High School Chemistry/The Wave Particle Duality

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In The Bohr Model of the Atom chapter you learned about the double-slit experiment. The double-slit experiment proved that light passing through two closely spaced slits diffracted into circular waves, which then interfered with each other and made interesting patterns on the opposite wall. From the double-slit experiment, it was obvious that light behaved like a wave. But then along came a new set of results from black body radiation and the photoelectric effect. Both of these experiments could only be understood by assuming that light was a particle! Eventually, scientists had no choice but to accept the fact that light was actually both a wave and a particle – hence the wave-particle duality of light. Now, you probably didn't find the arguments in the previous chapter all that hard to accept. You probably thought to yourself "Fine. If Einstein says light is a wave and a particle, I'll believe him. He's the super-genius scientist, and I've never understood light anyhow."

Now let's talk about something that you probably do understand (or at least think you do). Let's talk about matter. Before reading any further, answer the following question – "Does matter behave like a wave, or a particle?" Obviously it behaves like a particle, right? Didn't we already decide that matter was made up of tiny particles called atoms, and that those atoms were, themselves, made up of even tinier particles called electrons, protons, and neutrons? Don’t worry. Everything that you've learned so far is absolutely true. Atoms, electrons, protons and neutrons do behave like particles. But that's not the whole story. Atoms, electrons, protons, and neutrons also behave like waves! In other words, matter is just like light in that it has both wave-like and particle-like properties.

Lesson Objectives[edit]

  • Explain the wave-particle duality of matter.
  • Define the de Broglie relationship and give a general description of how it was derived.
  • Use the de Broglie relationship to calculate the wavelength of an object given the object's mass and velocity.

Electrons Were First Only Considered to Have Particle Properties[edit]

Do you remember how electrons were first discovered? It was all thanks to J. J. Thomson, with his clever cathode ray tube experiment. Briefly, J. J. Thomson cut a small hole into the anode of a cathode ray tube. This allowed cathode rays to pass through the anode and strike phosphor-coated glass on the other side. Since the phosphor glowed when it was struck by the cathode rays J. J. Thomson could actually visualize how the cathode ray traveled through the cathode ray tube.

What J. J. Thomson noticed, of course, was that the cathode rays hit the glass of the cathode ray tube directly opposite the hole in the anode. In other words, the cathode rays traveled in a straight line from the cathode, through the anode, to the glass at the end of the cathode ray tube. That sounds pretty much like a particle, doesn't it? After all, if the electrons in the cathode ray were wave-like, wouldn't they have diffracted when they passed through the hole in the anode? In that case, a huge glowing circle should have appeared on the phosphor painted glass, rather than just a tiny glowing spot.

Cathode ray diffraction.png

Early experiments didn't show any evidence of electron diffraction, so most scientists believed that electrons were particles. It turned out, though, that electrons really do have wave-like properties. The wavelengths were just too small to diffract in the cathode ray experiment.

In addition, by experimenting with magnets, J. J. Thomson had proven that the electron had mass. Up until the 1920s, scientists had never observed wave-like behavior from an object with mass. In fact, you probably can't think of anything with mass that exhibits obvious wave properties. Baseballs don’t diffract; bullets fired from neighboring rifles at a shooting range never form interference patterns.

You might ask about water waves, like the type you see in the ocean. Technically speaking, the water in the ocean doesn't behave like a wave. Instead, it's the energy traveling through the water that behaves like a wave. Water may have mass, but the energy traveling through the water is "mass-less". So far we've established the fact that you never see an object with mass exhibit wave-like properties. Why, then, would anyone suggest that matter, even matter as small as an electron, could be described in terms of a wave-particle duality?

What we know as ocean waves are actually energy waves passing through the water.

de Broglie Proposed That Electrons are Particles and Waves[edit]

Figure 6.1: Louis de Broglie.

We never see matter behaving like a wave, but in 1924 a French graduate student named Louis de Broglie (Figure 6.1) suggested that matter did in fact, have wave-like properties. It was a strange proposal on de Broglie's part, but even stranger was the fact that no one laughed at him! Instead, de Broglie was awarded a Nobel Prize in 1929. So how did de Broglie convince the scientists of his day to believe his theory? In order to argue against de Broglie's wave-particle duality of matter, scientists would have had to argue with Einstein – and no one argues with Einstein!

Einstein is most famous for saying "mass is related to energy". Of course, this is usually written out as an equation, rather than as words:

E = m \times c \times c\,\!
\text{or}\,\!
E\,\! =\,\! m\,\! \times\,\! c^2\,\!
\text{Energy}\,\!
\text{in Joules}\,\!
  \text{Mass}\,\!
\text{in kg}\,\!
  \text{Speed of Light}\,\!
(c = 3.00 \times 10^8\,\text{m/s})\,\!

(In The Science of Chemistry chapter, you were told that you didn't need to know where the equation E = mc2 comes from. You still don't need to know where it comes from, but you do need to know what it means!) So Einstein was responsible for defining the amount of energy in any object with mass. Remember, though, that Einstein was also responsible for proposing the wave-particle duality of light. You have learned that, because of the wave-particle duality of light, the energy of a wave can be related to the wave's frequency by the equation:

E\,\! =\,\! h\,\! \times\,\! f\,\!
\text{Energy}\,\!
(\text{Joules})\,\!
  \text{Planck}'\text{s Constant}\,\!
(6.63 \times 10^{-34}\,\text{J}\cdot\text{s})\,\!
  \text{Frequency}\,\!
(\text{Hz or s}^{-1})\,\!

Louis de Broglie looked at Einstein’s first equation relating mass and energy. Then he looked at Planck's equation, relating energy and frequency (remember, in The Bohr Model of the Atom chapter you learned that frequency is a property associated with all waves, and is related to wavelength according to the equation c = ). After thinking for a while, de Broglie said to himself, "If mass is related to energy, and energy is related to frequency, then mass must be related to frequency!" That's really a very logical argument. It's like saying "If I'm related to my brother, and my brother is related to my sister, then I must be related to my sister." Even though de Broglie's argument was logical, it led to a very illogical result. If mass is related to frequency, then an object with mass must also have frequency. In other words, an object with mass must have wave-like properties! This led de Broglie to propose the wave-particle duality of matter. But where were these so called matter waves? Why hadn't anybody seen them?

de Broglie Derived an Equation for the Wavelength of a Particle[edit]

Using Einstein's two equations, E = mc2 and E = hf, along with the equation relating a wave's frequency and its wavelength, c = , de Broglie was able to derive the following relationship between the wavelength of an object to the object's mass:

\lambda\,\! =\,\! \frac{h}{m \times c}
\text{Wavelength}\,\!
(\text{m})\,\!
=\,\! \frac{\text{Planck}'\text{s Constant } (6.63 \times 10^{-34}\,\text{J}\cdot\text{s})}{\text{Mass } (\text{kg}) \times \text{Speed of Light } (3.00 \times 10^8\,\text{m/s})}

If you're good at math, you may be able to derive the exact same formula yourself. When de Broglie looked at his equation, though, he realized that something was wrong. Why should the wavelength of an object like an electron, or a baseball, depend on the speed of light? Neither baseballs nor electrons travel at the speed of light – only light travels at the speed of light! The problem, of course, was that de Broglie had derived his relationship using equations that applied to light, and not to matter. Luckily, the problem was easy to fix. All de Broglie needed to do was replace the speed of light, c, with some general speed, v.

\lambda\,\! =\,\! \frac{h}{m \times c}
\text{Wavelength}\,\!
(\text{m})\,\!
=\,\! \frac{\text{Planck}'\text{s Constant } (6.63 \times 10^{-34}\,\text{J}\cdot\text{s})}{\text{Mass } (\text{kg}) \times \text{Velocity } (\text{m/s})}

That way, the de Broglie relationship between mass and wavelength could be applied to any thing, traveling at any speed. It could be applied to light, traveling at the speed c = 3.00×108 m/s, it could be applied to a baseball, traveling at the speed v = 45 m/s, and it could be applied to a marathon runner traveling at the speed of v = 3 m/s. Let's try to find the wavelengths of a few common objects (Figure 6.2).

Example 1
Figure 6.2: The wavelength for a typical baseball is far, far too small to be detectable, even in a laboratory!

One of the fastest baseballs ever pitched traveled at a speed of 46.0 m/s. If the average baseball has a mass of 0.145 kg, what is the baseball's wavelength?


Solution:

Planck's constant, h = 6.63×10−34 J · s (Notice that you always know Planck's constant, even if the question doesn't give it to you.)

speed, v = 46.0 m/s

mass, m = 0.145 kg

\lambda = \frac{h}{m \times v} = \frac{(6.63 \times 10^{-34}\,\text{J}\cdot\text{s})}{(0.145\,\text{kg})(46.0\,\text{m/s})}
\lambda = \frac{6.63 \times 10^{-34}\,\text{J}\cdot\text{s}}{6.67\,\text{kg}\cdot\text{m/s}} = 9.94 \times 10^{-35} \frac{\text{J}\cdot\text{s}}{\text{kg}\cdot\text{m/s}}

Dividing by a fraction is the same as multiplying by its reciprocal. This applies to units as well as numbers, so dividing by (kg/m · s) is the same as multiplying by (s/kg · m).

\lambda = 9.94 \times 10^{-35} \frac{\text{J}\cdot\text{s}\cdot\text{s}}{\text{kg}\cdot\text{m}}

In order to account for units, you have to substitute basic units for compound units. The definition of a Joule is, 1 J = 1 kg · m2/s2. Substituting these basic units into the equation in place of Joules yields:

\lambda = 9.94 \times 10^{-35} \frac{\text{kg}\cdot\text{m}^2}{\text{s}^2}\times\frac{\text{s}\cdot\text{s}}{\text{kg}\cdot\text{m}}

After cancelling units, the resultant unit is meters, the correct unit for wavelength.

\lambda = 9.94 \times 10^{-35}\,\text{m}\,\!
Example 2

Stock cars typically race at around a speed of 77.0 m/s. If the average stock car has a mass of 1,312 kg, what is the stock car's wavelength?

The wavelength for a typical stock car is far, far too small to be detectable.

Solution:

Planck's constant, h = 6.63×10−34 J · s (Again, you always know Planck's constant, even if the question doesn't give it to you.)

speed, v = 77.0 m/s

mass, m = 1,312 kg

\lambda = \frac{h}{m \times v} = \frac{(6.63 \times 10^{-34}\,\text{J}\cdot\text{s})}{(1,310\,\text{kg})(77.0\,\text{m/s})}
\lambda = \frac{6.63 \times 10^{-34}\,\text{J}\cdot\text{s}}{100,870\,\text{kg}\cdot\text{m/s}} = 6.57 \times 10^{-39} \frac{\text{J}\cdot\text{s}}{\text{kg}\cdot\text{m/s}}

When Joules is replaced with (kg · m2)/s2

\lambda = 6.57 \times 10^{-39} \frac{\text{kg}\cdot\text{m}^2}{\text{s}^2} \times \frac{\text{s}\cdot\text{s}}{\text{kg}\cdot\text{m}}

After cancelling units, the resultant unit is meters, the correct unit for wavelength.

\lambda = 6.57 \times 10^{-39}\,\text{m}\,\!

What do you notice about the wavelength of a typical baseball and the wavelength of a typical stock car? They're both extremely small, aren't they? In fact, even the strongest microscope in the world today can't see down to sizes like 9.94×10−35 m or 6.57×10−39 m. Well, that explains why we've never seen a matter wave – matter waves are just too small.

If matter waves are too small to see, then how did scientists prove that they exist? After all, the scientific method requires experimental evidence before a theory is accepted – and certainly before a scientist is awarded a Nobel Prize for that theory! Luckily, you can see matter waves. The only trick is to pick matter with waves that are big enough to see. Take another look at the two example questions that we just did. Notice how the only two factors influencing the wavelength of an object are the object's mass and the object's speed. (Of course, you also need to know the value of Planck's constant, h = 6.63×10−34 J · s, but since it's a constant, it doesn't change). In our two examples, the car was more massive than the baseball, and it was traveling at a faster speed. What do you notice about the car's wavelength compared to the baseball's wavelength? The car's wavelength was a lot smaller, wasn't it?

It turns out that the larger the mass of an object, and the faster the speed at which the object is traveling, the smaller the object's wavelength. Similarly, the smaller the mass of an object, and the slower the speed at which the object is traveling, the larger the object's wavelength. This is a direct consequence of de Broglie's equation for the wavelength – because both mass, m, and speed, v, appear in the denominator (the lower part of the fraction), they must both be big if you want a small wavelength, and small if you want a big wavelength.

Obviously, in order to see matter waves experimentally, it would be best to have a big wavelength. That means we need an object with a small mass and a slow speed. What's the smallest object that you know? It's an electron of course! Let's see if an electron wave is large enough to detect in a laboratory.

Example 3

What is the wavelength of an electron traveling at 1.25×105 m/s if the mass of the electron is 9.11×10−31 kg?


Solution:

Planck's constant, h = 6.63×10−34 J · s (Remember, you always know Planck's constant, even if the question doesn't give it to you.)

speed, v = 1.25×105 m/s

mass, m = 9.11×10−31 kg

\lambda = \frac{h}{m \times v} = \frac{(6.63 \times 10^{-34}\,\text{J}\cdot\text{s})}{(9.11 \times 10^{-31}\,\text{kg})(1.25 \times 10^5\,\text{m/s})}
\lambda = \frac{6.63 \times 10^{-34}\,\text{J}\cdot\text{s}}{1.14 \times 10^{-25}\,\text{kg}\cdot\text{m/s}} = 5.82 \times 10^{-9} \frac{\text{J}\cdot\text{s}}{\text{kg}\cdot\text{m/s}}

When Joules is replaced with (kg · m2)/s2

\lambda = 5.82 \times 10^{-9} \frac{\text{kg}\cdot\text{m}^2}{\text{s}^2} \times \frac{\text{s}\cdot\text{s}}{\text{kg}\cdot\text{m}}

After cancelling units, the resultant unit is meters, the correct unit for wavelength.

\lambda = 5.82 \times 10^{-9}\,\text{m}\,\!

Now that wavelength is bigger! All right, it's still not huge, but it's big enough that, in the 1920s, scientists were able to find evidence of electron waves diffracting as electrons were forced through a thin metal film. But let's return to one of our initial questions – if electrons have wave-like properties, why didn't J. J. Thomson see electron diffraction in his cathode ray tube? The answer is simple – in that experiment, the hole that the electrons had to pass through was just too big to cause diffraction. Waves only diffract when their wavelengths are about the same size as the opening that they are forced through. Since electron waves are extremely small, there's no way that they will diffract unless they are forced through extremely small openings. As for your body when you walk through a door – well, if you're worried about your body diffracting, use your mass and the speed at which you walk through the door to figure out your wavelength. That should tell you about how narrow the door has to be to cause your body to diffract. From now on, you can avoid doors of that size! (If you do the calculation properly, you should get a number around 1×10−35 and, of course, you couldn't fit through a door of that size).

Lesson Summary[edit]

  • At first electrons were thought to behave only as particles; de Broglie stated that all matter has wave properties and used Einstein's E = mc2 formula and the formula E = hf to derive the formula: λ = h/(mv) to describe the wavelength λ of an object with mass m traveling at speed v.
  • Matter waves are impossible to detect for ordinary objects, like baseballs and cars, because they are extremely small. The larger the mass, and the faster the speed of an object, the smaller its wavelength.
  • Scientists found evidence of electrons diffracting when forced through a thin metal film.

Review Questions[edit]

  1. In the last chapter you learned that light has wave-like properties and particle-like properties. Can you think of anything else that might have both wave-like properties and particle-like properties?
  2. Decide whether each of the following statements is true or false.
    (a) Einstein was the first scientist to propose matter waves.
    (b) You can see baseballs diffract when you throw them.
    (c) The de Broglie's wave equation can only be applied to matter traveling at the speed of light.
    (d) Most matter waves are very small, and that is why scientists didn't realize matter had wave-like properties until the 1920s.
  3. Choose the correct word in each of the following statements.
    (a) The (more/less) massive an object is, the longer its wavelength is.
    (b) The (faster/slower) an object is traveling, the shorter its wavelength is.
    (c) A particle with a mass of 1.0 g has a (longer/shorter) wavelength than a particle with a mass of 3.0 g if both are traveling at the same speed.
    (d) A baseball moving at 10 m/s has a (longer/shorter) wavelength than a baseball moving at 4 m/s.
  4. Choose the correct word in each of the following statements
    (a) An electron has a (longer/shorter) wavelength than a proton if both are traveling at the same speed.
    (b) An electron wave has a (higher/lower) frequency than a proton wave if both particles are traveling at the same speed.
    (c) If you want to increase the wavelength of an electron, you should (slow the electron down/speed the electron up).
  5. Choose the correct statement from the options below. The factors that influence an object's wavelength are…
    (a) Only the speed of the object
    (b) Only the speed of light
    (c) The speed of light and the mass of the object
    (d) Only the mass of the object
    (e) The speed of the object and the mass of the object
  6. Choose the correct statement from the options below.
    (a) Light behaves only like a wave, and matter behaves only like a particle
    (b) Light behaves only like a wave, and matter behaves only like a wave
    (c) Light behaves only like a particle, and matter behaves only like a wave
    (d) Light behaves like a wave and like a particle, but matter only behaves like a particle
    (e) Light behaves only like a wave, but matter behaves like a wave and like a particle
    (f) Light behaves like a wave and like a particle, and matter behaves like a wave and like a particle as well
  7. Fill in each of the following blanks.
    (a) de Broglie used the equations ___________ and _________ to derive an equation for the wavelength of a matter wave.
    (b) Scientists first saw matter waves by looking for them in _________. This was a good idea, because ________ are small enough to have matter waves that can be observed in a laboratory.
  8. What is the wavelength of a 5.0 kg bowling ball that rolls down the lane at 2.0 m/s?
  9. If you walk through a door at 1.0 m/s, and you weight 120 lbs (or 54 kg), what is your wavelength? (This is also approximately the width of the door that would cause your body to diffract.)
  10. A car has a mass of 1250 kg. If the car's wavelength is 2.41×10−38 m, at what speed is the car traveling?
  11. A bobsled sliding down the run at 14.8 m/s has a wavelength of 1.79×10−37 m. What is the total mass of the bobsled?

Vocabulary[edit]

wave-particle duality of matter
Matter exhibits both particle-like and wave-like properties.


Quantum Mechanics Model of the Atom · Schrodinger's Wave Functions