1
Wave–Particle Duality

1.1 Exercises

1. E1.1 Consider a microwave source that is generating 2.0 GHz electromagnetic radiation. Compute the wavelength of the microwaves. If this microwave source was used in an oven of width 30 cm, how many wavelengths of the microwave can be included across the oven’s width?

   Compute the energy per photon for the 2.0 GHz frequency. If a cup containing 250 mL of water is irradiated by this source, how many photons must be absorbed to raise the temperature of the water from 25 °C to 80 °C (a nice temperature for a cup of tea). For simplicity, assume that the water density is 1.0 g/mL, that the heat capacity is 4.184 J/(g °C), and that they do not change over the temperature range.

   Solution

   First we calculate the wavelength of a 2.0 GHz microwave and then compare it to the oven’s width. The wavelength and frequency are related by with being the speed of light cm , so

   Hence, the oven is about wide.

   Here we calculate the energy in a 2.0 GHz photon and compare it to the energy needed to warm the water (assuming no extraneous losses). The energy and frequency are related by , so that the energy per photon is

   The amount of energy the water must absorb is , where is the mass of water (250 mL or 250 g), is the heat capacity, and is the change in temperature °C. Thus

   so that the number of 2.0 GHz photons will be

   Because we have ignored any extraneous losses (e.g., heat conduction to the container and convective cooling), this value is a lower bound.

2. E1.2 Consider an ultraviolet light source that generates 300 nm electromagnetic radiation. Compute the frequency of the ultraviolet light. If one photon of this light is absorbed by an organic molecule, how much energy does the molecule gain? Is this energy enough to break a carbon–carbon bond in the molecule? Use a “typical” carbon–carbon bond energy of J for your comparison. Perform the same calculations for a photon of wavelength 600 nm and a photon of wavelength 1200 nm. Perform your comparisons using the energy units of J and of eV.

   Solution

   The wavelength and frequency are related by with being the speed of light ( m ), so m m.

   The energy and frequency are related by , so the energy per photon J s) () J.

   This amount of energy is “just” sufficient to break a bond of J.

   The corresponding energies for 600 and 1200 nm photons are J and J, neither of which is sufficient to break the typical carbon-carbon double bond.

   The corresponding energies in eV ( J eV) are 4.12 eV (300 nm light), 2.06 eV (600 nm light), and 1.03 eV (1200 nm light).

3. E1.3 Consider an electron with a kinetic energy of 1.0 eV (i.e., it has been accelerated across a 1 V potential difference).

   Compute the momentum of this electron. Compare this momentum to that of a “typical” gas molecule at room temperature (consider the gas molecule to have a speed of 500 m/s).

   Compute this electron’s speed. At what fraction of light speed ( m/s) is the electron moving?

   Compute this electron’s wavelength. Compare this wavelength to the diameter of a hydrogen atom (ca. 128 pm). Perform this same calculation for a 10 eV electron and a 100 eV electron. Comment on the trends in your values. How many electron wavelengths can fit into a hydrogen atom at these different energies?

   Solution

   The momentum is related to the kinetic energy by , so we find the momentum by

   The momentum of a “typical” gas-phase nitrogen molecule () is

   which is about 43 times greater than the momentum of the electron.

   The electron’s speed is

   This value is 0.002, or 0.2%, of the speed of light! While this speed is significant, it is still small enough to neglect relativistic effects.

   The electron’s wavelength can be calculated using the de Broglie relationship, so that

   where we have used kg m for the momentum of the electron. This wavelength is 9 to 10 times larger than the characteristic size of an H atom.

   For 10 eV electrons nm, and for 100 eV electrons pm. The electron wavelength decreases as the square root of its kinetic energy and a 100 eV electron has a wavelength that is similar to the diameter of an H-atom.

4. E1.4 A typical value for a particle’s kinetic energy at 25 °C is J. Use this value of the kinetic energy to estimate the speed of spheres with different masses; i.e.,
   1. ping pong ball (2.60 g)
   2. a 10.0 diameter polystyrene bead (0.300 g/ kg/)
   3. a 50.0 nm radius colloidal particle of Ag (10.5 g/ )
   4. Buckminster fullerene () (0.720 kg/mol)
   5. He atom (4.0 g/mol kg/mol).

   Use these speeds and masses to estimate the de Broglie wavelength of these spheres. Comment on the trend in your wavelengths. For which, if any, of these particles would you expect their wave properties to be important? If the kinetic energy was decreased by 100 times, how would your wavelengths change? Do you think that wave properties would be important under these circumstances?

   Solution

   The speed and kinetic energy are related by

   Hence we find

   1. for the ping pong ball
   2. for the polystyrene bead we first compute its mass by and then find its speed by
   3. for the silver colloid particle we first compute its mass by and then find its speed by
   4. for the Buckminsterfullerene we first compute its mass by and then find its speed by
   5. for the He atom we first compute its mass by and then find its speed by To find the de Broglie wavelengths , we use the fundamental relation By way of example, we consider the Ag colloid particle and calculate

   Proceeding in a like manner for each of the cases above we find

   particle

   /m

   ping-pong ball

   polystyrene bead

   Ag particle

   fullerene,

   He atom

   These numbers suggest that it is not necessary to consider the wave nature of these particles under these conditions; i.e., the wavelength is small compared to the size of structures from which it might collide so that diffraction is not important.

5. E1.5 Describe the photoelectric effect experiment.
   1. Provide a sketch of the apparatus.
   2. State the implications of the experiment.
   3. Describe what is observed in the experiment and how it relates to the experiment’s implications.

   Solution

   1. Fig. 1.2a of the textbook gives a schematic of the photoelectric effect apparatus.
   2. The principal implication is that light can behave has a particle.
   3. The two observations are that the stopping potential depends on the light frequency and not on intensity, while the number of photoelectrons depends on light intensity and not frequency. These results are exactly the opposite of the behavior that one expects for a classical wave and are exactly what would be expected if the light behaved as a particle.
6. E1.6 Consider the diffraction of photons, electrons, and neutrons from an aperture with diameter . Consider the case where is 1 cm and the case where it is cm.
   1. If you direct a light beam onto the aperture, how large must the wavelength be so that diffraction can be observed? What is the frequency of the light you found?
   2. If you direct an electron beam onto the aperture, how large must the speed of the electrons be so that diffraction can be observed?
   3. We assume that the de Broglie relationship holds not only for electrons, but also for any particle. How large must the speed of the neutrons be for the aperture to diffract a neutron beam?

   Do not be disturbed if the answers to these exercises are not experimentally feasible. The goal is to clarify the content of Equations (1.7) and (1.9)

   Solution

   Diffraction occurs when the wavelength of the wave is approximately the same as the size of the aperture. Considering the size of the aperture as 1 cm and cm,

   1. For an aperture of 1 cm, the wavelength is 1 cm, and the corresponding frequency is cm / 1 cm . For a cm aperture, the wavelength is cm, and the corresponding frequency is cm cm .
   2. We need to find the electron’s wavelength through the de Broglie relationship, . For a 1 cm wavelength J s) kg0.01 m) cm . For a cm wavelength J s) kg) ( m) cm .
   3. We need to find the neutron’s wavelength through the de Broglie relationship, For a 1 cm wavelength J s) kg0.01 m) cm . For a cm wavelength J s) kg m) cm .
7. E1.7 If photons are particles they have momentum. Compute the momentum of a 590 nm photon. Compare this momentum to that of a Na atom moving at a speed of 900 m/s, which is a typical value at 1200 °C. Assume that 590 nm photons collide head on with the sodium atom so that the momentum exchange is twice the photon momentum, how many photons are needed to ’stop’ the sodium atom?

   Solution

   Again we...