P8.36.f^Chapter 36 Ending^490^1205^,,^49236^49816%
WHAT WE HAVE LEARNED
EXAM STUDY GUIDE
KEY TERMS
spectral emittance, p. 1172 Wien's law, p. 1173 Wien displacement law, p. 1173 Planck's constant, p. 1173 Planck's radiation law, p. 1173 spectral brightness, p. 1173 photon, p. 1177 photoelectric effect, p. 1177 work function, p. 1178 wave-particle duality, p. 1179 photomultiplier tube, p. 1180 Bremsstrahlung, p. 1182 quantum electrodynamics, p. 1182 Compton scattering, p. 1182 Compton wavelength, p. 1183 matter waves, p. 1185 de Broglie wavelength, p. 1185 Heisenberg uncertainty relation, p. 1189 spin, p. 1192 fermions, p. 1192 bosons, p. 1192 Pauli exclusion principle, p. 1193 partition function, p. 1193 Maxwell-Boltzmann distribution, p. 1193 Bose-Einstein distribution, p. 1194 Fermi-Dirac distribution, p. 1194
Page 1199
NEW SYMBOLS AND EQUATIONS

h = 6.62606876(52) · 10−34 J s, Planck's constant

φ, work function in photoelectric effect, eV0 = hfφ

, Compton scattering formula

, Compton wavelength of an electron

, Heisenberg uncertainty relation for position and momentum

, Heisenberg uncertainty relation for energy and time

ANSWERS TO SELF-TEST OPPORTUNITIES
  • 36.1 Kmax = (ffmin)h = fhφ.

  • 36.2 148 million km = 148 · 106 · 103 m = 1.48 · 1011 m

    In visible spectrum Pvisible = P/4 = 9.43 · 1025 W.

    In 1 second, 9.43 · 1025 J of energy is emitted by the Sun.

    Assume the average wavelength of the visible photons is λ = 550 nm. The energy of each photon is then

  • 36.3 Momentum of 1.00 eV electron: mec2 = 511 keV

    Momentum of 100 keV X-ray:

    Momentum of electron is 1% of X-ray, negligible.

  • 36.4 For visible light, the energy of the photon ranges from 1.59 eV to 3.27 eV.

    For a 2.26 eV photon, the momentum is

    Momemtum of 1.00 eV electron: mec2 = 511 keV

    (pc)photon/(pc)electron = (2.26 eV)/(1010 eV) = 2.24 · 10−3

    The momentum of the visible photon is 0.224% of the momentum of the electron, negligible.

  • 36.5

    The de Broglie wavelength for an electron with a kinetic energy of 1000 eV is

    λ = 0.0387879 nm.

    Calculating the momentum nonrelativistically gives us

    λ = 0.0388068 nm.

    The difference is small.

  • 36.6 Δv = 2.00 km/h = 0.556 m/s

  • 36.7 Numbers of states are 18, 16, 495, respectively.

Page 1200
PROBLEM-SOLVING PRACTICE
Problem-Solving Guidelines
  1. The starting point for most calculations involving photons or matter waves is to relate the particle properties of energy E and momentum p to the wave properties of wavelength λ and frequency f. The key relations are E = hf, p = E/c = hf/c = h/λ, and λ = h/p.

  2. Be careful and consistent in applying units. Often, converting all units to meters and kilograms will help keep track of unit exponents. Use of electron-volts will often simplify your calculations, but be sure to use Planck's constant with the appropriate units: h = 6.626 · 10−34 J s or h = 4.136 · 10−15 eV s.

  3. For checking your work, it is helpful to keep in mind some rough orders of magnitude: the size of an atom is 10−10 m; the mass of an electron is 10−30 kg; charge of a proton or electron is 10−19 C; at room temperature,

SOLVED PROBLEM36.1 Rubidium Bose-Einstein Condensate
PROBLEM

What is the minimum density of rubidium atoms needed in a magnetic trap at a temperature of 200 nK in order to have a chance to observe the onset of Bose-Einstein condensation? (Hint: The mass of a rubidium-87 atom is 1.5 · 10−25 kg.)

SOLUTION
THINK

In our section on the Bose-Einstein condensate, we stated that the rubidium atoms must be close enough to one another that they can overlap in a quantum mechanical sense. There we quoted this criterion as ρ · λ3 > 2.61. Thus our task amounts to finding the de Broglie wavelength of rubidium atoms in thermal motion at a temperature of 200 nK.

SKETCH
FIGURE 36.27
Sketch of overlapping rubidium atoms in the trap.

A sketch is perhaps not really needed in this case, but in Figure 36.27 we at least try to visualize the relationship of the quantum mechanical extension of the atomic wave function, the de Broglie wavelength, to the nearest-neighbor spacing of the atoms in the trap.

RESEARCH

We have already stated that the criterion for the density is given as

(i)

The de Broglie wavelength λ was given as

(ii)

in equation 36.22. For the momentum p of a rubidium atom we use the fact that the kinetic energy of an atom in a gas is given by the thermal energy at the given temperature,

(iii)
SIMPLIFY

Inserting equation (iii) into equation (ii), we find for the de Broglie wavelength in this case

(iv)

Solving equation (i) for the density and inserting our formula for the de Broglie wavelength in equation (iv) results in

Page 1201
CALCULATE

Inserting the values of the constants (h = 6.63 · 10−34 J s, kB = 1.38 · 10−23 J/K) and the given temperature (T = 200 nK = 2 · 10−7 K), we obtain for the de Broglie wavelength

and thus for the density

ROUND

Since the temperature was only given to one significant digit, our answer for the de Broglie wavelength is λ = 6 · 10−7 m. Note that this is approximately a factor 1000 larger than the atomic diameter of rubidium, which is approximately 5 · 10−10 m.

For the minimum density needed to have a chance to observe the BEC, our appropriately rounded result is ρ >1 · 1019 m−3.

DOUBLE-CHECK

Is a density of 1019atoms/m3 a large or a small number? Let us compare it to the density of water molecules and air molecules. One cubic meter of liquid water contains 3 · 1028 water molecules, and one cubic meter of air contains 3 · 1025 nitrogen and oxygen molecules. Thus, the gas density of rubidium atoms in the trap is approximately a million times lower than the density of air at normal atmospheric conditions.

If we have one million atoms in a trap at a density of 1019 atoms/m3 in an approximately spherical configuration, what is the radius of this sphere? The total volume occupied by the million atoms is

Because the volume of a sphere is , we find that the radius of the sphere containing one million atoms at this density is

This makes it clear that the Bose-Einstein condensate could not be imaged directly in the experiment conducted by Cornell and Wieman. This is why they needed to turn off the trap and let its contents expand by at least a factor of 10 in each direction before they were able to produce the images of the kind shown in Figure 36.26.

MULTIPLE-CHOICE QUESTIONS
QUESTIONS
PROBLEMS

A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One and two •• indicate increasing level of problem difficulty.

Section 36.2
  • 36.19 Calculate the peak wavelengths of

    1. the solar light received by Earth, and

    2. light emitted by the Earth.

    Assume the surface temperatures of the Sun and the Earth are 5800. K and 300. K, respectively.

    Answer

  • 36.20 Calculate the range of temperatures for which the peak emission of the blackbody radiation from a hot filament occurs within the visible range of the electromagnetic spectrum. Take the visible spectrum as extending from 380 nm to 780 nm. What is the total intensity of the radiation from the filament at these two temperatures?

  • 36.21 Ultra-high-energy gamma rays are found to come from the Equator of our galaxy, with energies up to 3.5 · 1012 eV. What is the wavelength of this light? How does the energy of this light compare to the rest mass of a proton?

    Answer

  • Page 1203
  • 36.22 Consider an object at room temperature (20. °C) and the radiation it emits. For radiation at the peak of the spectral energy density, calculate

    1. the wavelength,

    2. the frequency, and

    3. the energy of one photon.

  • •36.23 The temperature of your skin is approximately 35.0 °C.

    1. Assuming that it is a blackbody, what is the peak wavelength of the radiation it emits?

    2. Assuming a total surface area of 2.00 m2, what is the total power emitted by your skin?

    3. Based on your answer in (b), why don't you glow as brightly as a light bulb?

    Answer

  • 36.24 A pure, defect-free semiconductor material will absorb the electromagnetic radiation incident on that material only if the energy of the individual photons in the incident beam is larger than a threshold value known as the “band-gap” of the semiconductor. The known room-temperature band-gaps for germanium, silicon, and gallium-arsenide, three widely used semiconductors, are 0.66 eV, 1.12 eV, and 1.42 eV, respectively.

    1. Determine the room-temperature transparency range of these semiconductors.

    2. Compare these with the transparency range of ZnSe, a semiconductor with a band-gap of 2.67 eV, and explain the yellow color observed experimentally for the ZnSe crystals.

    3. Which of these materials could be used for a light detector for the 1550-nm optical communications wavelength?

  • 36.25 The mass of a dime is 2.268 g, its diameter is 17.91 mm, and its thickness is 1.350 mm. Determine

    1. the radiant energy coming out from a dime per second at room temperature,

    2. the number of photons leaving the dime per second, and

    3. the volume of air to have energy equal to 1 second of radiation from the dime.

    Answer

Section 36.3
  • 36.26 The work function of a certain material is 5.8 eV. What is the photoelectric threshold for this material?

  • 36.27 What is the maximum kinetic energy of the electrons ejected from a sodium surface by light of wavelength 470 nm?

    Answer

  • 36.28 The threshold wavelength for the photoelectric effect in a specific alloy is 400. nm. What is the work function in eV?

  • 36.29 In a photoelectric effect experiment, a laser beam of unknown wavelength is shined on a cesium cathode (work function φ = 2.100 eV). It is found that a stopping potential of 0.310 V is needed to eliminate the current. Next, the same laser is shined on a cathode made of an unknown material, and a stopping potential of 0.110 V is found to be needed to eliminate the current.

    1. What is the work function for the unknown cathode?

    2. What would be a possible candidate for the material of this unknown cathode?

    Answer

  • 36.30 You illuminate a zinc surface with 550-nm light. How high do you have to turn up the stopping voltage to squelch the photoelectric current completely?

  • 36.31 White light, λ = 400. to 750. nm, falls on barium (φ = 2.48 eV).

    1. What is the maximum kinetic energy of electrons ejected from it?

    2. Would the longest-wavelength light eject electrons?

    3. What wavelength of light would eject electrons with zero kinetic energy?

    Answer

  • 36.32 To determine the work function of the material of a photodiode, you measured the maximum kinetic energy of 1.50 eV corresponding to a certain wavelength. Later on you cut down the wavelength by 50.0% and found the maximum kinetic energy of the photoelectrons to be 3.80 eV. From this information determine

    1. work function of the material, and

    2. the original wavelength.

Section 36.4
  • 36.33 X-rays of wavelength λ = 0.120 nm are scattered from carbon. What is the Compton wavelength shift for photons detected at 90.0° angle relative to the incident beam?

    Answer

  • 36.34 A 2.0-MeV X-ray photon is scattered from a free electron at rest into an angle of 53°. What is the wavelength of the scattered photon?

  • 36.35 A photon with wavelength of 0.30 nm collides with an electron that is initially at rest. If the photon rebounds at an angle of 160°, how much energy did it lose in the collision?

    Answer

  • 36.36 X-rays having energy of 400.0 keV undergo Compton scattering from a target. The scattered rays are detected at 25.0° relative to the incident rays. Find

    1. the kinetic energy of the scattered X-ray, and

    2. the kinetic energy of the recoiling electron.

  • 36.37 Consider the equivalent of Compton scattering, but the case in which a photon scatters off of a free proton.

    1. If 140.-keV X-rays bounce off of a proton at 90.0°, what is their fractional change in energy (E0E)/E0?

    2. What energy of photon would be necessary to cause a 1.00% change in energy at 90.0° scattering?

    Answer

  • 36.38 An X-ray photon with an energy of 50.0 keV strikes an electron that is initially at rest inside a metal. The photon is scattered at an angle of 45°. What is the kinetic energy and momentum (magnitude and direction) of the electron after the collision? You may use the nonrelativistic relationship connecting the kinetic energy and momentum of the electron.

Section 36.5
  • 36.39 Calculate the wavelength of

    1. a 2.00 eV photon, and

    2. an electron with kinetic energy 2.00 eV.

    Answer

  • Page 1204
  • 36.40 What is the de Broglie wavelength of a 2.000 · 103-kg car moving at a speed of 100.0 km/h?

  • 36.41 A nitrogen molecule of mass m = 4.648 · 10−26 kg has a speed of 300.0 m/s.

    1. Determine its de Broglie wavelength.

    2. How far apart are the double slits if a nitrogen molecule fringe pattern, with fringes 0.30 cm apart, is observed on a screen 70.0 cm in front of the slits?

    Answer

  • 36.42 Alpha particles are accelerated through a potential difference of 20.0 V. What is their de Broglie wavelength?

  • 36.43 Consider an electron whose de Broglie wavelength is equal to the wavelength of green light (about 550 nm).

    1. Treating the electron nonrelativistically, what is its speed?

    2. Does your calculation confirm that a nonrelativistic treatment is sufficient?

    3. Calculate the kinetic energy of the electron in eV.

    Answer

  • 36.44 After you told him about de Broglie's hypothesis that particles of momentum p have wave characteristics with wavelength λ = h/p, your 60.0-kg roommate starts thinking of his fate as a wave and asks you if he could be diffracted when passing through the 90.0-cm-wide doorway of your dorm room.

    1. What is the maximum speed at which your roommate can pass through the doorway in order to be significantly diffracted?

    2. If it takes one step to pass through the doorstep, how long should it take your roommate to make that step (assume the length of his step is 0.75 m) in order for him to be diffracted?

    3. What is the answer to your roommate's question? Hint: Assume that significant diffraction occurs when the width of the diffraction aperture is less that 10.0 times the wavelength of the wave being diffracted.

  • ••36.45 Consider de Broglie waves for a Newtonian particle of mass m, momentum p = mv, and energy E = p2/(2m), that is, waves with wavelength λ = h/p and frequency f = E/h.

    1. Calculate the dispersion relation ω = ω(k) for these waves.

    2. Calculate the phase and group velocities of these waves. Which of these corresponds to the classical velocity of the particle?

    Answer

  • ••36.46 Now consider de Broglie waves for a (relativistic) particle of mass m, momentum p = mvγ, and total energy E = mc2γ, with γ = [1 − (v/c)2]−½. The waves have wavelength λ = h/p and frequency f = E/h as before, but with the relativistic momentum and energy.

    1. Calculate the dispersion relation for these waves.

    2. Calculate the phase and group velocities of these waves. Now which corresponds to the classical velocity of the particle?

Section 36.6
  • 36.47 A 50.0-kg particle has a deBroglie wavelength of 20.0 cm.

    1. How fast is the particle moving?

    2. What is the smallest speed uncertainty of the particle if its position uncertainty is 20.0 cm?

    Answer

  • 36.48 During the period of time required for light to pass through a hydrogen atom (r = 0.53 · 10−10 m), what is the least uncertainty in energy for the atom? Express your answer in electron volts.

  • 36.49 A free neutron (m = 1.67 · 10−27 kg) has a mean life of 900. s. What is the uncertainty in its mass (in kg)?

    Answer

  • 36.50 Suppose that Fuzzy, a quantum-mechanical duck, lives in a world in which Planck's constant ħ = 1.00 Js. Fuzzy has a mass of 0.500 kg and initially is known to be within a 0.750-m-wide pond. What is the minimum uncertainty in Fuzzy's speed? Assuming that this uncertainty prevails for 5.00 s, how far away could Fuzzy be from the pond after 5.00 s?

  • 36.51 An electron is confined to a box with a dimension of 20.0 μm. What is the minimum speed the electron can have?

    Answer

  • 36.52 A dust particle of mass 1.00 · 10−16 kg and diameter 5.00 μm is confined to a box of length 15.0 μm.

    1. How will you know whether the particle is at rest?

    2. If the particle is not at rest, what will be the range of its velocity?

    3. Using the lower range of velocity, how long will it take to move a distance of 1.00 mm?

Section 36.8
  • ••36.53 Consider a quantum state of energy E, which can be occupied by any number n of some bosonic particles, including n = 0. At absolute temperature T, the probability of finding n particles in the state is given by Pn = N exp(−nE/kBT), where kB is Boltzmann's constant and the normalization factor N is determined by the requirement that all the probabilities sum to unity. Calculate the mean or expected value of n, that is, the occupancy, of this state, given this probability distribution.

    Answer

  • 36.54 Consider the same quantum state as the preceding problem, with a probability distribution of the same form, but with fermionic particles, so that the only possible occupation numbers are n = 0 and n = 1. Calculate the mean occupancy 〈n〉 of the state in this case.

  • ••36.55 Consider a system made up of N particles. The average energy per particle is given by where Z is the partition function defined in equation 36.29. If this is a two-state system with E1 = 0 and E2 = E and g1 = g2 = 1, calculate the heat capacity of the system, defined as and approximate its behavior at very high and very low temperatures (that is, kBT ≫ 1 and kBT ≪ 1).

    Answer

Additional Problems
  • 36.56 Given that the work function of tungsten is 4.55 eV, what is the stopping potential in an experiment using tungsten cathodes at 360 nm?

  • 36.57 Find the ratios of de Broglie wavelengths of a 100-MeV proton to a 100-MeV electron.

    Answer

  • 36.58 An Einstein (E) is a unit of measurement equal to Avogadro's number (6.02 · 1023) of photons. How much energy is contained in 1 Einstein of violet light (λ = 400. nm)?

  • Page 1205
  • 36.59 In baseball, a 100.-g ball can travel as fast as 100. mph. What is the de Broglie wavelength of this ball? The Voyager spacecraft, with a mass of about 250. kg, is currently travelling at 125,000 km/h. What is its de Broglie wavelength?

    Answer

  • 36.60 What is the minimum uncertainty in the velocity of a 1.0-nanogram particle that is at rest on the head of a 1.0-mm-wide pin?

  • 36.61 A photovoltaic device uses monochromatic light of wavelength 700. nm that is incident normally on a surface of area 10.0 cm2. Calculate the photon flux rate if the light intensity is 0.300 W/cm2.

    Answer

  • 36.62 The Solar Constant measured by Earth satellites is roughly 1400. W/m2. Though the Sun emits light of different wavelengths, the peak of the wavelength spectrum is at 500. nm.

    1. Find the corresponding photon frequency.

    2. Find the corresponding photon energy.

    3. Find the number flux of photons arriving at Earth, assuming that all light emitted by the Sun has the same peak wavelength.

  • 36.63 Two silver plates in vacuum are separated by 1.0 cm and have a potential difference of 20.0 kV between them. What is the largest wavelength of light that can be shined on the cathode to produce a current through the anode?

    Answer

  • 36.64 How many photons per second must strike a surface of area 10.0 m2 to produce a force of 0.100 N on the surface, if the photons are in a monochromatic light wave of wavelength 600. nm? Assume the photons are absorbed.

  • 36.65 Suppose the wave function describing an electron predicts a statistical spread of 1.00 · 10−4 m/s in the electron's velocity. What is the corresponding statistical spread in its position?

    Answer

  • 36.66 What is the temperature of a blackbody whose peak emitted wavelength is in the X-ray portion of the spectrum?

  • 36.67 A nocturnal bird's eye can detect monochromatic light of frequency 5.8 · 1014 Hz with a power as small as 2.333 · 10−17 W. What is the corresponding number of photons per second a nocturnal bird's eye can detect?

    Answer

  • 36.68 A particular ultraviolet laser produces radiation of wavelength 355 nm. Suppose this is used in a photoelectric experiment with a calcium sample. What will the stopping potential be?

  • 36.69 What is the wavelength of an electron that is accelerated from rest through a potential difference of 1.00 · 10−5 V?

    Answer

  • 36.70 Compton used photons of wavelength 0.0711 nm.

    1. What is the wavelength of the photons scattered at θ = 180.°?

    2. What is energy of these photons?

    3. If the target were a proton and not an electron, how would your answer in (a) change?

  • •36.71 Calculate the number of photons originating at the Sun that are received in the Earth's upper atmosphere per year.

    Answer

  • 36.72 A free electron in a gas is struck by an 8.5-nm X-ray, which experiences an increase in wavelength of 1.5 pm. How fast is the electron moving after the interaction with the X-ray?

  • 36.73 An accelerator boosts a proton's kinetic energy so that the de Broglie wavelength of the proton is 3.5 · 10−15 m. What are the momentum and energy of the proton?

    Answer

  • 36.74 Scintillation detectors for gamma rays transfer the energy of a gamma-ray photon to an electron within a crystal, via the photoelectric effect or Compton scattering. The electron transfers its energy to atoms in the crystal, which re-emit it as a light flash detected by a photomultiplier tube. The charge pulse produced by the photomultiplier tube is proportional to the energy originally deposited in the crystal; this can be measured so an energy spectrum can be displayed. Gamma rays absorbed by the photoelectric effect are recorded as a photopeak in the spectrum, at the full energy of the gammas. The Compton-scattered electrons are also recorded, at a range of lower energies known as the Compton plateau. The highest-energy of these form the Compton edge of the plateau. Gamma-ray photons scattered 180.° by the Compton effect appear as a backscatter peak in the spectrum. For gamma-ray photons of energy 511 KeV, calculate the energies of the Compton edge and the backscatter peak in the spectrum.