asenath6477

09.09.2019 • 

Physics

Consider a beam of electrons in a vacuum,passing through a very narrow slit of width 2.00 \; \mu{\rm m} . the electrons then head toward an array ofdetectors a distance 0.9750 m away. these detectors indicate a diffraction pattern,with a broad maximum of electron intensity (i.e., the number ofelectrons received in a certain area over a certain period of time)with minima of electron intensity on either side, spaced0.493 cm from the center of the pattern. what is the wavelengthlambda of one of the electrons in this beam? recall that thelocation of the first intensity minima in a single slit diffractionpattern for light is y=l \lambda /a , where lis the distance to the screen (detector) and a is the width of the slit. the derivation of this formulawas based entirely upon the wave nature of light, so by debroglie's hypothesis it will also apply to the case of electronwaves.
express your answer in meters tothree significant figures.

Ответ:

anglebaby559p0nxvi

11.09.2021

Physics

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1.5 m/s2

Explanation:

U = 0 m/s ( from rest)

t = 20 s

S = 300 m

S = ut + 1/2 * a * t2

300 = 0 + 1/2 * a * 20*20

a = 1.5m /s2

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