Calculate the momentum of a particle which has a de Broglie wavelength of `2 Å,(h = 6.6 xx 10^(-34) kg m^(2) s^(-1))`

Answer

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

Explanation

`lamda = (h)/(p) or p = (h)/(lamda) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/(10^(-10)m) = 6.6 xx 10^(-24) kg m s^(-1) (1 Å = 10^(-10)m, 1nm = 10^(-9) m)`