Why doesn't calculating the de Broglie wavelength work with h in eV·s? I'm trying to calculate the

lasquiyas5loaa

lasquiyas5loaa

Answered question

2022-05-20

Why doesn't calculating the de Broglie wavelength work with h in eV·s?
I'm trying to calculate the de Broglie wavelength of a particle with a known momentum p = 980.93 G e V / s . The de Broglie relation is:
λ = h p
but I notice that if I use the value of 4.136 × 10 15 e V s for the Planck's constant, the answer that I get from the equation is in units of s 2 , not m. Specifically:
λ = 4.136 × 10 15 e V s 980.93 G e V / s = 4.25 × 10 27 s 2
which is not at all equivalent to the value that one gets when using h = 6.626 × 10 34 J s instead (since J s can be converted into k g m 2 s 2 s ).
λ = 6.626 × 10 34 k g m 2 s 2 s 5.237 × 10 16 k g m s = 1.27 × 10 18 m
What's going on here?

Answer & Explanation

cegielnikmzjkf

cegielnikmzjkf

Beginner2022-05-21Added 14 answers

You're running into trouble because in order to give momentum units of energy, you're setting the speed of light equal to 1, c = 1. If you keep the units of c the momentum should be given in units of eV / c. By dimensional analysis you can check for yourself that eV/s does not have units of momentum (kg m/s).
Therefore, in your case the momentum is actually given by p = 980.93   GeV / c which yields
λ = 4.136 × 10 15 eV s 980.93 GeV / c = 4.136 × 10 15 eV s 2.99 × 10 8 m/s 980.93 GeV = 1.27 × 10 18 m .
motorinum6fh9v

motorinum6fh9v

Beginner2022-05-22Added 3 answers

The momentum P is not homogeneous to eV/s but to eV.s/m which one can remember using the equation :
P 0 = E c
Where P 0 is the time component of the quadri-momentum, E the energy of the particle and c the speed of light.

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