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

Answer 1

The correct option is C $\frac{\lambda }{3}\sqrt{\frac{m}{M}}$

De Broglie wavelength $\lambda =\frac{h}{\sqrt{2mK}}$

where $K$ is the kinetic energy.

Kinetic energy $K=eV$

We get $\lambda =\frac{h}{\sqrt{2meV}}$

$⟹$ $\lambda \propto \frac{1}{\sqrt{mV}}$ ........(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 $\frac{{\lambda }_2}{\lambda }=\sqrt{\frac{mV}{M\left(9V\right)}}$

$⟹$ ${\lambda }_2=\frac{\lambda }{3}\sqrt{\frac{m}{M}}$