If the velocity of the electron in Bohr's first orbit is `2.19xx10^(6) m s^(-1)`, calculate the de Broglie wavelength associated with it.

Explanation

`lamda = (h)/(mv) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/((9.11 xx 10^(-31) kg) (2.19 xx 10^(6) ms^(-1))) = 3.32 xx 10^(-10) m = 332` pm