When an electron is accelerated through a particular potential field,it attains a spped if 9.47×10^6 m/s. 1)what is the characteristic wavelength of this electron? 2)Is the wavelength comparable to the size of atoms?

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Answer 1 · 2018-03-31T07:30:45

#7.69times 10^-11" m" # - of the same order as the typical size of atoms.

The characteristic de Broglie wavelength is

#lambda = h/(mv) = {6.63 times 10^(-34)" Kg"\ "m"^2"s"^-1}/{(9.11times 10^{-31}" Kg" )times (9.47times10^6" m/s")} = 7.69times 10^-11" m" #

Note that we have used the non-relativistic formula "mv" for the momentum because the speed is quite low compared to the speed of light.

This is of the same order as the size of atoms. (The first Bohr radius for Hydrogen is #5.29times 10^-11" m"#)