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In 1924, Louis de Broglie postulated that particles such as electrons and protons might exhibit wavelike properties. His thinking was guided by the notion that light has both wave and particle characteristics, so he postulated that particles such as electrons and protons would obey the same wavelength-momentum relation as that obeyed by light: =h/p, where is the wavelength, p the momentum, and hPlanck's constant.

A: Find the de Broglie wavelength for an electron moving at a speed of 1.00×106m/s. (Note that this speed is low enough that the classical momentum formula p=mv is still valid.) Recall that the mass of an electron isme=9.11×1031kg, and Planck's constant is h=6.626×1034Js. answer= 7.270×1010

B: Find the de Broglie wavelength of a baseball pitched at a speed of 43.3 m/s . Assume that the mass of the baseball is 0.143kg. answer= 1.07×1034

C: Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00m. The electrons then head toward an array of detectors a distance 1.051 m away. These detectors indicate a diffraction pattern, with a broad maximum of electron intensity (i.e., the number of electrons received in a certain area over a certain period of time) with minima of electron intensity on either side, spaced 0.522 cm from the center of the pattern. What is the wavelength of one of the electrons in this beam? Recall that the location of the first intensity minima in a single slit diffraction pattern for light is y=L/a, where L is the distance to the screen (detector) and a is the width of the slit. The derivation of this formula was based entirely upon the wave nature of light, so by de Broglie's hypothesis it will also apply to the case of electron waves. answer is 9.933X10^-9

D: What is the momentum p of one of these electrons?

Explanation / Answer

A.
p =mv = h/lambda
(9.11*10^-31*1*10^6) = (6.626*10^-34 / lambda)

lambda = 7.27 * 10^-10 m

B.
p =mv = h/lambda
(9.11*10^-31*0.143*10^-3) = (6.626*10^-34 / lambda)

lambda = 1.07 * 10^-34 m

C.
deBorglie relationship is:

l = h/p where l =wavelength and p = momentum

You are given xmin = L l/a and the values for xmin, L, and a

Solve for l

l = a*xmin/L = (2e-6*0.522x10^-2)/1.051 = 9.933 * 10^-9 m

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