Calculate De Broglie Wavelength
Now that we know what the wavelength equation is, let's put it to use, shall we?
Calculate the De Broglie wavelength of a proton moving at 1.20 x 106 m/s. The rest mass of a proton is 1.67 x 10-27 kg.
Firstly, we need to calculate the momentum:
$$p=m*v$$
$$p=(1.67x10^{-27}\,kg)(1.20x10^6\frac{m}{s})$$
$$p=2.004x10^{-21}\frac{kg*m}{s}$$
Now we can plug this into the wavelength equation:
$$lambda=\frac{h}{p}$$
$$\lambda=\frac{6.626x10^{-34}\frac{kg*m^2}{s}}{2.004x10^{-21}\frac{kg*m}{s}}$$
$$\lambda=3.31x10^{-7}\,m$$
$$1x10{-9}\,m=1\,nm$$
$$3.31x10^{-7}\,m*\frac{1\,nm}{1x10^{-9}\,m}=331\,nm$$
For reference, purple light is about 380 nm, so this wavelength is a bit too small to be visible by the human eye.
De Broglie's Wavelength for Electrons
When electrons orbit the nucleus, the De Broglie waves form a closed loop. The electrons can only exist as standing waves that "fit" in the electron cloud and are "allowed", as shown below:
Fig.2: An allowed energy level (left) and one that is not allowed (right)
On the left, the standing wave fits along the electron cloud, so it is at an allowed energy level. On the right, the wave doesn't fit and is therefore not at an allowed energy level.
Basically, electrons are allowed to exist at set, quantized energy levels called "shells":
Fig.3: Allowed energy levels, also called shells
The closer an electron is to the nucleus, the lower in energy it is. The electrons at the n=1 state are called ground-state electrons, since they are at the lowest energy level.
Now that we've covered how electron waves behave, let's calculate the wavelength!
A ground state electron (n=1) is orbiting a hydrogen nucleus at 2.18 x 106 m/s. The mass of an electron is 9.11 x 10-31 kg. What is the wavelength of this electron?
Like before, we need to first calculate the momentum:
$$p=m*v$$
$$p=(9.11x10^{-31}\,kg)(2.18x10^{6}\frac{m}{s})$$
$$p=1.99x10^{-24}\frac{kg*m}{s}$$
$$\lambda=\frac{h}{p}$$
$$\lambda=\frac{6.626x10^{-34}\frac{kg*m^2}{s}}{1.99x10^{-24}\frac{kg*m}{s}}$$
$$\lambda=3.33x10^{-10}\,m$$
$$\lambda=0.333\,nm$$
Applications of the De Broglie Wavelength
1. The wave properties of matter can only be seen in very small things. Using electrons as the source, a de Broglie wavelength interference pattern can be made. The average energy of an electron in an electron microscope is 10 eV, so the de Broglie wavelength is 3.9 x 10-10 m.
This is similar to how far apart atoms are. So, a crystal acts as an electron diffraction grating. The crystal structure can be figured out by looking at the diffraction pattern.
2. The wavelength used in a microscope limits the size of the smallest things we can see. The shortest wavelength of visible light is 400 nm, which is equal to 4 x 10-7 m. Most electron microscopes use wavelengths that are 1,000 times smaller and can study very small details.
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