Question

An electron of mass $m$ has de-Broglie wavelength $\lambda$ when accelerated through potential difference $V$. When proton of mass $M$, is accelerated through potential difference $9V$, 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) $E=\frac{1}{2}m{v}_0^2=\frac{p^2}{2m}=\frac{h^2}{2m\cdot {\lambda }^2}\dots \dots$ (ii) $\therefore eV=\frac{h^2}{2m\cdot {\lambda }^2}\Rightarrow \lambda =\frac{h}{\sqrt{2meV}}\dots \dots$ (iii) When proton of mas $M$ is accelerated through a potential difference of $9V$, then the de - Broglie wavelength obtained is \lambda^{\prime}=\frac{h}{\sqrt{2 M e(9 V)}}=\frac{h}{3 \times \sqrt{2 M e V}} \times \frac{\sqrt{m}}{\sqrt{m}} $\therefore {\lambda }^{\prime }=\frac{\lambda }{3}\cdot \sqrt{\frac{m}{M}}$