Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope, often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is $1.6 \times 10^{6} \,ms^{-1}$, calculate the de Broglie wavelength associated with this electron.

According to de Broglie's equation, $\lambda = \frac{h}{mv}$.
Given:
$h = 6.626 \times 10^{-34} \,Js$
$m = 9.109 \times 10^{-31} \,kg$
$v = 1.6 \times 10^{6} \,ms^{-1}$
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34} \,Js}{(9.109 \times 10^{-31} \,kg)(1.6 \times 10^{6} \,ms^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34}}{1.4574 \times 10^{-24}} \,m$
$\lambda = 4.546 \times 10^{-10} \,m \approx 4.55 \times 10^{-10} \,m$
Converting to picometers:
$\lambda = 455 \,pm$.