Q. A proton and an $\alpha$-particle are accelerated through the same potential difference. The ratio of their de-Broglie wavelength $(\lambda_{p}/\lambda_{\alpha})$ is
A) $1/2\sqrt{2}$
B) 1
C) 2
D) $2\sqrt{2}$
Explanation
(d) $\frac{\lambda_{p}}{\lambda_{\alpha}} = \frac{\frac{h}{\sqrt{2em_{p}\ V}}}{\sqrt{2 \times 2e4\ m_{p}\ V}} = 2\sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.