1
COMEDK 2020
MCQ (Single Correct Answer)
+1
-0

A proton accelerated through a potential V has de-Broglie wavelength $$\lambda$$. Then, the de-Broglie wavelength of an $$\alpha$$-particle, when accelerated through the same potential V is

A
$$\frac{\lambda}{2}$$
B
$$\frac{\lambda}{\sqrt2}$$
C
$$\frac{\lambda}{2\sqrt2}$$
D
$$\frac{\lambda}{8}$$
COMEDK Subjects
EXAM MAP
Medical
NEET
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
CBSE
Class 12