Question

A proton and electron are accelerated by same potential difference starting from the rest have de-Broglie wavelength ${\lambda }_p$ and ${\lambda }_e$.

• A. ${\lambda }_e={\lambda }_p$
• B. ${\lambda }_e<{\lambda }_p$
• C. ${\lambda }_e>{\lambda }_p$
• D. none of these

Answer 1

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

Let $m_e$ and $m_p$ be the mass of electron and proton respectively. Let the applied potential different be $V$.

Thus, the de-Broglie wavelength of the $e^{-}$,

${\lambda }_e=\frac{h}{\sqrt{2m_{e}eV}}⟶\left(1\right)$

and de - Broglie wavelength of proton

${\lambda }_p=\frac{h}{\sqrt{2m_{p}eV}}⟶\left(2\right)$

Dividing,

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{\sqrt{m_e}}{\sqrt{m_p}}$

$m_e<m_p$

$\therefore$ $\frac{{\lambda }_p}{{\lambda }_e}<1$

$\Rightarrow {\lambda }_p{\lambda }_e$