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

NTA AbhyasNTA Abhyas 2020

Solution:

$\lambda =\frac{h}{\sqrt{2 m E}}\Rightarrow \lambda \propto \frac{1}{\sqrt{m}}\Rightarrow \frac{\lambda _{P}}{\lambda _{\alpha }}=\sqrt{\frac{m_{\alpha }}{m_{P}}}=\frac{2}{1}$