Question
An electron, a doubly ionized helium ion $\left( He ^{++}\right)$ 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
✓
Answer
$\lambda=\frac{ h }{ P }=\frac{ h }{\sqrt{2 m ( KE )}}$
$\lambda \propto \frac{1}{\sqrt{ m }} \Rightarrow \lambda=\frac{ C }{\sqrt{ m }}$
$m _{ He ^{++}}> m _{ P }> m _{ e }$
$\therefore \lambda_{ He ^{++}}<\lambda_{ P }<\lambda_{ e }$
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