We are asked to determine DeBroglie's wavelength of a bullet. Let's remember that DeBroglie's wavelength is the ratio of Plank's constant and the momentum of the particle, therefore, we have the following formula:
[tex]\lambda=\frac{h}{p}[/tex]Where:
[tex]\begin{gathered} \lambda=\text{ wavelength }\lbrack m\rbrack \\ h=\text{ Plank's constant }\lbrack Js\rbrack \\ p=\text{ momentum }\lbrack kg\frac{m}{s}\rbrack \end{gathered}[/tex]The momentum of the particle is the product of the mass and the velocity, therefore, we have:
[tex]p=mv[/tex]Substituting in the formula for the wavelength:
[tex]\lambda=\frac{h}{mv}[/tex]Now, we substitute the values:
[tex]\lambda=\frac{6.63\times10^{-34}J\cdot s}{(0.075\operatorname{kg})(350\frac{m}{s})}[/tex]Now we solve the operations:
[tex]\lambda=2.5\times10^{-35}m[/tex]Therefore, DeBroglie's wavelength is 2.5 x 10 ^-35 meters.
