An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)
Solution
For photon, E = ${{hc} \over \lambda }$<br><br>${\lambda _p} = {{hc} \over E}$ .... (i)<br><br>For electron, ${\lambda _e} = {{hc} \over {\sqrt {2mE} }}$ .... (ii)<br><br>$${{{\lambda _e}} \over {{\lambda _p}}} = {{{{hc} \over {\sqrt {2mE} }}} \over {{{hc} \over E}}} = \sqrt {{E \over {2m{c^2}}}} = {1 \over c}{\left( {{E \over {2m}}} \right)^{1/2}}$$
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%.