Question

The accelerating potential (in volt) that must be imparted to a proton beam to give it an effective wavelength of $0.005nm$ is________ .

Answer 1

Answer: 32.85

$u=\frac{h}{m\lambda }$
$\therefore$ mass of proton $=1.67\times 10^{-27}kg$
$u=\frac{6.625\times 10^{-34}}{1.67\times 10^{-27}\times 0.005\times 10^{-9}}$
$=7.94\times 10^4$ metre ${\sec }^{-1}$
Now accelerating potential is $V$, then velocity $(u)$ acquired by the charge particle having charge $Q$ and mass $m$.
$\because Q.V=\frac{1}{2}m{u}^2$
$u=\sqrt{\left(\frac{2QV}{m}\right)}=\sqrt{\left(\frac{2\times 1.602\times 10^{-9}\times V}{1.67\times 10^{-27}}\right)}$
or $7.94\times 10^4=\sqrt{\left(\frac{2\times 1.602\times 10^{-19}\times V}{1.67\times 10^{-27}}\right)}$
$V=32.85$ volt