Calculate de Broglie wavelength of an electron travelling at `1 %` of the speed of light.

Explanation

de-Broglie wavelength can be calculate as,
`lambda = (h)/(mv)`......(1)
where, `h = 6.626 xx 10^(-34)Js`
`m = 9.1 xx 10^(-31)kg`
`v = (1)/(100) xx 3xx 10^(8) = 3 xx 10^(6)m//sec`
substituting these values in eq(1), we get
`lambda = (6.626 xx 10^(-34))/(9.1 xx 10^(-31)xx 3 xx 10^(6))`
`= 242.7 xx 10^(-12)m = 242.7 pm`