Find the ratio of de-Broglie wavelengths of deuteron having energy E and \(\alpha\)-particle having energy 2E :
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Remember that an \(\alpha\)-particle is a Helium nucleus (\(4m_p\)) and a deuteron is an isotope of Hydrogen (\(2m_p\)).
The ratio depends on \(\frac{1}{\sqrt{mK}}\), so \(\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 K_2}{m_1 K_1}}\).