The velocity associated with a proton moving in a potential difference of $1000 \, V$ is $4.37 \times 10^{5} \, ms^{-1}$. If a hockey ball of mass $0.1 \, kg$ is moving with this velocity,calculate the wavelength associated with this velocity.

According to de Broglie's expression,$\lambda = \frac{h}{mv}$.
Substituting the given values:
$h = 6.626 \times 10^{-34} \, J \cdot s$
$m = 0.1 \, kg$
$v = 4.37 \times 10^{5} \, ms^{-1}$
$\lambda = \frac{6.626 \times 10^{-34} \, J \cdot s}{(0.1 \, kg)(4.37 \times 10^{5} \, ms^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34}}{4.37 \times 10^{4}} \, m$
$\lambda = 1.516 \times 10^{-38} \, m$