Question

If proton, electron and $\alpha$-particle are moving with same kinetic energy then the order of their de-broglie's wavelength.

• A. ${\lambda }_p>{\lambda }_e>{\lambda }_{\alpha }$
• B. ${\lambda }_{\alpha }>{\lambda }_p>{\lambda }_e$
• C. ${\lambda }_{\alpha }<{\lambda }_p<{\lambda }_e$
• D. ${\lambda }_e={\lambda }_p<{\lambda }_{\alpha }$

Answer 1

The correct option is C ${\lambda }_{\alpha }<{\lambda }_p<{\lambda }_e$

We know that de-broglie's wavelength is given by
$\lambda =\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mKE}}$
if KE same then $\lambda \propto \frac{1}{\sqrt{m}}m\uparrow \lambda \downarrow$
the order of mass of electron, proton and alpha particle is $\alpha >p>e^{-}$ hence, order of de-broglie's wavelength is ${\lambda }_{\alpha }<{\lambda }_p<{\lambda }_{e^{-}}$