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
Solution
$\lambda = {h \over p} = {h \over {\sqrt {2km} }}$<br><br>$${{{\lambda _e}} \over {{\lambda _p}}} = \sqrt {{{{k_p}{m_p}} \over {{k_e}{m_e}}}} = \sqrt {{{{m_p}} \over {{m_e}}}} $$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
This question is part of PrepWiser's free JEE Main question bank. 145 more solved questions on Dual Nature of Matter and Radiation are available — start with the harder ones if your accuracy is >70%.