"An $\alpha$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $({\lambda _\alpha }:{\lambda _{C12}})$ is :"

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${K_\alpha } = {K_C}$

${{p_\alpha ^2} \over {2{m_\alpha }}} = {{p_C^2} \over {2{m_C}}}$

${{{p_\alpha }} \over {{p_C}}} = \sqrt {{{{m_\alpha }} \over {{m_C}}}}$

So $${{{\lambda _\alpha }} \over {{\lambda _C}}} = {{h/{p_\alpha }} \over {h/{p_C}}} = \sqrt {{{{m_C}} \over {{m_\alpha }}}} $$

So ${{{\lambda _\alpha }} \over {{\lambda _C}}} = \sqrt 3$

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An $\alpha$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $({\lambda _\alpha }:{\lambda _{C12}})$ is :

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An $\alpha$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $({\lambda _\alpha }:{\lambda _{C12}})$ is :

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