The Particle-Wave Duality Reviewed

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 [link].)

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 = 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 = h / λ

. 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 [link].)

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:

Integrated Concepts

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:

[link] illustrates how these strategies are applied to an integrated-concept problem.

Recoil of a Dust Particle after Absorbing a Photon

The following topics are involved in this integrated concepts worked example:

Topics
Photons (quantum mechanics)
Linear Momentum

A 550-nm photon (visible light) is absorbed by a $1 . 00-μg$

particle of dust in outer space. (a) Find the momentum of such a photon. (b) What is the recoil velocity of the particle of dust, assuming it is initially at rest?

Strategy Step 1

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 (a) of this example asks for the momentum of a photon, a topic of the present chapter. Part (b) considers recoil following a collision, a topic of Linear Momentum and Collisions.

Strategy Step 2

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.

Solution for (a)

The momentum of a photon is related to its wavelength by the equation:

p = \frac{h}{λ} .

Entering the known value for Planck’s constant $h$

and given the wavelength $λ$

, we obtain

\begin{array}{l} p & = & \frac{6.63 \times 10^{- 34} J \cdot s}{550 \times 10^{–9} m} \\ = & 1.21 \times 10^{- 27} kg \cdot m/s . \end{array}

Discussion for (a)

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.

Solution for (b)

Conservation of momentum in the absorption of this photon by a grain of dust can be analyzed using the equation:

p_{1} + p_{2} = {p'}_{1} + {p'}_{2} (F_{net} = 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

p_{1} = {p'}_{2} = mv,

where $p_{1}$

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 = \frac{p}{m},

where $p$

is the photon momentum found in part (a). Entering known values (noting that a microgram is $10^{- 9} kg$

) gives

\begin{array}{l} v & = & \frac{1.21 \times 10^{- 27} kg \cdot m/s}{1.00 \times 10^{- 9} kg} \\ = & 1.21 \times 10^{–18} 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

Conceptual Questions

Problems & Exercises

Integrated Concepts

(a) Calculate the amount of microwave energy in joules needed to raise the temperature of 1.00 kg of soup from $20 . 0º C$

to $100 ºC$

. (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?

(a) $3 . 35 \times 10^{5} J$

(b) $1 . 12 \times 10^{–3} kg \cdot m/s$

(d) $6.23 \times 10^{–7} J$

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.

(a) $1 . 62 \times 10^{3} m/s$

(b) $4 . 42 \times 10^{- 19} J$

for photon, $1 . 19 \times 10^{- 24} J$

for electron, photon energy is $3 . 71 \times 10^{5}$

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.

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 $1 . {00 kW/m}^{2}$

reflects from a mirror of area $2 . {00 m}^{2}$

. (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?

(a) 2.00 kJ

(b) $1 . 33 \times 10^{- 5} kg \cdot m/s$

(d) yes