Question

An electron of mass has de-Broglie wavelength when accelerated through potential difference. When proton of mass, is accelerated through potential difference 9 , the de-Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage)

• A. $\frac{\lambda }{3}\sqrt{\frac{M}{m}}$
• B. $\frac{\lambda }{3}\cdot \frac{M}{m}$
• C. $\frac{\lambda }{3}\sqrt{\frac{m}{M}}$
• D. $\frac{\lambda }{3}\cdot \frac{m}{M}$

Answer 1

Answer: (C)

When electron or any charged particle is accelerated through potential difference V, then kinetic energy gained is given by $E$
$=eV\dots \dots$ (i)
$E=\frac{1}{2}m{v}_0^2=\frac{p^2}{2m}=\frac{h^2}{2m\cdot {\lambda }^2}\dots \dots \text{ (ii) }$
$\therefore eV=\frac{h^2}{2m\cdot {\lambda }^2}\Rightarrow \lambda =\frac{h}{\sqrt{2meV}}\dots \dots (iii)$
When proton of $\mathrm{mas}M$ is accelerated through a potential difference of $9V$, then the de - Broglie wavelength obtained is
${\lambda }^{\prime }=\frac{h}{\sqrt{2Me9V}}=\frac{h}{3\times \sqrt{2MeV}}\times \frac{\sqrt{m}}{\sqrt{m}}$
$\therefore {\lambda }^{\prime }=\frac{\lambda }{3}\cdot \sqrt{\frac{m}{M}}$