Question

A proton in a linear accelerator has a de Broglie wavelength of (1.43x10²) pm. What is the speed of the proton (in m/s)?

Answer 1

From de Brogile wavelength, $\lambda=\frac{h}{mv}$

$velocity, v=h/\lambda m$

mass of proton, $m=1.673\times10^{-27}kg$

Planck's constant, $h=6.626\times 10^{-34}m^2kg/s$

$wavelength, \lambda=1.143\times 10^{-7}m$

velocity, $v=\frac{6.626\times 10^{-34}}{1.43\times 10^{-7} \times 1.673\times 10^{-27}}=2.7696m/s$