Question

What will be the ratio of de-Broglie wavelengths of proton and $\alpha$ -particle of same energy :-

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

Answer 1

Answer: (A)

$\lambda =\frac{h}{\sqrt{2mE}}\Rightarrow \lambda \propto \frac{1}{\sqrt{m}}\Rightarrow \frac{{\lambda }_P}{{\lambda }_{\alpha }}=\sqrt{\frac{m_{\alpha }}{m_P}}=\frac{2}{1}$