Question

A proton and an electron are accelerated by the same potential difference. Let ${\lambda }_e$ and ${\lambda }_p$ denote the de Broglie wavelength of the electron and the proton respectively, then

• A. ${\lambda }_e={\lambda }_p$
• B. ${\lambda }_e<{\lambda }_p$
• C. ${\lambda }_e>{\lambda }_p$
• D. ${\lambda }_e\ge {\lambda }_p$

Answer 1

The correct option is C ${\lambda }_e>{\lambda }_p$

$\frac{1}{2}m{v}^2=eV$
$mv=\sqrt{2meV}$
So de Broglie wavelength
$\lambda =\frac{h}{mv}$
$\lambda =\frac{h}{\sqrt{2meV}}$
So ${\lambda }_e=\frac{h}{\sqrt{2m_{e}eV}}$
${\lambda }_p=\frac{h}{\sqrt{2m_{p}eV}}$
So, ${\lambda }_e>{\lambda }_p$