Question

The waves associated with matter is called matter waves. Let ${\lambda }_e$ and ${\lambda }_p$ be the de-Broglie wavelengths associated with electron and proton respectively. If they are accelerated by same potential, then

• A. ${\lambda }_e>{\lambda }_p$
• B. ${\lambda }_p>{\lambda }_e$
• C. ${\lambda }_p={\lambda }_e$
• D. ${\lambda }_e=\frac{1}{{\lambda }_p}$

Answer 1

Answer: (A)

${\lambda }_e>{\lambda }_p$

The de-Broglie wavelength of a charged particle ($e$) of mass $m$ , accelerated through potential $V$ is given by

$\lambda =h/\sqrt{2meV}$

For electron ,

${\lambda }_e=h/\sqrt{2m_e{e}_{e}V}$ ............................eq1

For proton ,

${\lambda }_p=h/\sqrt{2m_p{e}_{p}V}$ ...........................eq2

Dividing eq1 by eq2 ,

$\frac{{\lambda }_e}{{\lambda }_p}=\frac{\sqrt{m_p{e}_p}}{\sqrt{m_e{e}_e}}$

Now , $m_e=m_p$

Hence , $\frac{{\lambda }_e}{{\lambda }_p}=\frac{\sqrt{m_p}}{\sqrt{m_e}}$

As $m_p>m_e$

therefore ${\lambda }_e>{\lambda }_p$