Question

An electron diffraction experiment was performed with a beam of electrons accelerated by a potential difference of 10.0 K eV.If the wave length of the electron beam in angstorm is $123\times {10}^{-x}$, then what is the value of x?

Answer 1

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

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

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

$\lambda$ = $\frac{h}{mv}$ = $\frac{6.63\times {10}^{-34}Js}{(9.1\times {10}^{-31}kg)(5.93\times {10}^{7}m{s}^{-1})}$= $\frac{1.23\times {10}^{-11}kg{m}^2{s}^{-1}}{kgm{s}^{-1}}=1.23\times {10}^{-11}=0.123A$