# Particle velocity

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1. Feb 2, 2016

### ognik

1. The problem statement, all variables and given/known data
To what velocity would an electron (neutron) have to be slowed down, if its wavelength
is to be I meter? Are matter waves of macroscopic dimensions a real possibility?

2. Relevant equations

I have assumed this could apply to pretty much any free particle of mass m, and is an introductory question only in nature.

3. The attempt at a solution

I have added the units below

I took $p=\frac{h}{\lambda}$, with $\lambda=1m$, so that in general for particles, $v = \frac{h}{m} = \frac{6.626 \times 10^{-34} J.s} {1m \times 9.1 \times 10^{-31}kg} = 7.28 \times 10^{-4} m.s^{-2}$?

For the 2nd part of the question, it seems to me that the velocity is unrealistically slow for a free particle, and would be more so for those with higher mass. I find the 2nd part of the question a little ambiguous, but assuming they are referring to wavelengths of the order of 1m, it would seem from this that waves of macroscopic dimensions are unlikely?

Last edited: Feb 2, 2016
2. Feb 2, 2016

### Staff: Mentor

Well, depends on "macroscopically". Double-slit experiments with electrons, atoms and even small molecules have been performed. Bose-Einstein condensates allow to reach even lower energies, with wavelengths of millimeters.

3. Feb 2, 2016

### ognik

I had read about that, but supposed that for this question they wanted a reply based on the velocity I found; I really can't read much into that velocity myself. I wonderede about quantisation of energy, so used $\Delta E = h \nu = \frac{hv}{\lambda} = 4.8 \times 10^{-37}J = 3 \times10^{-18} eV$ - this looks too low to me, wouldn't there be a minimum energy ?

4. Feb 2, 2016

### Staff: Mentor

There is no minimal energy - unless you constrain the spread of the particle (e.g. to 1 meter).
Be careful with the energy calculation here: for massive particles, the product of frequency and wavelength is not the speed of the particle. You need a different way to calculate it.

5. Feb 2, 2016

### ognik

Thanks mfb, point to remember.

Firstly, is my velocity calculation OK?

In the absence of any known potential, we can use the KE, where $T=\frac{1}{2}mv^2 = 5.03 \times 10^{-38} J = 3.15 \times 10^{-19} eV$ ? Still seems very low?

On a side point - for a bound particle (eg electron in an atom) - then there would be a minimum energy corresponding to the lowest energy level?

6. Feb 2, 2016

### Staff: Mentor

Should be m/s, but apart from that it looks fine.
Well, it is very low.
Sure. On the other hand, the zero is a bit arbitrary for the potential. But you have a minimal kinetic energy.

7. Feb 2, 2016

### ognik

Oops on the m/s.
I was expecting to find the energy of the order of 1eV, if an electron got knocked out of a ground state orbital for example, wouldn't it have at least that amount of energy? I'm wondering if the value I got is reasonable (given that we are using more classical physics)?

8. Feb 3, 2016

### Staff: Mentor

Well, typically electrons have way shorter wavelengths - that's what the calculation is showing.