Question

by doing a nuclear diffraction experiment, you measure the de broglie wavelength of a proton to be 8.29 fm. (a) what is the speed of the proton? (b) through what potential difference must it be accelerated to achieve that speed?

Answer 1

By the de Broglie wavelength equation, the speed of proton is 0.47 × $10^{8}$ m/s.

λ = h / mv

λ = wavelength, given = 8.29 fm

h = plank's constant, given = 6.626 × $10^{-34}$

m = mass, given = 1.672 × $10^{-27}$ kg

v = velocity

Now, v = h / mλ

Put the values in formula,  v = h / mλ

v = 0.47 × $10^{8}$ m/s

Therefore, the speed of proton is 0.47 × $10^{8}$ m/s.

Learn more about de Broglie wavelength here:- brainly.com/question/15330461

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