Question

An electron diffraction experiment was performed with a beam of electrons accelerated by a potential difference of 10.0 K eV. The wave length of the electron beam in $A^o$ is:

Answer 1

$K.E=\frac{1}{2}m{v}^2$

$v=(\frac{2\times KE}{m}{)}^{\frac{1}{2}}$

$10.0keV=(10\times {10}^{3}eV)(1.602\times {10}^{-19}Je{V}^{-1})=1.602\times {10}^{-15}J=1.602\times {10}^{-15}kg{m}^2{s}^{-2}$

$\therefore v=(\frac{2\times 1.602\times {10}^{-15}Kg{m}^2{s}^{-2}}{9.1\times {10}^{-31}Kg}{)}^{\frac{1}{2}}=(35.17\times {10}^{-14}{)}^{\frac{1}{2}}=5.93\times {10}^{7}m{s}^{-1}$

$\lambda =\frac{h}{mv}=\frac{6.63\times {10}^{-34}Js}{(9.1\times {10}^{-31}kg)(5.93\times {10}^{7}m{s}^{-1})}=\frac{1.23\times {10}^{-8}kg{m}^2{s}^{-1}}{kgm{s}^{-1}}=123\times {10}^{-10}=123A^o$