Q. The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton of same kinetic energy is
A) 1:2
B) 1:1
C) 1:$\sqrt{2}$
D) 4:1
Explanation
(a) de-Broglie wavelength, $\lambda = \frac{h}{\sqrt{2mE_{k}}}$
$\frac{\lambda_{\alpha}}{\lambda_{p}} = \sqrt{\frac{m_{p}}{m_{\alpha}}}$
=$\sqrt{\frac{1}{4}}\ $=$\frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.