An electron moving with speed v and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is :
Solution
${\lambda _e} = {\lambda _{Ph}}$<br><br>${h \over {{p_e}}} = {h \over {{p_{ph}}}}$<br><br>$\sqrt {2m{k_e}} = {{{E_{ph}}} \over c}$<br><br>$2m{k_e} = {{{{({E_{ph}})}^2}} \over {{c^2}}}$<br><br>$${{{k_e}} \over {{E_{ph}}}} = {{{E_{ph}}} \over {{c^2}}}\left( {{1 \over {2m}}} \right)$$<br><br>$= {{{p_{ph}}} \over c}\left( {{1 \over {2m}}} \right)$<br><br>$= {{{p_e}} \over c}\left( {{1 \over {2m}}} \right)$<br><br>$= {{mv} \over c}{1 \over {2m}}$<br><br>$= {v \over {2c}}$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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