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

Answer

`6.626xx10^(-24)"kg "m s^(-1)`

Explanation

According to de Broglie equation, `lambda=(h)/(m v)=(h)/(p) or p=(h)/(lambda)`
`p=((6.626xx10^(-34) "kg" m^(2)s^(-1)))/((0.1xx10^(-9)m))=6.626xx10^(-24) "kg"m s^(-1)`.