Question

Calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005nm.

Answer 1

For proton, $u=\frac{h}{m\lambda }$

$\therefore u=\frac{6.626\times {10}^{-34}}{1.672\times {10}^{-27}\times 0.005\times {10}^{-9}}=7.93\times {10}^{4}m/s$

If accelerating potential is 'V' , then velocity acquired by the charged particle having charge Q and mass m,
$\frac{1}{2}m{u}^2=QV$
$Q=1.6\times {10}^{-19}C$

$m=1.672\times {10}^{-27}kg$

$u=7.93\times {10}^{4}m/s$

By putting all the values we get,
$V=32.82$ Volt