Calculate the momentum of a particle which has de Broglie wavelength of 0.1 nm.

Answer

`6.63 xx 10^(-24) kg ms^(-1)`

Explanation

`mv = h/lambda`
`= (6.63 xx 10^(-34) kg m^2 s^(-1))/(0.1 xx 10^(-9) m)`
`= 6.63 xx 10^(-24) kg m s^(-1)`.