A photon and an electron each hae an energy of 6.0×10^3 eV. To calculate their respective wavelengths. Which of these could be used to probe atomic structures?
1
Expert's answer
2019-12-27T08:57:09-0500
The concept that matter behaves like a wave was proposed by Louis de Broglie in 1924. The de Broglie wavelength is
(1) λ=ph​ ,
where p - the momentum of particle, and h is the Planck constant. Wave-like behavior of matter was first experimentally demonstrated by George Paget Thomson's thin metal diffraction experiment, and independently in the Davisson–Germer experiment both using electrons, and it has also been confirmed for other elementary particles, neutral atoms and even molecules [1]. The rest energy of an electron Ee​=0.511MeV, and especially a proton Ep​=0.938GeV , is much larger than the kinetic energy Ek​=6.0KeV specified in the problem; therefore, to estimate the wavelength, one can use the formulas of nonrelativistic mechanics.
Ek​=2mv2​ ; p=mv
We can rewrite these as
Ek​=2mp2​ ; and p=2mEk​​
Recalling Einstein's formula for the energy of rest E=mc2 of a body with mass m, we find
In the calculations we used the value hc=1.24eV⋅μm [2]
Answer: λe​=0.16⋅10−10m ; λp​=0.36⋅10−12m. We can see that for a proton the wavelength is much shorter than that of electron, thus the proton is more usefull to probe atomic structures.
"assignmentexpert.com" is professional group of people in Math subjects! They did assignments in very high level of mathematical modelling in the best quality. Thanks a lot
Comments