Calculate the deBroglie wavelength of an electron travilling at `1%` of the speed of the light

Explanation

The mass of electron is `9.11 xx 10^(-31) kg ` One percent of the speed of light is
`s = (1.100)(3.00 xx 10^(8) m s^(-1))`
`= 3.00 xx 10^(6) m s^(-1)`
The momentum of the electron is given by
`p = mv = (9.11 xx 10^(-31) kg) (3.00 xx 10^(6) m s^(-1))`
` = 2.733 xx 10^(-24) kg m s^(-1)`
The de Broglie wavelength of this electron is
`lambda = (h)/(p) = (6.626 xx 10^(-34) Js)/(2.733 xx 10^(-24) kg m s^(-1))`
`= 2.424 xx 10^(-10) m = 242 .4 "pm" `
The wavelength is of atomic dimenssion