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:

• A. $\lambda_{e}<\lambda_{ P }<\lambda_{ He ^{+ +}}$
• B. $\lambda_{e} < \lambda_{He ^{+ +}}=\lambda_{ P }$
• C. $\lambda_{ e }>\lambda_{ He ^{+ +}}>\lambda_{ P }$
• D. $\lambda_{e}>\lambda_{ P }>\lambda_{ He ^{+ +}}$

Answer 1

Answer: (D)

$\lambda=\frac{h}{P}=\frac{h}{\sqrt{2 m(K E)}}$
$\lambda \propto \frac{1}{\sqrt{ m }} \Rightarrow \lambda=\frac{ C }{\sqrt{ m }}$
$m _{ He ^{++}} > m _{ p }> m _{ e }$
$\therefore \lambda_{ He^{ ++}}<\lambda_{ P }<\lambda_{ c }$