What will be the uncertainty in position if the uncertainty in momentum of electron is 1/10 kgm s given H 6.62 10 kg/m sec?

By the end of this section, you will be able to:

Use both versions of Heisenberg’s uncertainty principle in calculations.
Explain the implications of Heisenberg’s uncertainty principle for measurements.

Matter and photons are waves, implying they are spread out over some distance. What is the position of a particle, such as an electron? Is it at the center of the wave? The answer lies in how you measure the position of an electron. Experiments show that you will find the electron at some definite location, unlike a wave. But if you set up exactly the same situation and measure it again, you will find the electron in a different location, often far outside any experimental uncertainty in your measurement. Repeated measurements will display a statistical distribution of locations that appears wavelike. (See Figure 1.)

Figure 1. The building up of the diffraction pattern of electrons scattered from a crystal surface. Each electron arrives at a definite location, which cannot be precisely predicted. The overall distribution shown at the bottom can be predicted as the diffraction of waves having the de Broglie wavelength of the electrons.

After de Broglie proposed the wave nature of matter, many physicists, including Schrödinger and Heisenberg, explored the consequences. The idea quickly emerged that, because of its wave character, a particle’s trajectory and destination cannot be precisely predicted for each particle individually. However, each particle goes to a definite place (as illustrated in Figure 1). After compiling enough data, you get a distribution related to the particle’s wavelength and diffraction pattern. There is a certain probability of finding the particle at a given location, and the overall pattern is called a probability distribution. Those who developed quantum mechanics devised equations that predicted the probability distribution in various circumstances.

It is somewhat disquieting to think that you cannot predict exactly where an individual particle will go, or even follow it to its destination. Let us explore what happens if we try to follow a particle. Consider the double-slit patterns obtained for electrons and photons in Figure 2. First, we note that these patterns are identical, following d sin θ = mλ, the equation for double-slit constructive interference developed in Photon Energies and the Electromagnetic Spectrum, where d is the slit separation and λ is the electron or photon wavelength.

Figure 2. Double-slit interference for electrons (a) and photons (b) is identical for equal wavelengths and equal slit separations. Both patterns are probability distributions in the sense that they are built up by individual particles traversing the apparatus, the paths of which are not individually predictable.

Both patterns build up statistically as individual particles fall on the detector. This can be observed for photons or electrons—for now, let us concentrate on electrons. You might imagine that the electrons are interfering with one another as any waves do. To test this, you can lower the intensity until there is never more than one electron between the slits and the screen. The same interference pattern builds up! This implies that a particle’s probability distribution spans both slits, and the particles actually interfere with themselves. Does this also mean that the electron goes through both slits? An electron is a basic unit of matter that is not divisible. But it is a fair question, and so we should look to see if the electron traverses one slit or the other, or both. One possibility is to have coils around the slits that detect charges moving through them. What is observed is that an electron always goes through one slit or the other; it does not split to go through both. But there is a catch. If you determine that the electron went through one of the slits, you no longer get a double slit pattern—instead, you get single slit interference. There is no escape by using another method of determining which slit the electron went through. Knowing the particle went through one slit forces a single-slit pattern. If you do not observe which slit the electron goes through, you obtain a double-slit pattern. How does knowing which slit the electron passed through change the pattern? The answer is fundamentally important—measurement affects the system being observed. Information can be lost, and in some cases it is impossible to measure two physical quantities simultaneously to exact precision. For example, you can measure the position of a moving electron by scattering light or other electrons from it. Those probes have momentum themselves, and by scattering from the electron, they change its momentum in a manner that loses information. There is a limit to absolute knowledge, even in principle.

Figure 3. Werner Heisenberg was one of the best of those physicists who developed early quantum mechanics. Not only did his work enable a description of nature on the very small scale, it also changed our view of the availability of knowledge. Although he is universally recognized for his brilliance and the importance of his work (he received the Nobel Prize in 1932, for example), Heisenberg remained in Germany during World War II and headed the German effort to build a nuclear bomb, permanently alienating himself from most of the scientific community. (credit: Author Unknown, via Wikimedia Commons)

It was Werner Heisenberg who first stated this limit to knowledge in 1929 as a result of his work on quantum mechanics and the wave characteristics of all particles. (See Figure 3). Specifically, consider simultaneously measuring the position and momentum of an electron (it could be any particle). There is an uncertainty in position Δx that is approximately equal to the wavelength of the particle. That is, Δx ≈ λ.

As discussed above, a wave is not located at one point in space. If the electron’s position is measured repeatedly, a spread in locations will be observed, implying an uncertainty in position Δx. To detect the position of the particle, we must interact with it, such as having it collide with a detector. In the collision, the particle will lose momentum. This change in momentum could be anywhere from close to zero to the total momentum of the particle,

$p=hλp=\frac{h}{\lambda}\\$

. It is not possible to tell how much momentum will be transferred to a detector, and so there is an uncertainty in momentum Δp, too. In fact, the uncertainty in momentum may be as large as the momentum itself, which in equation form means that

$Δp≈hλ\Delta{p}\approx\frac{h}{\lambda}\\$

The uncertainty in position can be reduced by using a shorter-wavelength electron, since Δx ≈ λ. But shortening the wavelength increases the uncertainty in momentum, since

$p=hλp=\frac{h}{\lambda}\\$

. Conversely, the uncertainty in momentum can be reduced by using a longer-wavelength electron, but this increases the uncertainty in position. Mathematically, you can express this trade-off by multiplying the uncertainties. The wavelength cancels, leaving ΔxΔp ≈ h.

So if one uncertainty is reduced, the other must increase so that their product is ≈h.

With the use of advanced mathematics, Heisenberg showed that the best that can be done in a simultaneous measurement of position and momentum is

$ΔxΔp≥h4π\Delta{x}\Delta{p}\ge\frac{h}{4\pi}\\$

This is known as the Heisenberg uncertainty principle. It is impossible to measure position x and momentum p simultaneously with uncertainties Δx and Δp that multiply to be less than

$h4π\frac{h}{4\pi}\\$

. Neither uncertainty can be zero. Neither uncertainty can become small without the other becoming large. A small wavelength allows accurate position measurement, but it increases the momentum of the probe to the point that it further disturbs the momentum of a system being measured. For example, if an electron is scattered from an atom and has a wavelength small enough to detect the position of electrons in the atom, its momentum can knock the electrons from their orbits in a manner that loses information about their original motion. It is therefore impossible to follow an electron in its orbit around an atom. If you measure the electron’s position, you will find it in a definite location, but the atom will be disrupted. Repeated measurements on identical atoms will produce interesting probability distributions for electrons around the atom, but they will not produce motion information. The probability distributions are referred to as electron clouds or orbitals. The shapes of these orbitals are often shown in general chemistry texts and are discussed in The Wave Nature of Matter Causes Quantization.

If the position of an electron in an atom is measured to an accuracy of 0.0100 nm, what is the electron’s uncertainty in velocity?
If the electron has this velocity, what is its kinetic energy in eV?

The uncertainty in position is the accuracy of the measurement, or Δx = 0.0100 nm. Thus the smallest uncertainty in momentum Δp can be calculated using

$ΔxΔp≥h4π\Delta{x}\Delta{p}\ge\frac{h}{4\pi}\\$

. Once the uncertainty in momentum Δp is found, the uncertainty in velocity can be found from Δp = mΔv. Using the equals sign in the uncertainty principle to express the minimum uncertainty, we have

$ΔxΔp=h4π\displaystyle\Delta{x}\Delta{p}=\frac{h}{4\pi}\\$

Solving for Δp and substituting known values gives

$m/s\displaystyle\Delta{p}=\frac{h}{4\pi\Delta{x}}=\frac{6.63\times10^{-34}\text{ J}\cdot\text{ s}}{4\pi\left(1.00\times10^{-11}\text{ m}\right)}=5.28\times10^{-24}\text{ kg}\cdot\text{ m/s}\\$

Thus, Δp = 5.28 × 10−24 kg · m/s = mΔv.

Solving for Δv and substituting the mass of an electron gives

$m/s\displaystyle\Delta{v}=\frac{\Delta{p}}{m}=\frac{5.28\times10^{-24}\text{ kg}\cdot\text{ m/s}}{9.11\times10^{-31}\text{ kg}}=5.79\times10^6\text{ m/s}\\$

Solution for Part 2

Although large, this velocity is not highly relativistic, and so the electron’s kinetic energy is

$eV\begin{array}{lll}KE_{e}&=&\frac{1}{2}mv^2\\\text{ }&=&\frac{1}{2}\left(9.11\times10^{-31}\text{ kg}\right)\left(5.79\times10^6\text{ m/s}\right)^2\\\text{ }&=&\left(1.53\times10^{-17}\text{ J}\right)\left(\frac{1\text{ eV}}{1.60\times10^{-19}\text{ J}}\right)=95.5\text{ eV}\end{array}\\$

Discussion

Since atoms are roughly 0.1 nm in size, knowing the position of an electron to 0.0100 nm localizes it reasonably well inside the atom. This would be like being able to see details one-tenth the size of the atom. But the consequent uncertainty in velocity is large. You certainly could not follow it very well if its velocity is so uncertain. To get a further idea of how large the uncertainty in velocity is, we assumed the velocity of the electron was equal to its uncertainty and found this gave a kinetic energy of 95.5 eV. This is significantly greater than the typical energy difference between levels in atoms (see Table 1 in Photon Energies and the Electromagnetic Spectrum), so that it is impossible to get a meaningful energy for the electron if we know its position even moderately well.

Why don’t we notice Heisenberg’s uncertainty principle in everyday life? The answer is that Planck’s constant is very small. Thus the lower limit in the uncertainty of measuring the position and momentum of large objects is negligible. We can detect sunlight reflected from Jupiter and follow the planet in its orbit around the Sun. The reflected sunlight alters the momentum of Jupiter and creates an uncertainty in its momentum, but this is totally negligible compared with Jupiter’s huge momentum. The correspondence principle tells us that the predictions of quantum mechanics become indistinguishable from classical physics for large objects, which is the case here. There is another form of Heisenberg’s uncertainty principle for simultaneous measurements of energy and time. In equation form,

$ΔEΔt≥h4π\Delta{E}\Delta{t}\ge\frac{h}{4\pi}\\$

, where ΔE is the uncertainty in energy and Δt is the uncertainty in time. This means that within a time interval Δt, it is not possible to measure energy precisely—there will be an uncertainty ΔE in the measurement. In order to measure energy more precisely (to make ΔE smaller), we must increase Δt. This time interval may be the amount of time we take to make the measurement, or it could be the amount of time a particular state exists, as in the next Example 2.

An atom in an excited state temporarily stores energy. If the lifetime of this excited state is measured to be 1.0 × 10−10 s, what is the minimum uncertainty in the energy of the state in eV? The minimum uncertainty in energy ΔE is found by using the equals sign in

$ΔEΔt≥h4π\Delta{E}\Delta{t}\ge\frac{h}{4\pi}\\$

and corresponds to a reasonable choice for the uncertainty in time. The largest the uncertainty in time can be is the full lifetime of the excited state, or Δt = 1.0 × 10−10 s. Solving the uncertainty principle for ΔE and substituting known values gives

$J\displaystyle\Delta{E}=\frac{h}{4\pi\Delta{t}}=\frac{6.63\times10^{-34}\text{ J}\cdot\text{ s}}{4\pi\left(1.0\times10^{-10}\text{ s}\right)}=5.3\times10^{-25}\text{ J}\\$

Now converting to eV yields

$eV\displaystyle\Delta{E}=\left(5.3\times10^{-25}\text{ J}\right)\left(\frac{1\text{ eV}}{1.6\times10^{-19}\text{ J}}\right)=3.3\times10^{-6}\text{ eV}\\$

Discussion

The lifetime of 10−10 s is typical of excited states in atoms—on human time scales, they quickly emit their stored energy. An uncertainty in energy of only a few millionths of an eV results. This uncertainty is small compared with typical excitation energies in atoms, which are on the order of 1 eV. So here the uncertainty principle limits the accuracy with which we can measure the lifetime and energy of such states, but not very significantly.

The uncertainty principle for energy and time can be of great significance if the lifetime of a system is very short. Then Δt is very small, and ΔE is consequently very large. Some nuclei and exotic particles have extremely short lifetimes (as small as 10−25 s), causing uncertainties in energy as great as many GeV (109 eV). Stored energy appears as increased rest mass, and so this means that there is significant uncertainty in the rest mass of short-lived particles. When measured repeatedly, a spread of masses or decay energies are obtained. The spread is ΔE. You might ask whether this uncertainty in energy could be avoided by not measuring the lifetime. The answer is no. Nature knows the lifetime, and so its brevity affects the energy of the particle. This is so well established experimentally that the uncertainty in decay energy is used to calculate the lifetime of short-lived states. Some nuclei and particles are so short-lived that it is difficult to measure their lifetime. But if their decay energy can be measured, its spread is ΔE, and this is used in the uncertainty principle

$(ΔEΔt≥h4π)\left(\Delta{E}\Delta{t}\ge\frac{h}{4\pi}\right)\\$

to calculate the lifetime Δt.

There is another consequence of the uncertainty principle for energy and time. If energy is uncertain by ΔE, then conservation of energy can be violated by ΔE for a time Δt. Neither the physicist nor nature can tell that conservation of energy has been violated, if the violation is temporary and smaller than the uncertainty in energy. While this sounds innocuous enough, we shall see in later chapters that it allows the temporary creation of matter from nothing and has implications for how nature transmits forces over very small distances.

Finally, note that in the discussion of particles and waves, we have stated that individual measurements produce precise or particle-like results. A definite position is determined each time we observe an electron, for example. But repeated measurements produce a spread in values consistent with wave characteristics. The great theoretical physicist Richard Feynman (1918–1988) commented, “What there are, are particles.” When you observe enough of them, they distribute themselves as you would expect for a wave phenomenon. However, what there are as they travel we cannot tell because, when we do try to measure, we affect the traveling.

Matter is found to have the same interference characteristics as any other wave.
There is now a probability distribution for the location of a particle rather than a definite position.
Another consequence of the wave character of all particles is the Heisenberg uncertainty principle, which limits the precision with which certain physical quantities can be known simultaneously. For position and momentum, the uncertainty principle is
$ΔxΔp≥h4π\Delta{x}\Delta{p}\ge\frac{h}{4\pi}\\$
, where Δx is the uncertainty in position and Δp is the uncertainty in momentum.
For energy and time, the uncertainty principle is
$ΔEΔt≥h4π\Delta{E}\Delta{t}\ge\frac{h}{4\pi}\\$
whereΔE is the uncertainty in energy andΔt is the uncertainty in time.
These small limits are fundamentally important on the quantum-mechanical scale.

What is the Heisenberg uncertainty principle? Does it place limits on what can be known?

(a) If the position of an electron in a membrane is measured to an accuracy of 1.00 μm, what is the electron’s minimum uncertainty in velocity? (b) If the electron has this velocity, what is its kinetic energy in eV? (c) What are the implications of this energy, comparing it to typical molecular binding energies?
(a) If the position of a chlorine ion in a membrane is measured to an accuracy of 1.00 μm, what is its minimum uncertainty in velocity, given its mass is 5.86 × 10−26 kg? (b) If the ion has this velocity, what is its kinetic energy in eV, and how does this compare with typical molecular binding energies?
Suppose the velocity of an electron in an atom is known to an accuracy of 2.0 × 103 m/s (reasonably accurate compared with orbital velocities). What is the electron’s minimum uncertainty in position, and how does this compare with the approximate 0.1-nm size of the atom?
The velocity of a proton in an accelerator is known to an accuracy of 0.250% of the speed of light. (This could be small compared with its velocity.) What is the smallest possible uncertainty in its position?
A relatively long-lived excited state of an atom has a lifetime of 3.00 ms. What is the minimum uncertainty in its energy?
(a) The lifetime of a highly unstable nucleus is 10−20 s What is the smallest uncertainty in its decay energy? (b) Compare this with the rest energy of an electron.
The decay energy of a short-lived particle has an uncertainty of 1.0 MeV due to its short lifetime. What is the smallest lifetime it can have?
The decay energy of a short-lived nuclear excited state has an uncertainty of 2.0 eV due to its short lifetime. What is the smallest lifetime it can have?
What is the approximate uncertainty in the mass of a muon, as determined from its decay lifetime?
Derive the approximate form of Heisenberg’s uncertainty principle for energy and time, ΔEΔt ≈ h, using the following arguments: Since the position of a particle is uncertain by Δx ≈ λ, where λ is the wavelength of the photon used to examine it, there is an uncertainty in the time the photon takes to traverse Δx. Furthermore, the photon has an energy related to its wavelength, and it can transfer some or all of this energy to the object being examined. Thus the uncertainty in the energy of the object is also related to λ. Find Δt and ΔE; then multiply them to give the approximate uncertainty principle.

Heisenberg’s uncertainty principle: a fundamental limit to the precision with which pairs of quantities (momentum and position, and energy and time) can be measured

uncertainty in energy: lack of precision or lack of knowledge of precise results in measurements of energy

uncertainty in time: lack of precision or lack of knowledge of precise results in measurements of time

uncertainty in momentum: lack of precision or lack of knowledge of precise results in measurements of momentum

uncertainty in position: lack of precision or lack of knowledge of precise results in measurements of position

probability distribution: the overall spatial distribution of probabilities to find a particle at a given location

1. (a) 57.9 m/s; (b) 9.55 × 10−9 eV; (c) From Table 1 in Photon Energies and the Electromagnetic Spectrum, we see that typical molecular binding energies range from about 1eV to 10 eV, therefore the result in part (b) is approximately 9 orders of magnitude smaller than typical molecular binding energies. 3. 29 nm; 290 times greater

5. 1.10 × 10−13 eV

7. 3.3 × 10−22 s

9. 2.66 × 10−46 kg

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Page 2

By the end of this section, you will be able to:

Explain the concept of particle-wave duality, and its scope.

Figure 1. On a quantum-mechanical scale (i.e., very small), particles with and without mass have wave properties. For example, both electrons and photons have wavelengths but also behave as particles.

Particle-wave duality—the fact that all particles have wave properties—is one of the cornerstones of quantum mechanics. We first came across it in the treatment of photons, those particles of EM radiation that exhibit both particle and wave properties, but not at the same time. Later it was noted that particles of matter have wave properties as well. The dual properties of particles and waves are found for all particles, whether massless like photons, or having a mass like electrons. (See Figure 1.)

There are many submicroscopic particles in nature. Most have mass and are expected to act as particles, or the smallest units of matter. All these masses have wave properties, with wavelengths given by the de Broglie relationship

$λ=hp\lambda=\frac{h}{p}\\$

. So, too, do combinations of these particles, such as nuclei, atoms, and molecules. As a combination of masses becomes large, particularly if it is large enough to be called macroscopic, its wave nature becomes difficult to observe. This is consistent with our common experience with matter.

Some particles in nature are massless. We have only treated the photon so far, but all massless entities travel at the speed of light, have a wavelength, and exhibit particle and wave behaviors. They have momentum given by a rearrangement of the de Broglie relationship,

$p=hλp=\frac{h}{\lambda}\\$

. In large combinations of these massless particles (such large combinations are common only for photons or EM waves), there is mostly wave behavior upon detection, and the particle nature becomes difficult to observe. This is also consistent with experience. (See Figure 2.)

Figure 2. On a classical scale (macroscopic), particles with mass behave as particles and not as waves. Particles without mass act as waves and not as particles.

The particle-wave duality is a universal attribute. It is another connection between matter and energy. Not only has modern physics been able to describe nature for high speeds and small sizes, it has also discovered new connections and symmetries. There is greater unity and symmetry in nature than was known in the classical era—but they were dreamt of. A beautiful poem written by the English poet William Blake some two centuries ago contains the following four lines:

To see the World in a Grain of Sand And a Heaven in a Wild Flower Hold Infinity in the palm of your hand

And Eternity in an hour

The problem set for this section involves concepts from this chapter and several others. Physics is most interesting when applied to general situations involving more than a narrow set of physical principles. For example, photons have momentum, hence the relevance of Linear Momentum and Collisions. The following topics are involved in some or all of the problems in this section:

Dynamics: Newton’s Laws of Motion
Work, Energy, and Energy Resources
Linear Momentum and Collisions
Heat and Heat Transfer Methods
Electric Potential and Electric Field
Electric Current, Resistance, and Ohm’s Law
Wave Optics
Special Relativity

Identify which physical principles are involved.
Solve the problem using strategies outlined in the text.

The following topics are involved in this example:

Photons (quantum mechanics)
Linear Momentum

A 550-nm photon (visible light) is absorbed by a 1.00-μg particle of dust in outer space.

Find the momentum of such a photon.
What is the recoil velocity of the particle of dust, assuming it is initially at rest?

To solve an integrated-concept problem, such as those following this example, we must first identify the physical principles involved and identify the chapters in which they are found. Part 1 of this example asks for the momentum of a photon, a topic of the present chapter. Part 1 considers recoil following a collision, a topic of Linear Momentum and Collisions. The following solutions to each part of the example illustrate how specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so on. The momentum of a photon is related to its wavelength by the equation:

$p=hλp=\frac{h}{\lambda}\\$

Entering the known value for Planck’s constant h and given the wavelength λ, we obtain

$m/s\begin{array}{lll}p&=&\frac{6.63\times10^{-34}\text{ J }\cdot\text{ s}}{550\times10^{-9}\text{ m}}\\\text{ }&=&1.21\times10^{-27}\text{ kg }\cdot\text{ m/s}\end{array}\\$

Discussion for Part 1

This momentum is small, as expected from discussions in the text and the fact that photons of visible light carry small amounts of energy and momentum compared with those carried by macroscopic objects. Conservation of momentum in the absorption of this photon by a grain of dust can be analyzed using the equation:

p1 + p2 = p′1 + p′2(Fnet=0).

The net external force is zero, since the dust is in outer space. Let 1 represent the photon and 2 the dust particle. Before the collision, the dust is at rest (relative to some observer); after the collision, there is no photon (it is absorbed). So conservation of momentum can be written p1 = p′2 = mv, where p1 is the photon momentum before the collision and p′2 is the dust momentum after the collision. The mass and recoil velocity of the dust are m and v, respectively. Solving this for v, the requested quantity, yields

$v=pmv=\frac{p}{m}\\$

, where p is the photon momentum found in Part 1. Entering known values (noting that a microgram is 10−9 kg) gives

$m/s\begin{array}{lll}v&=&\frac{1.21\times10^{-27}\text{ kg }\cdot\text{ m/s}}{1.00\times10^{-9}\text{ kg}}\\\text{ }&=&1.21\times10^{-18}\text{ m/s}\end{array}\\$

Discussion

The recoil velocity of the particle of dust is extremely small. As we have noted, however, there are immense numbers of photons in sunlight and other macroscopic sources. In time, collisions and absorption of many photons could cause a significant recoil of the dust, as observed in comet tails.

Section Summary

The particle-wave duality refers to the fact that all particles—those with mass and those without mass—have wave characteristics.
This is a further connection between mass and energy.

In what ways are matter and energy related that were not known before the development of relativity and quantum mechanics?

Integrated Concepts. The 54.0-eV electron in Table 1 in The Wave Nature of Matter has a 0.167-nm wavelength. If such electrons are passed through a double slit and have their first maximum at an angle of 25.0º, what is the slit separation d?
Integrated Concepts. An electron microscope produces electrons with a 2.00-pm wavelength. If these are passed through a 1.00-nm single slit, at what angle will the first diffraction minimum be found?
Integrated Concepts. A certain heat lamp emits 200 W of mostly IR radiation averaging 1500 nm in wavelength. (a) What is the average photon energy in joules? (b) How many of these photons are required to increase the temperature of a person’s shoulder by 2.0ºC, assuming the affected mass is 4.0 kg with a specific heat of 0.83 kcal/kg · ºC. Also assume no other significant heat transfer. (c) How long does this take?
Integrated Concepts. On its high power setting, a microwave oven produces 900 W of 2560 MHz microwaves. (a) How many photons per second is this? (b) How many photons are required to increase the temperature of a 0.500-kg mass of pasta by 45.0ºC, assuming a specific heat of 0.900 kcal/kg · ºC? Neglect all other heat transfer. (c) How long must the microwave operator wait for their pasta to be ready?
Integrated Concepts. (a) Calculate the amount of microwave energy in joules needed to raise the temperature of 1.00 kg of soup from 20.0ºC to 100ºC. (b) What is the total momentum of all the microwave photons it takes to do this? (c) Calculate the velocity of a 1.00-kg mass with the same momentum. (d) What is the kinetic energy of this mass?
Integrated Concepts. (a) What is γ for an electron emerging from the Stanford Linear Accelerator with a total energy of 50.0 GeV? (b) Find its momentum. (c) What is the electron’s wavelength?
Integrated Concepts. (a) What is γ for a proton having an energy of 1.00 TeV, produced by the Fermilab accelerator? (b) Find its momentum. (c) What is the proton’s wavelength?
Integrated Concepts. An electron microscope passes 1.00-pm-wavelength electrons through a circular aperture 2.00 μm in diameter. What is the angle between two just-resolvable point sources for this microscope?
Integrated Concepts. (a) Calculate the velocity of electrons that form the same pattern as 450-nm light when passed through a double slit. (b) Calculate the kinetic energy of each and compare them. (c) Would either be easier to generate than the other? Explain.
Integrated Concepts. (a) What is the separation between double slits that produces a second-order minimum at 45.0º for 650-nm light? (b) What slit separation is needed to produce the same pattern for 1.00-keV protons.
Integrated Concepts. A laser with a power output of 2.00 mW at a wavelength of 400 nm is projected onto calcium metal. (a) How many electrons per second are ejected? (b) What power is carried away by the electrons, given that the binding energy is 2.71 eV? (c) Calculate the current of ejected electrons. (d) If the photoelectric material is electrically insulated and acts like a 2.00-pF capacitor, how long will current flow before the capacitor voltage stops it?
Integrated Concepts. One problem with x rays is that they are not sensed. Calculate the temperature increase of a researcher exposed in a few seconds to a nearly fatal accidental dose of x rays under the following conditions. The energy of the x-ray photons is 200 keV, and 4.00 × 1013 of them are absorbed per kilogram of tissue, the specific heat of which is 0.830 kcal/kg · ºC. (Note that medical diagnostic x-ray machines cannot produce an intensity this great.)
Integrated Concepts. A 1.00-fm photon has a wavelength short enough to detect some information about nuclei. (a) What is the photon momentum? (b) What is its energy in joules and MeV? (c) What is the (relativistic) velocity of an electron with the same momentum? (d) Calculate the electron’s kinetic energy.
Integrated Concepts. The momentum of light is exactly reversed when reflected straight back from a mirror, assuming negligible recoil of the mirror. Thus the change in momentum is twice the photon momentum. Suppose light of intensity 1.00 kW/m2 reflects from a mirror of area 2.00 m2. (a) Calculate the energy reflected in 1.00 s. (b) What is the momentum imparted to the mirror? (c) Using the most general form of Newton’s second law, what is the force on the mirror? (d) Does the assumption of no mirror recoil seem reasonable?
Integrated Concepts. Sunlight above the Earth’s atmosphere has an intensity of 1.30 kW/m2. If this is reflected straight back from a mirror that has only a small recoil, the light’s momentum is exactly reversed, giving the mirror twice the incident momentum. (a) Calculate the force per square meter of mirror. (b) Very low mass mirrors can be constructed in the near weightlessness of space, and attached to a spaceship to sail it. Once done, the average mass per square meter of the spaceship is 0.100 kg. Find the acceleration of the spaceship if all other forces are balanced. (c) How fast is it moving 24 hours later?

1. 0.395 nm

3. (a) 1.3 × 10−19 J; (b) 2.1 × 1023;(c) 1.4 × 102 s

5. (a) 3.35 × 105 J; (b) 1.12 × 10−3 kg · m/s; (c) 1.12 × 10−3 m/s; (d) 6.23 × 10−7 J

7. (a) 1.06 × 103; (b) 5.33 × 10−16 kg · m/s; (c) 1.24 × 10−18 m

9. (a) 1.62 × 103 m/s; (b) 4.42 × 10−19 J for photon, 1.19 × 10−24 J for electron, photon energy is 3.71 × 105 times greater; (c) The light is easier to make because 450-nm light is blue light and therefore easy to make. Creating electrons with 7.43 μeV of energy would not be difficult, but would require a vacuum.

10. (a) 2.30 × 10−6 m; (b) 3.20 × 10−12 m

12. 3.69 × 10−4 ºC

14. (a) 2.00 kJ; (b) 1.33 × 10−5 kg · m/s; (c) 1.33 × 10−5 N; (d) yes