Question

Electron and proton are accelerated through one volt of potential difference, what will be the ratio of their wavelengths?

• A. $\frac{m_p}{m_e}$
• B. $\frac{m_p{v}_e}{m_e{v}_p}$
• C. $\frac{m_p^2}{m_e^2}$
• D. $\sqrt{\frac{m_p}{m_e}}$

Answer 1

The correct option is D $\sqrt{\frac{m_p}{m_e}}$

As we know,
$\frac{{\lambda }_e}{{\lambda }_p}=\frac{\frac{h}{m_e{v}_e}}{\frac{h}{m_p{v}_p}}=\frac{m_p{v}_p}{m_e{v}_e}$
$\therefore$ Kinetic energy of electron = kinetic energy of proton
$\therefore \frac{1}{2}{m}_e{v}_e^2=\frac{1}{2}{m}_p{v}_p^2$
$\frac{v_p^2}{v_e^2}=\frac{m_e}{m_p}\therefore \frac{v_p}{v_e}=\sqrt{\frac{m_e}{m_p}}\therefore \frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}$