Question

The kinetic energy of electron and proton is ${10}^{-32}J$ . Then the relation between their de-Broglie wavelengths is

• A. ${\lambda }_p$ < ${\lambda }_e$
• B. ${\lambda }_p$ > ${\lambda }_e$
• C. ${\lambda }_p$ = ${\lambda }_e$
• D. ${\lambda }_p$ < $2{\lambda }_e$

Answer 1

The correct option is A ${\lambda }_p$ < ${\lambda }_e$

By using $\lambda =\frac{h}{\sqrt{2mE}}$. Hence $\lambda \propto \frac{1}{\sqrt{m}}$
Since $m_p$ > $m_e$ so ${\lambda }_p$ < ${\lambda }_e$.