The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1.
Question:

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

Solution:

According to de Broglie’s expression,

$\lambda=\frac{\mathrm{h}}{m v}$

Substituting the values in the expression,

$\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{(0.1 \mathrm{~kg})\left(4.37 \times 10^{5} \mathrm{~ms}^{-1}\right)}$

$\lambda=1.516 \times 10^{-38} \mathrm{~m}$