JEE MAIN - Physics (2021 - 22th July Evening Shift - No. 14)
An electron of mass me and a proton of mass mp are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is
$${{{m_e}} \over {{m_p}}}$$
1
$${{{m_p}} \over {{m_e}}}$$
$$\sqrt {{{{m_p}} \over {{m_e}}}} $$
Explanation
$$\lambda = {h \over p} = {h \over {\sqrt {2km} }}$$
$${{{\lambda _e}} \over {{\lambda _p}}} = \sqrt {{{{k_p}{m_p}} \over {{k_e}{m_e}}}} = \sqrt {{{{m_p}} \over {{m_e}}}} $$
$${{{\lambda _e}} \over {{\lambda _p}}} = \sqrt {{{{k_p}{m_p}} \over {{k_e}{m_e}}}} = \sqrt {{{{m_p}} \over {{m_e}}}} $$
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