Question

Through how many volts of potential difference must an electron be accelerated to achieve a wavelength of 0.27 nm?

Answer 1

if V (volts) is the potential diff by which "e" was accelerated, gain in its kinetic energy

$KE=\frac{mv^{2}}{2}=eV$
$v=\sqrt{\frac{2eV}{m}}$

$mv=\sqrt{2eVm}$
  
wavelength associated with electron

$\lambda =\frac{h}{mv}$

$\lambda =\frac{h}{\sqrt{2eVm}}$
$\sqrt{2eVm}=\frac{h}{\lambda }$

$V=\frac{(\frac{h}{\lambda })^{2}}{2em}$
$V=\frac{(\frac{6.6\cdot 10^{-34}}{0.27\cdot 10^{-9} })^{2}}{2\cdot 9.1\cdot 10^{-31}\cdot 1.6\cdot 10^{-19}}$

$V=20.51\, V$