Question

The accelerating potential that must be imparted to a proton beam to given it an effective wavelength of 0.05 nm is
(At. of proton = 1.008 g)

• A. 0.325 V
• B. 5.205 V
• C. 52.05 V
• D. 3.25 V

Answer 1

The correct option is A 0.325 V

Wavelength of proton, $\lambda =0.05\times {10}^{-9}m$

Mass of proton $m=1.67\times {10}^{-27}kg$

Charge on proton, $q=1.6\times {10}^{-19}C$

Using, $p=mv=\frac{h}{\lambda }$

$v=\frac{6.6\times {10}^{-34}}{1.67\times {10}^{-27}\times 0.05\times {10}^{-9}}=7.904\times {10}^{3}m/s$

Now potential energy due to accelerating potential will provide kinetic energy to proton,

$qV=\frac{1}{2}m{v}^2$

$V=\frac{1.67\times {10}^{-27}\times (7.904\times {10}^3{)}^2}{2\times 1.6\times {10}^{-19}}$

$V=0.326$ $V$