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
$$\frac{\lambda}{8}$$
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