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

Answer

According to de Broglie’s expression,
$\lambda=\frac{\text{h}}{\text{mv}}$
Substituting the values in the expression,
$\lambda=\frac{6.626\times10^{-34}\text{Js}}{(0.1\text{kg})(4.37\times10^5\text{ms}^{-1})}$
$\lambda=1.516\times10^{-38}\text{m}$

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