Calculate the momentum of a particle which has a de Broglie wavelength of `10 nm`

Answer

A, B

Explanation

`1 J = 1 kg m^(2) s^(2) = 10^(7)erg`
`1 eV = 1.60 xx 10^(-19) J`
`p = (h)/(lambda) = 6.62 xx 10^(-34)( js)10 xx 10^(-9)(m)`
`= 6.626 xx 10^(-26) J s m^(-1)`
`= 6.626 xx 10^(-26) kg m s^(-1)`