\[\lambda = \frac{h}{p} = \frac{h}{mv}\quad[\text{Eq.} (8.5)]\]
Note: If you feel uncomfortable with working with variables, fractions, prefixes, or scientific notation, you should review the following tutorials before working through this tutorial:
Sections
Introduction
Light behaves as both a wave (Maxwell’s equations; Section 6.5 of your text) and a particle (Planck’s photon energy equation; Sections 8.1 and 8.5), a phenomenon known as wave-particle duality. Louis de Broglie argued that since photons (massless particles) can exhibit wave-particle duality, then perhaps massive particles share the same behavior. Using an equation that Einstein developed for the linear momentum (\(p\)) of a photon given its wavelength, \(p = h/\lambda\), it would seem logical to assume that by solving Einstein’s equation for wavelength, the wavelength of a massive particle should be given by \(\lambda = h/p\). Since the magnitude of the linear momentum vector is \(p = mv\) [Equation (6.2) in your text], then the wavelength of matter should be given by \(\lambda = h/mv\), which is Equation (8.5).
Since Planck’s constant, \(h\), never changes, Equation (8.5) tells us that as the momentum of a particle increases, the associated particle wavelength decreases, and vice versa. But momentum can increase by increasing the product of the particle’s mass (\(m\)) and its speed (\(v\)). We are not familiar with massive particles having wavelike characteristics simply because the wavelengths are incredibly small. This is because the value of Planck’s constant is extremely small: \(h = 6.626\times 10^{-34}\text{ J s}\). Assuming that a man with a mass of 70 kg is strolling at 1 m/s, then his momentum would be \(p = mv = 70\text{ km m/s}.\) Substituting that value into de Broglie’s equation gives \[\lambda = 6.626\times 10^{-34}\text{ J s}/70\text{ kg m/s} = 9.4\times10^{-36}\text{ m}.\] For comparison, the typical diameter of the nucleus of an atom is about \(1\times10^{-15}\text{ m} = 1\text{ fm}\). In other words, the wavelength of the man is about one hundred millionth of one trillionth of the diameter of an atom’s nucleus! Chances are you won’t notice yourself behaving like a wave.
The situation changes dramatically if, instead of calculating the wavelength of a walking 70-kg man, a \(9.1\times 10^{-31}\text{ kg}\) electron is considered instead. In that case, if the electron is moving at 1 m/s (extremely unlikely unless specifically constrained!) then the wavelength would be \(\lambda = 0.72\text{ mm}\). Now we are getting into the size range that we can relate to.
Speeds of electrons are often more in the range of \(10^5\text{ m/s}\) to near the speed of light. An electron with a kinetic energy of 1 eV has a speed of 600 km/s or \(6\times 10^5\text{ m/s}\). The corresponding wavelength for the electron at that speed would be about \(1.2\text{ nm}\). (Again for reference, the diameter of an atom is roughly \(1\times 10^{-10}\text{ m} = 0.1\text{ nm}.\))
Examples
- Referring back to the last paragraph in the introduction, what would the particle wavelength be of a proton moving at 600 km/s? (The mass of a proton is about 1800 times greater than the mass of an electron.)
If the speed of the proton is the same as the electron from the last paragraph, but its mass is 1800 times greater, then the denominator of Equation (8.5) would be 1800 times greater, and so the wavelength would 1800 times shorter, or \[\lambda = 1.2\text{ nm}/1800 = 1.2\times10^{-9}\text{ m}/1.8\times10^3 = 6.7\times10^{-13}\text{ m} = 0.67\text{ pm}.\] (pm is a picometer. Recall the prefixes table.) - Suppose an electron is moving at 1800\text{ km/s}. Based on the information in the last paragraph of the introduction, what would the electron’s wavelength be?
The electron’s speed is \(1800/600 = 3\) times greater than the example in the introduction, implying that its momentum is also 3 times greater. Since momentum is in the denominator of de Broglie’s equation, the wavelength would be 3 times shorter, or \(1.2\text{ nm}/3 = 0.4\text{ nm}.\)
Quiz Yourself
Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.
(Answers are available below.)
- If the speed of an electron is increased by a factor of 100, does its wavelength increase or decrease, and by what factor?
- An atom of the most common type of carbon has a mass that is about 12 times greater than the mass of a proton. If a carbon atom is traveling at 600 km/s, what would its wavelength be?
- An electron microscope uses electrons instead of photons to probe very small structures. The resolution of the microscope is roughly the same as the wavelength of the particles being used as probes. If a beam of electrons is accelerated to \(1\times 10^7\text{m/s}\), what would the wavelengths of the electrons be?
Answers
- Since the electron’s speed is in the denominator of Equation (8.5), if its speed increased by a factor of 100, its wavelength would decrease by a factor of 100.
- 0.056 pm or 56 fm.
- 0.072 nm. (This is slightly smaller than the size of a hydrogen atom.)