Q. What will be the ratio of de-Broglie wavelengths of proton and $\alpha$-particle of same energy
A) 2 :1
B) 1 :2
C) 4 :1
D) 1 :4
Explanation
(a) $\lambda = \frac{h}{\sqrt{2mE}} \Rightarrow \lambda \propto \frac{1}{\sqrt{m}} \Rightarrow \frac{\lambda_{p}}{\lambda_{\alpha}} = \sqrt{\frac{m_{a}}{m_{p}}} = \frac{2}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.