An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)
Solution
$\lambda$<sub>e</sub> = ${h \over {{p_e}}}$ = ${h \over {\sqrt {2mE} }}$
<br><br>E = ${{hc} \over {{\lambda _{photon}}}}$
<br><br>$\Rightarrow$ ${{\lambda _{photon}}}$ = ${{hc} \over E}$
<br><br>$\therefore$ ${{{\lambda _e}} \over {{\lambda _{photon}}}}$ = ${1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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