Question

The energies of a proton and an $\alpha$ particle are equal. Then the ratio of their de-Broglie wavelengths is

• A. 1 : 2
• B. 2 : 1
• C. 1 : 4
• D. 4 : 1

Answer 1

The correct option is B 2 : 1

By using $\lambda =\frac{h}{\sqrt{2mE}}\Rightarrow \lambda \propto \frac{1}{\sqrt{m}}\Rightarrow \frac{{\lambda }_{proton}}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }}{m_p}}=\frac{2}{1}$.