An electron, a doubly ionized helium ion (He++) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda$e, $\lambda$He++ and $\lambda$p is :
Solution
$\lambda$ = ${h \over P}$ = ${h \over {\sqrt {2m\left( {KE} \right)} }}$
<br><br>$\therefore$ $\lambda$ $\propto$ ${1 \over {\sqrt m }}$
<br><br>m<sub>He<sup>++</sup></sub> > m<sub>p</sub> > m<sub>e</sub>
<br><br>$\therefore$ $\lambda$<sub>e</sub> > $\lambda$<sub>p</sub> > $\lambda$<sub>He<sup>++</sup></sub>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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