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 $\lambda =\frac{h}{\sqrt{2mK}}$
where $K$ is the kinetic energy.
Kinetic energy $K=eV$
We get $\lambda =\frac{h}{\sqrt{2meV}}$
$\Rightarrow \lambda \propto \frac{1}{\sqrt{mV}}$....(1)
Given : $V_1=V{m}_1=m{m}_2=M{V}_2=9V$
From equation (1), we get
$\frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{m_1{V}_1}{m_2{V}_2}}$
Or $\frac{{\lambda }_2}{\lambda }=\sqrt{\frac{mV}{M(9V)}}$
$\Rightarrow {\lambda }_2=\frac{\lambda }{3}\sqrt{\frac{m}{M}}$