JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 13)
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 :
$$1:\sqrt 3 $$
$$\sqrt 3 :1$$
$$3:1$$
$$2:\sqrt 3 $$
Explanation
$${K_\alpha } = {K_C}$$
$${{p_\alpha ^2} \over {2{m_\alpha }}} = {{p_C^2} \over {2{m_C}}}$$
$${{{p_\alpha }} \over {{p_C}}} = \sqrt {{{{m_\alpha }} \over {{m_C}}}} $$
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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