Question

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

Answer 1

The associated wavelength is obtained from de-Broglie equation.

Here

$h=6.626\times {10}^{-34}Kg{m}^2{s}^{-1}$

$m=0.1kgv=4.37\times {10}^{5}m{s}^{-1}$

Substituting the above values in de broglie equation , we get

$\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-34}}{0.1\times 4.37\times {10}^5}=1.516\times {10}^{-28}m$