Question

If alpha particle, proton and electron move with the same momentum, then their respective de-Broglie wavelengths ${\lambda }_{\alpha },{\lambda }_p,{\lambda }_e$ are related as

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

Answer 1

The correct option is A ${\lambda }_{\alpha }={\lambda }_p={\lambda }_e$

Wavelength $\lambda =\frac{h}{p}$
Since, momentum is same for all particles.
So, wavelength will be same for all
i.e., ${\lambda }_{\alpha }={\lambda }_p={\lambda }_e$