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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

Step-by-step explanation:

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

Mass of the electron
m=9.11* 10^(-31)kg

Plank's constant
h=6.624* 10^(-34)Js

We have to find the De Broglie wavelength

De Broglie wavelength is given by
\lambda =(h)/(mv)=(6.624* 10^(-34))/(9.11* 10^(-31)* 10^6)=0.727* 10^(-9)m

So wavelength will be 0.727 nm

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