Question

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4V$. The de-Broglie wavelengths are ${\lambda }_e$ and ${\lambda }_p$ for electrons and protons respectively. The ratio of $\frac{{\lambda }_e}{{\lambda }_p}$ is given by : (given me is mass of electron an($m_p$ is mass of proton).

• A. $\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}$
• B. $\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_e}{m_p}}$
• C. $\frac{{\lambda }_e}{{\lambda }_p}=\frac{1}{2}\sqrt{\frac{m_e}{m_p}}$
• D. $\frac{{\lambda }_e}{{\lambda }_p}=2\sqrt{\frac{m_p}{m_e}}$

Answer 1

Answer: (D)

Energy in joule (E)
= charge x potential diff. in volt
$E_{lectron}=q_{e}V$ and $E_{roton}=q_{p}4V$
de-Broglie wavelength \lambda $=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$
${\lambda }_e=\frac{h}{\sqrt{2m_{e}eV}}$ and ${\lambda }_p=\frac{h}{\sqrt{2m_{e}4eV}}\left(\because q_e=q_p\right)$
$\therefore \frac{{\lambda }_e}{{\lambda }_p}=\frac{\frac{h}{\sqrt{2m_{e}eV}}}{\frac{h}{\sqrt{2m_{e}4eV}}}=\sqrt{\frac{2m_{e}4eV}{2m_{e}eV}}$
$=2\sqrt{\frac{m_p}{m_e}}$