Question

The de-Broglie wavelength of an electron, accelerated by potential V is $\lambda$. What will be the de-Broglie wavelength of the electron which is accelerated by 4V potential?

Answer 1

The de-Broglie wavelength associated with moving electron is

$\lambda =\frac{h}{mv}=\frac{h}{m\sqrt{\frac{2eV}{m}}}=\frac{h}{\sqrt{2meV}}$.........(i)

Now, V' = 4V

${\lambda }^{\prime }=\frac{h}{\sqrt{2me\left(4V\right)}}$

${\lambda }^{\prime }=\frac{h}{2\sqrt{2meV}}=\frac{\lambda }{2}$.........From (i).