Answer
$$v = 3960m/s $$
$$V = 8.19\times 10^{-2}V$$
Work Step by Step
According to De Broglie, wavelength of particle moving at a velocity v is given by
$$\lambda = \frac{h}{mv}$$
where m is the mass of the particle.
$$\therefore v = \frac{h}{m_{proton}\lambda}$$
$$v = \frac{6.626\times 10^{-34}}{1.673\times 10^{-27}
\times 100\times 10^{-12} } = 3960m/s $$
Say that to acquire such velocity, the proton is passed through a potential V.
Energy provided by Potential V = K.E of the proton
$$eV = \frac{1}{2}m_{proton}v^2$$
$$V = \frac{1}{2e}m_{proton}v^2$$
$$V = \frac{1}{2\times 1.602\times 10^{-19}}(1.673\times 10^{-27})(3960)^2$$
$$V = 0.0819V = 8.19\times 10^{-2}V$$
