two angles, along with the index of refraction of air, to find that the index of refraction of the liquid from Eq. Combining, nL n2 L 1. Let x be the distance between the points on the surface where the vertical ray crosses and the bent ray crosses. E Use the results of Ex. This makes a huge difference for part c!
E a The critical angle is given by Eq. The diameter is twice this radius, or 2 0. Since part of the light undergoes total internal reflection while the other part does not, then the angle of incidence must be approximately equal to the critical angle.
As the wavelength increases the index of refraction decreases. In short, red light has a larger critical angle than blue light. If the angle of incidence is midway between the critical angle of red and the critical angle of blue, then the blue component of the light will experience total internal reflection while the red component will pass through as a refracted ray.
So yes, the light can be made to appear bluish. See above. Getting the effect to work will require considerable sensitivity. E a There needs to be an opaque spot in the center of each face so that no refracted ray emerges. The radius of the spot will be large enough to cover rays which meet the surface at less than the critical angle. Each side has an area of E The sun rotates once every 26 days at the equator, while the radius is 7.
E a No relative motion, so every 6 minutes. The shift is The frequency observed by the detector from the second source is Eq. Then 0. E P Consider the triangle in Fig. The true position corresponds to the speed of light, the opposite side corresponds to the velocity of earth in the orbit. Water waves travel more slowly in shallower water, which means they always bend toward the normal as they approach land.
Once in the water the problem is identical to Sample Problem The reflected ray in the water is parallel to the incident ray in the water, so it also strikes the water normal, and is transmitted normal. Once the ray is in the water then the problem is identical to Sample Problem It is mildly entertaining to note that the value of n0 is unim- portant, only the value of a!
We will ignore the ay term in the denominator because it will always be small compared to 1. P a The fraction of light energy which escapes from the water is dependent on the critical angle. Light radiates in all directions from the source, but only that which strikes the surface at an angle less than the critical angle will escape. We never needed to know the depth h.
P Consider the two possible extremes: a ray of light can propagate in a straight line directly down the axis of the fiber, or it can reflect off of the sides with the minimum possible angle of incidence. Start with the harder option. The actual answer for the speed of the airplane is half this because there were two Doppler shifts: once when the microwaves struck the plane, and one when the reflected beam was received by the station.
Hence, the plane approaches with a speed of E You are 30 cm from the mirror, the image is 10 cm behind the mirror. You need to focus 40 cm away.
E Sketch a line from Sarah through the right edge of the mirror and then beyond. Sarah can see any image which is located between that line and the mirror. E The images are fainter than the object. Several sample rays are shown. E The image is displaced. The eye would need to look up to see it.
E Three. There is a single direct image in each mirror and one more image of an image in one of the mirrors. By similar triangles the diameter of the pupil and the diameter of the part of the mirror d which reflects light into the eye are related by d 5. E a Seven; b Five; and c Two. This is a problem of symmetry. E Seven. Three images are the ones from Exercise 8. But each image has an image in the ceiling mirror. That would make a total of six, except that you also have an image in the ceiling mirror look up, eh?
So the total is seven! E A point focus is not formed. The envelope of rays is called the caustic. You can see a similar effect when you allow light to reflect off of a spoon onto a table. E The image is magnified by a factor of 2.
An important question to ask is whether or not the image is real or virtual. If it is a virtual image it is behind the mirror and someone looking at the mirror could see it. If it were a real image it would be in front of the mirror, and the man, who serves as the object and is therefore closer to the mirror than the image, would not be able to see it.
So we shall assume that the image is virtual. The image distance is then a negative number. The focal length is half of the radius of curvature, so we want to solve Eq. All dimensioned variables below f, r, i, o are measured in centimeters. Substitute Eq. E a Consider the point A. Light from this point travels along the line ABC and will be parallel to the horizontal center line from the center of the cylinder. Since the tangent to a circle defines the outer limit of the intersection with a line, this line must describe the apparent size.
All dimensioned variables below r, i, o are measured in centimeters. E b If the beam is small we can use Eq. Parallel incoming rays correspond to an object at infinity.
Starting with Eq. The lens is converging since f is positive. Since m is positive and greater than one the lens is converging. Then f is positive. Since m is positive and less than one the lens is diverging. Then f is negative. Upright images have positive magnification. Real images have negative magnification. E a Real images from real objects are only produced by converging lenses. E Step through the exercise one lens at a time. Since i is positive it is a real image, and it is located to the right of the converging lens.
This image becomes the object for the diverging lens. This would mean the image formed by the diverging lens would be a virtual image, and would be located to the left of the diverging lens. The image is virtual, so it is upright.
E a The parallel rays of light which strike the lens of focal length f will converge on the focal point. This point will act like an object for the second lens. This real image is 1. It acts as an object for the mirror. 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. 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. 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.
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