2 years ago
When a quantum wavelength is smaller than a particle it behaves like a particle. What determines the quantum wave length size?
You can leave an optional "tip" with Mahalo's virtual currency, Mahalo Dollars. If you are asking a difficult question that might require some research, or if you'd like a wide variety of feedback, a higher tip often leads to more answers to your question.
M$1 Answer
The quantum wave-length of a particle is also known as its deBroglie wavelength (pronounced "debroy"). The de Broglie wavelength is defined as:
lambda = (h c)/(p c)
where h is Plank's constant, c is the speed of light in vacuum, as a result, hc = 1239.84 eV nm and pc is expressed in electron volts.
In the limit where the kinetic energy of a massive particle is much lower than that particle's rest mass, we can make the approximation:
(p c) = ~ sqrt(2 * E_k * m_o * c^2) where m_0 is the particle rest mass and E_k its kinetic energy.
For example, a 1 eV photon would have a de Broglie wavelength of 1240 nm, whereas an electron with 1 eV kinetic energy, and a rest mass of 0.511 MeV, would have a de Broglie wavelength of 1.23 nm, or about 1000 times smaller than a photon with the same kinetic energy. This is why electron microscopes have much better resolution than optical ones.
For more details, read the page at http://hyperphysics.phy-astr.gsu.edu/HBASE/quantum/debrog2.html .
lambda = (h c)/(p c)
where h is Plank's constant, c is the speed of light in vacuum, as a result, hc = 1239.84 eV nm and pc is expressed in electron volts.
In the limit where the kinetic energy of a massive particle is much lower than that particle's rest mass, we can make the approximation:
(p c) = ~ sqrt(2 * E_k * m_o * c^2) where m_0 is the particle rest mass and E_k its kinetic energy.
For example, a 1 eV photon would have a de Broglie wavelength of 1240 nm, whereas an electron with 1 eV kinetic energy, and a rest mass of 0.511 MeV, would have a de Broglie wavelength of 1.23 nm, or about 1000 times smaller than a photon with the same kinetic energy. This is why electron microscopes have much better resolution than optical ones.
For more details, read the page at http://hyperphysics.phy-astr.gsu.edu/HBASE/quantum/debrog2.html .
You can leave an optional "tip" with Mahalo's virtual currency, Mahalo Dollars. If you are asking a difficult question that might require some research, or if you'd like a wide variety of feedback, a higher tip often leads to more answers to your question.
M$
Lambda is the wavelength. For massive particles, the de Broglie wavelength is describing the probability distribution of finding the particle in different locations.
The de Broglie equation provides the matter wavelength for any energy/momentum, so yes, that would be included in it. There is no need for a diagram for h c as those are two scalar constants - the Plank constant and the speed of light in vacuum. When you multiply the two values you get the 1239.4 eV nm. The units of eV are energy, and those of nm are distance, which could be wavelength. Once you divide by p c in units of eV, you're left with only the unit of nm, which is what you expect for a wavelength.
To the other comment, photons have no rest mass, so their only energy is their "kinetic" energy. The energy of a photon is related to its wavelength by the equation:
E = h c / lambda (see e.g. http://en.wikipedia.org/wiki/Photon ).
Since h c = 1239.84 eV nm, swapping E and lambda you get:
lambda_photon = 1239.84 eV nm/ E = 1239.84 nm for E = 1 eV.
Lambda is the wavelength, measured in units of distance.
Now I understand the equations. Thanks. Lambda is the quantum wavelength.
What does lambda describe? Does lambda describe kinetic energy of a particle with mass in motion?
Are you saying that 1 eV photon with only kinetic energy would have a 1240 nm wavelength? Will you show how the wavelength was determined? Is lambda a force or a distance?
Is E=MC^2 converted to kinetic energy expressed in the deBroglie equation?
Will you show in a diagram what hc means 1239.84 eV nm? Is the eV the amplitude of the wave and the distance crest to crest of the wave one nanometer?