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

Velocity of proton is calculated as:

$\begin{array}{ccc}\mathrm{v}& =& \frac{\mathrm{h}}{\mathrm{m\lambda }}\\ & =& \frac{6.626\times {10}^{-34}}{1.627\times {10}^{-27}\times {10}^{-9}}\\ & =& 7.93\times {10}^4 \mathrm{meter} {\sec }^{-1} ...\left(\mathrm{I}\right)\end{array}$

Relation between Q (charge), Mass (m), accelerating potential (V), and velocity (v).

$\begin{array}{ccc}\frac{1}{2}{\mathrm{mv}}^2& =& \mathrm{Q}.\mathrm{V}\\ \mathrm{v}& =& \sqrt{\frac{2.\mathrm{Q}.\mathrm{V}}{\mathrm{m}}}\\ & =& \sqrt{\frac{2\times 1.602\times {10}^{-19}}{1.627\times {10}^{-27}}\times \mathrm{V}} ...\left(\mathrm{II}\right)\end{array}$

Equate (I) and (II)

$\begin{array}{ccc}\left(7.93\times {10}^4 \mathrm{meter} {\sec }^{-1}\right)& =& \frac{2\times 1.602\times {10}^{-19}}{1.627\times {10}^{-27}}\times \mathrm{V}\\ \mathrm{V}& =& 32.82 \mathrm{volt}\end{array}$