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In relating a particle's energy to its wavelength, two equations are used. The first is the kinetic energy equation:
Equation Number One: KE = (1/2) mv2
(The second equation is down the page a bit.)
There are three symbols in this equation:
a) KE stands for kinetic energy
b) m stands for mass
c) v stands for velocity
I would like to emphasize that these are symbols, which are standing in place of the actual numerical values. Another example of an equation using symbols is PV = nRT. I would also like to emphasize that these symbols are not the units. For example, the symbol P stands for the pressure and the unit on the numerical value for pressure is atm.
Next, some comments on the units attached to the numerical values represented by the symbols in the kinetic energy equation.
The unit on mass is kilograms. That means, if you get a mass in grams given in the problem, you must convert it to kilograms.The unit on velocity is meters per second, most usually written m s¯1 (it can also be written m/s). One possible point of confusion: The unit m means meters, the symbol m means mass. You have to keep the two separate.
The unit on KE is kg m2 s¯2. Remember that the velocity (symbol = v) is squared. That means the unit (m/s) gets both the length (unit = m, meaning meters) and the time (unit = s, meaning seconds) get squared.
Here's one last confusing thing to make you feel better: the name (as opposed to the symbol) of the KE unit is Joule.
You really do need to keep the difference between a symbol and a unit clear in your mind.
Now, let's manipulate the KE equation. I'm going to use E instead of KE:
E = (1/2) mv22E = mv2
2Em = m2v2
2Em = (mv)2
Here's a key point from physics: mv is a particle's momentum. The standard symbol for momentum is p:
2Em = p2p = √(2Em)
Now comes the second equation. It is one of two related equations called the de Broglie equations. You can read more about de Broglie's work here. He received the 1929 Nobel Prize in Physics for this work. (I will discuss the second de Broglie equation below the following example problems.)
Equation Number Two: λ = h/p
There are three symbols in this equation:
a) λ stands for the wavelength of the particle
b) h stands for Planck's Constant
c) p stands for the momentum of the particle
Just above, we developed that p = √(2Em). We can now substitute that into the de Broglie equation:
λ = h/√(2Em)
We will use this result in the example problems that follow.
You may skip the following discussion of the second de Broglie equation. I do not use it in any of the examples problems to follow. it is here in case you migt be interested in it.
The second de Broglie equation is this: ν = E/h
There are three symbols in this equation:
a) ν stands for frequency (sometimes ν is replaced by f)
b) E stands for kinetic energy
c) h stands for Planck's Constant
Suppose an electron has momentum equal to p, then its wavelength is λ = h/p and its frequency is f = E/h. (ChemTeam comment: note the use of the two de Broglie equations.)
The velocity of a de Broglie wave is:
v = λν
Substituting, we obtain:
v = (h/p) (E/h)v = E/p
Substituting for E, we obtain:
v = [(1/2) p2/m] / pv = [(1/2) p] / m
ChemTeam comment (including the indented equations just below): remember that p is momentum and that p = mv. Look at (1/2) p2/m and substitute mv for the p:
E = (1/2) (mv)2/mE = (1/2) m2v2/m
E = (1/2) mv2
Now comes the question: Should not the velocity of an electron be v = p/m? (ChemTeam comment: replace p with mv to get mv/m = v.) The posted answer was this:
No, the v that you are calculating is not the classical momentum of the particle: it is the phase velocity of the wave. The classical momentum of the particle is the group velocity of a wave packet, which turns out to be p/m, but the analysis takes about one week of a QM class.
At this point, we are beyond what the ChemTeam understands. To those of you reading this, best wishes with your continued studies!
Now, on to the example problems.
Problem #1: What is the wavelength of an electron (mass = 9.11 x 10¯31 kg) traveling at 5.31 x 106 m/s?
1) The first step in the solution is to calculate the kinetic energy of the electron:
KE = (1/2)mv2x = (1/2) (9.11 x 10¯31 kg) (5.31 x 106 m/s)2
x = 1.28433 x 10¯17 kg m2 s¯2 (I kept some guard digits)
When I use this value just below, I will use J (for Joules).
2) Next, we will use the de Broglie equation to calculate the wavelength:
λ = h/pλ = h/√(2Em)
x = 6.626 x 10¯34 J s / √[(2) (1.28433 x 10¯17 J) (9.11 x 10¯31 kg)]
Just to be sure about two things: (1) the unit on Planck's Constant is Joule-seconds, both are in the numerator and (2) there are three values following the radical in the denominator. All three of them are under the radical sign.
The answer:
x = 1.37 x 10¯10 m
I'd like to compare this wavelength to ultraviolet light, if I may. Let's use 4000 Å = 4000 x 10¯8 cm = 4 x 10¯7 m.
Our electron's wavelength is almost 3000 times shorter than our ultraviolet example and its wavelength puts it in the X-ray region of the electromagnetic spectrum.
This turned out to be very important because one could then take a beam of electrons and perform experiments with detectable results. You can't do that with the short wavelengths of heavier particles (see examples below).
In 1926, de Broglie predicted that matter had wave-like properties. In 1927, experiments were done that showed electrons behaved as a wave (by showing the property of diffraction and interference patterns). In 1937, the Nobel Prize in Physics was awarded to Clint Davisson and George Thomson (son of J.J. Thomson) for this work.
I would like to pause briefly to analyze the units in the above problem. Specifically this equation:
λ = h/√(2Em)
The units on Planck's Constant are J ⋅ s, but I would like to use this instead:
kg m2 s-2 ⋅ s
I realize that seconds would cancel (leaving a s-1). Set that aside for the moment, but remember that a unit of seconds does cancel in the above set of units.
In the denominator, there are units on E and m, as follows:
kg m2 s-2 ⋅ kg
I realize that kilograms would be a square, but I want to emphasize the energy times mass formulation.
Remembering that the units in the denominator are all under the radical sign, we apply the radical sign to the units in the denominator, arriving at this:
kg m s-1
So, we now have this:
(kg m2 s-1) / (kg m s-1)
Everything cancels, except for a meter in the numerator, which is exactly what we want.
Problem #2: What is the wavelength in meters of a proton traveling at 255,000,000 m/s (which is 85% of the speed of light)? (Assume the mass of the proton to be 1.673 x 10¯27 kg.)
1) Calculate the kinetic energy of the proton:
KE = (1/2)mv2x = (1/2) (1.673 x 10¯27 kg) (2.55 x 108 m/s)2
x = 5.43934 x 10¯11 J
2) Use the de Broglie equation:
λ = h/pλ = h/√(2Em)
x = 6.626 x 10¯34 J s / √[(2) (5.43934 x 10¯11 J) (1.673 x 10¯27 kg)]
x = 1.55 x 10¯15 m
This wavelength is comparable to the radius of the nuclei of atoms, which range from 1 x 10¯15 m to 10 x 10¯15 m (or 1 to 10 fm).
Problem #3: Calculate the wavelength (in nanometers) of a H atom (mass = 1.674 x 10-27 kg) moving at 698 cm/s
1) Convert cm/s to m/s:
698 cm/s = 6.98 m/s
2) Calculate the kinetic energy of the proton:
KE = (1/2)mv2x = (1/2) (1.674 x 10¯27 kg) (6.98 m/s)2
x = 5.84226 x 10¯27 J
3) Use the de Broglie equation:
λ = h/pλ = h/√(2Em)
x = 6.626 x 10¯34 J s / √[(2) (5.84226 x 10¯27 J) (1.674 x 10¯27 kg)]
x = 1.50 x 10¯7 m
4) Convert m to nm:
1.50 x 10¯7 m = 150. nm
5) Comment: the absolute temperature of the H atom moving at 6.98 m/s can be calculated:
KE = (3/2) kTT = 2KE / 3k
T = [(2) (5.84226 x 10¯27 J) / [(3) (1.38065 x 10¯23 J/K)]
T = 0.000304 K
Pretty cold, if you ask me!
Problem #4: What is the wavelength of a 5.00 ounce baseball traveling at 100.0 miles per hour? (5.00 oz = 0.14175 kg and 100 mph = 44.70 m/s)
1) Calculate the kinetic energy of the baseball:
KE = (1/2)mv2x = (1/2) (0.14175 kg) (44.70 m/s)2
x = 141.6146 J (as always, some guard digits)
2) Use the de Broglie equation:
λ = h/pλ = h/√(2Em)
x = 6.626 x 10¯34 J s / √[(2) (141.6146 J) (0.14175 kg)]
x = 1.046 x 10¯34 m
Yep, pretty short wavelength! Compare it to the Planck length.
Problem #5: An atom of helium has a de Broglie wavelength of 4.30 x 10¯12 meter. What is its velocity?
Solution:
1) Use the de Broglie equation to determine the energy (not momentum) of the atom [note the appearence of the mass (in kg) of a He atom]:
λ = h/pλ = h/√(2Em)
4.30 x 10¯12 m = 6.626 x 10¯34 J s / √[(2) (x) (6.646632348 x 10¯27 kg)]
I dropped the units.
4.30 x 10¯12 times √[(2) (x) (6.646632348 x 10¯27) = 6.626 x 10¯34]
√[(2) (x) (6.646632348 x 10¯27)] = 6.626 x 10¯34 / 4.30 x 10¯12
I divided the right side and then squared both sides.
(2) (x) (6.646632348 x 10¯27) = 2.374466 x 10¯44
x = 1.786217333 x 10¯18 J
2) Use the kinetic energy equation to get the velocity:
KE = (1/2)mv21.786217333 x 10¯18 = (1/2) (6.646632348 x 10¯27) v2
v2 = 5.3748 x 108
v = 2.32 x 104 m/s
Problem #6: Calculate the velocity of an electron (mass = 9.10939 x 10¯31 kg) having a de Broglie wavelength of 269.7 pm
Solution:
1) Convert pm to m:
269.7 pm = 269.7 x 10-12 m = 2.697 x 10-10 m
2) Use the de Broglie equation to determine the energy (not momentum) of the atom:
λ = h/pλ = h/√(2Em)
2.697 x 10¯10 m = 6.626 x 10¯34 J s / √[(2) (x) (9.10939 x 10¯31 kg)]
I dropped the units.
2.697 x 10¯10 times √[(2) (x) (9.10939 x 10¯31) = 6.626 x 10¯34]
√[(2) (x) (9.10939 x 10¯31)] = 6.626 x 10¯34 / 2.697 x 10¯10
I divided the right side and then squared both sides.
(2) (x) (9.10939 x 10¯31) = 6.035885 x 10¯48
x = 3.313 x 10¯18 J
3) Use the kinetic energy equation to get the velocity:
KE = (1/2)mv23.313 x 10¯18 = (1/2) (9.10939 x 10¯31) v2
v2 = 7.2738 x 1012
v = 2.697 x 106 m/s
Problem #7: Calculate the velocity of a neutron with a wavelength of 65 pm:
Solution:
1) Convert 65 pm to m:
65 pm = 65 x 10¯12 m = 6.5 x 10¯11 m
2) Use the de Broglie equation to determine the energy (not momentum) of the atom (the mass of the neutron is in example #8):
λ = h/pλ = h/√(2Em)
6.5 x 10-11 m = 6.626 x 10¯34 J s / √[(2) (x) (1.67493 x 10¯27 kg)]
Algebra!
x = 3.102055 x 10¯20 J
3) Use the kinetic energy equation to get the velocity (I dropped the units since I know m/s results):
KE = (1/2)mv23.102055 x 10¯20 = (1/2) (1.67493 x 10¯27) (v)2
v = 6086 m/s
Problem #8: Calculate the de Broglie wavelength of a neutron (mass = 1.67493 x 10¯27 kg) moving at one five-hundredth of the speed of light (c/500).
Solution:
1) Determine the speed of the neutron:
3.00 x 108 m/s divided by 500 = 6.00 x 105 m/s
2) Calculate the kinetic energy of the neutron (I used Joule for the energy unit):
KE = (1/2)mv2KE = (1/2) (1.67493 x 10¯27 kg) (6.00 x 105 m/s)2
KE = 5.02479 x 10 ¯22 J
3) Use the de Broglie equation:
λ = h/pλ = h/√(2Em)
λ = 6.626 x 10¯34 J s / √[(2) (5.02479 x 10 ¯22 J) (1.67493 x 10¯27 kg)]
λ = 5.107 x 10¯10 m
Problem #9: Calculate the wavelength of an object weighing 100.0 kg and moving at 160 km/hour.
Solution:
1) Convert km/hr to m/s:
160 km/hr = 160,000 m/hr160,000 m/hr = 160,000 m/3600 s = 44.44 m/s
2) Solve for the kinetic energy:
K.E. = (1/2) (100.0 kg) (44.44 m/s)2K.E. = 9.87654 x 104 J
3) Use λ = h/p:
λ = h/√(2Em)λ = 6.626 x 10¯34 J s / √[(2) (9.87654 x 104 J) (9.10939 x 10¯31 kg)]
λ = 1.49 x 10¯37 m
Problem #10: What is the de Broglie wavelength (in nm) of a molecule of buckminsterfullerene (C60), moving at a speed of 100.0 m/s?
Solution:
1) We need the mass of one C60 molecule in kilograms:
a) calculate molar mass:
60 x 12.011 = 720.66 g/molb) calculate mass of one molecule:
720.66 g/mol divided by 6.022 x 1023 molecules/mol = 1.1967 x 10¯21 g/moleculec) convert to kilograms:
1.1967 x 10¯21 g/molecule divided by 1000 g/kg = 1.1967 x 10¯24 kg
2) Solve for the kinetic energy:
K.E. = (1/2) (1.1967 x 10¯24 kg) (100.0 m/s)2K.E. = 5.9835 x 10¯21 J
3) Use λ = h/p:
λ = h/√(2Em)λ = 6.626 x 10¯34 J s / √[(2) (5.9835 x 10¯21 J) (1.1967 x 10¯24 kg)]
λ = 5.5369 x 10¯12 m
4) Convert to nm:
5.5369 x 10¯12 m times 109 nm / m = 5.537 x 10¯3 nm (to four sig figs)