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 $9 V$, 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 $E=e V \ldots \ldots$ (i) $E=\frac{1}{2} m v_{0}^{2}=\frac{p^{2}}{2 m}=\frac{h^{2}}{2 m \cdot \lambda^{2}} \ldots \ldots$ (ii) $\therefore e V=\frac{h^{2}}{2 m \cdot \lambda^{2}} \Rightarrow \lambda=\frac{h}{\sqrt{2 m e V}} \ldots \ldots$ (iii) When proton of mas $M$ is accelerated through a potential difference of $9 V$, then the de - Broglie wavelength obtained is $$ \lambda^{\prime}=\frac{h}{\sqrt{2 M e(9 V)}}=\frac{h}{3 \times \sqrt{2 M e V}} \times \frac{\sqrt{m}}{\sqrt{m}} $$ $\therefore \lambda^{\prime}=\frac{\lambda}{3} \cdot \sqrt{\frac{m}{M}}$