Answer
$\lambda=4.6\times10^{-34}\,m$
Since wavelength isch small and can be neglected, the wave nature of matter is not significant for bullets.
Work Step by Step
Recall that de Broglie wavelength is $\lambda=\frac{h}{mv}$ where $h$ is Planck's constant, $m$ is the mass and $v$ is the velocity.
Substituting values, we have
$ \lambda=\frac{6.626\times10^{-34}\,J\cdot s}{(1.9\times10^{-3}\,kg)(765\,m/s)}$
$=4.6\times10^{-34}\,m$
Since wavelength is small and can be neglected, the wave nature of matter is not significant for bullets.
