Question

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4\, V$. 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 $\lambda _{p}=\frac{h}{\sqrt{2m_{e}4eV}}\left(\because q_{e}=q_{p}\right)$
$\therefore \frac{\lambda_{e}}{\lambda_{p}}=\frac{\frac{h}{\sqrt{2m_{e}eV}}}{\frac{h}{\sqrt{2m_{e}4eV}}}=\sqrt{\frac{2m_{e}4eV}{2m_{e}eV}}$
$=2\sqrt{\frac{m_{p}}{m_{e}}}$