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)

De Broglie wavelength $\lambda=\frac{h}{\sqrt{2 m K}}$
where $K$ is the kinetic energy.
Kinetic energy $K = eV$
We get $\lambda=\frac{ h }{\sqrt{2 meV }}$
$\Rightarrow \lambda \propto \frac{1}{\sqrt{ mV }}$....(1)
Given : $V _{1}= V \,\,\, m _{1}= m \,\,\, m _{2}= M \,\,\, V _{2}=9 V$
From equation (1), we get
$\frac{\lambda_{2}}{\lambda_{1}}=\sqrt{\frac{ m _{1} V _{1}}{ m _{2} V _{2}}}$
Or $\frac{\lambda_{2}}{\lambda}=\sqrt{\frac{ mV }{ M (9 V )}}$
$\Rightarrow \lambda_{2}=\frac{\lambda}{3} \sqrt{\frac{ m }{ M }}$