De Broglie wavelength computable by just fixing speed of light ?

In summary, according to this summarizer, it is possible to compute the deBroglie wavelength of a 1s electron in units of Planck length, or alternatively in terms of ct, lightspeed time.
  • #1
xortdsc
98
0
Hi,

I was wondering if it is possible to compute e.g. bohr radii for a metric system whose correspondence to real units is unknown and only the speed of light is known.

Let's say the only thing I know is that lightwaves travel x spaceunits in t timeunits, therefore defining the speed of light. The obvious problem is that I lack correspondance to real units such as meters for space and seconds for time.
So the question is: Is it possible to compute e.g. the bohr radius of an electron in groundstate in the hydrogen in units of x (or alternatively ct, lightspeed time) from just this defined lightspeed ? I've read that sometimes physisists set c=1 and hbar=1, but I don't get how to convert the dependent constants appropriately. Would this help solve my problem at all ? And if so how would this be done exactly ?

Thanks and cheers.
 
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  • #2
I think I figured it out finally. Turned out that system I meant was the Planck unit system. ;)

To compute the deBroglie wavelength of a 1s electron:

Code:
-> c                 = 1
-> h                 = 2pi
-> M (planck-mass)   = 2,1765e−8 kg
-> alpha             = 1/137
-> v                 = Z/n * alpha
-> v                 = 1/1 * 1/137
-> v                 = 1/137
-> m                 = m_e(in kg) / M
-> m                 = 9,10938291e-31 kg / 2,1765e−8 kg
-> m                 = 4,1853355892487939352170916609235e-23
-> lambda            = h / (m * v)
-> lambda            = 2pi / (4,1853355892487939352170916609235e-23 * 1/137)
-> lambda            = 20566962164152375997583578,037296
So this lambda-value can be interpreted as the distance a lightwave travels in 20566962164152375997583578 Planck-time units, right ?

so in meters this becomes
Code:
-> L (planck length) = 1,616199e−35 m
-> lambda_m (in m)   = lambda * L
-> lambda_m (in m)   = 20566962164152375997583578,037296 * 1,616199e−35 m
-> lambda_m (in m)   = 3,32403036827409059349185812403e-10 m
that seems fitting.
 
Last edited:
  • #3
Well now that I have everything in Planck units I still don't see how that gives me result, because I don't know how many Planck time units my timeunit is. Don't I understand something here or is it really impossibe to compute because some necessary constant (t/T or x/L) is unknown ?
 

1. What is De Broglie wavelength?

The De Broglie wavelength is a concept in quantum mechanics that describes the wave-like behavior of particles, such as electrons or protons. It is named after French physicist Louis de Broglie, who proposed that all particles have a wavelength associated with them.

2. How is De Broglie wavelength calculated?

The De Broglie wavelength can be calculated using the equation λ = h/mv, where λ is the wavelength, h is Planck's constant, m is the mass of the particle, and v is the velocity of the particle.

3. Can the De Broglie wavelength be computed by just fixing the speed of light?

Yes, the De Broglie wavelength can be computed by just fixing the speed of light. This is because in the equation λ = h/mv, the speed of light (c) is already included in Planck's constant (h).

4. How is the De Broglie wavelength related to the uncertainty principle?

The De Broglie wavelength is related to the uncertainty principle, which states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. The De Broglie wavelength is used to describe the momentum of a particle, and thus is related to the uncertainty in its position.

5. Why is the De Broglie wavelength important in quantum mechanics?

The De Broglie wavelength is important in quantum mechanics because it helps us understand the wave-particle duality of matter. It also plays a crucial role in the development of the Schrödinger equation, which is used to describe the behavior of quantum particles.

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