An electron, a doubly ionized helium ion ( He ^ ++ ) and a proton… - Vidyadip

MCQ

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

A
$\lambda_{ e }<\lambda_{ P }<\lambda_{ He ^{++}}$
B
$\lambda_{e}<\lambda_{H e^{++}}=\lambda_{P}$
C
$\lambda_{e}>\lambda_{H e^{++}}>\lambda_{P}$
✓
$\lambda_{e}>\lambda_{P}>\lambda_{H e^{++}}$

✓

Answer

Correct option: D.

$\lambda_{e}>\lambda_{P}>\lambda_{H e^{++}}$

d
$\lambda=\frac{ h }{ P }=\frac{ h }{\sqrt{2 m ( KE )}}$

$\lambda \propto \frac{1}{\sqrt{ m }} \Rightarrow \lambda=\frac{ C }{\sqrt{ m }}$

$m _{ He ^{++}}> m _{ P }> m _{ e }$

$\therefore \lambda_{ He ^{++}}<\lambda_{ P }<\lambda_{ e }$

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