Question

An electron, a doubly ionized helium ion $\left({\text{He}}^{++}\right)$ and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelenght ${\lambda }_e,{\lambda }_{{\text{He}}^{++}}$ and ${\lambda }_p$ is:

• A. ${\lambda }_e>{\lambda }_{{\text{He}}^{++}}>{\lambda }_p$
• B. ${\lambda }_e>{\lambda }_{{\text{He}}^{++}}={\lambda }_p$
• C. ${\lambda }_e>{\lambda }_p>{\lambda }_{{\text{He}}^{++}}$
• D. ${\lambda }_e<{\lambda }_p<{\lambda }_{{\text{He}}^{++}}$

Answer 1

The correct option is C ${\lambda }_e>{\lambda }_p>{\lambda }_{{\text{He}}^{++}}$

de-Broglie wavelength is given by,
$\lambda =\frac{h}{P}=\frac{h}{\sqrt{2m(K.E)}}$
Since, $h$ and kinetic energy are same for all,
$\therefore \lambda \propto \frac{1}{\sqrt{m}}$
As $m{}_{{\text{He}}^{++}}>m{}_p>m{}_e$
So, ${\lambda }_{{\text{He}}^{++}}<{\lambda }_p<{\lambda }_e\text{or}{\lambda }_e>{\lambda }_p>{\lambda }_{{\text{He}}^{++}}$
Hence, $\left(C\right)$ is correct option.