Question

A proton and an electron are accelerated by the same potential difference. Let ${\lambda }_e$ and ${\lambda }_p$ denote the de Broglie wavelengths of the electron and the proton respectively.

• A. ${\lambda }_e={\lambda }_p$
• B. ${\lambda }_e<{\lambda }_p$
• C. ${\lambda }_e>{\lambda }_p$
• D. ${\lambda }_e$ and ${\lambda }_p$ depend on the accelerating potential difference.

Answer 1

Answer: (C), (D)

The correct options are
C ${\lambda }_e>{\lambda }_p$
D ${\lambda }_e$ and ${\lambda }_p$ depend on the accelerating potential difference.

Charge present on both panticle are same
$\because \lambda =\frac{h}{mqV}$ so $q_1=q_2$
$\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_{p}v}{m_{e}v}}=>1$
$\because m_p>>m_e$
$\therefore {\lambda }_e>{\lambda }_p$.