Answer
$3.97\times 10^{6}$ $m/s$
Work Step by Step
The de Broglie wavelength $(\lambda)$ is express by the formula,
$\lambda=\frac{h}{p}$ ......................$(1)$
where $h$ is the planck's constant and $p$ is the momentum of the moving particle.
The relation between momentum $p$ and speed $v$ is expressed by
$p=mv$
where $m$ is the mass of a proton.
Now, eq. $1$ can be written as
$\lambda=\frac{h}{mv}$
or, $v=\frac{h}{m\lambda}$
The de Broglie wavelength ($\lambda$) of the proton is given: $\lambda=100$ $fm$
Thus, the speed of the proton is given by
$v=\frac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times100\times 10^{-15}}$ $m/s$
or, $v=3.97\times 10^{6}$ $m/s$
