Question

The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × ${10}^{5}m{s}^{–1}$. If the hockey ball of mass 0.1 Kg is moving with this velocity, calculate the wavelength associated with this velocity.

Answer 1

Step 1:

According to de Broglie’s expression,

$\lambda =\frac{h}{mv}$

Step 2:

The associated wavelength is obtained from de-Broglie equation.

Substituting the values in the expression,
$\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-34}}{0.1\times 4.37\times {10}^5}=1.516\times {10}^{-28}m$

The wavelength associated with this velocity is $1.516\times {10}^{-28}m$