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Find the de Broglie wavelength λ for an electron moving at a speed of 1.00 x 10^6m/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 is me=9.11 x 10^−31kg, and Planck's constant is h=6.626 x 10^−34J*s. Express your answer in meters to three significant figures.

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Answer:

De broglie wavelength will be 0.727 nm

Explanation:

We have given speed of the electron [tex]v=10^6m/sec[/tex]

Mass of the electron [tex]m=9.11\times 10^{-31}kg[/tex]

Plank's constant [tex]h=6.624\times 10^{-34}Js[/tex]

We have to find the De Broglie wavelength

De Broglie wavelength is given by [tex]\lambda =\frac{h}{mv}=\frac{6.624\times 10^{-34}}{9.11\times 10^{-31}\times 10^6}=0.727\times 10^{-9}m[/tex]

So wavelength will be 0.727 nm

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